Show that \(S\) is

(i)     not reflexive;

(ii)     symmetric;

(iii)     transitive.

Explain why there exists an element \(a \in A\) that is not related to itself.

Hence prove that there is at least one element of \(A\) that is not related to any other element of \(A\).

(i)     \(0S0\) is not true so \(S\) is not reflexive     A1AG

(ii)     \(aSb \Rightarrow ab > 0 \Rightarrow ba > 0 \Rightarrow bSa\) so \(S\) is symmetric     R1AG

(iii)     \(aSb\) and \(bSc \Rightarrow ab > 0\) and \(bc > 0 \Rightarrow a{b^2}c > 0 \Rightarrow ac > 0\)     M1

since \({b^2} > 0\) (as \(b\) could not be 0) \( \Rightarrow aSc\) so \(S\) is transitive     R1AG

Note: R1 is for indicating that \({b^2} > 0\).

[4 marks]

since \(R\) is not reflexive there is at least one element \(a\) belonging to \(A\) such that \(a\) is not related to \(a\)     R1AG

[1 mark]

argue by contradiction: suppose that \(a\) is related to some other element \(b\), ie, \(aRb\)     M1

since \(R\) is symmetric \(aRb\) implies \(bRa\)     R1A1

since \(R\) is transitive \(aRb\) and \(bRa\) implies \(aRa\)     R1A1

giving the required contradiction     R1

hence there is at least one element of \(A\) that is not related to any other member of \(A\)     AG

[6 marks]

Prove that \(f({e_G}) = {e_H}\).

Prove that \({\text{Ker}}(f)\) is a subgroup of \(\{ G,{\text{ }} * \} \).

let \(a \in G\) and \(f(a) \in H\)

\(f\) is a homomorphism so \(f(a * {e_G}) = f(a) \circ f({e_G})\)     (M1)

\(f(a) = f(a) \circ f({e_G})\)     A1

\({e_H} = f({e_G})\)     AG

[2 marks]

from part (a) \({e_G} \in {\text{Ker}}(f)\) and associativity follows from G    R1

let \(a,{\text{ }}b \in {\text{Ker}}(f)\)

\(f(a * b) = f(a) \circ f(b) = {e_H} \circ {e_H} = {e_H}\)     A1

hence closed since \(a * b \in {\text{Ker}}(f)\)

\({e_H} = f({a^{ - 1}} * a) = f({a^{ - 1}}) \circ f(a) = f({a^{ - 1}}) \circ {e_H} = f({a^{ - 1}})\)     M1A1

hence \({a^{ - 1}} \in {\text{Ker}}(f)\)     R1

hence \({\text{Ker}}(f)\) is subgroup of \(G\)     AG

[6 marks]

Represent the following set on a Venn diagram,

\(A\Delta B\), the symmetric difference of the sets \(A\) and \(B\);

Represent the following set on a Venn diagram,

\(A \cap (B \cup C)\).

For sets \(P\), \(Q\) and \(R\), verify that \(P \cup (Q\Delta R) \ne (P \cup Q)\Delta (P \cup R)\).

In the context of the distributive law, describe what the result in part (b)(i) illustrates.

     A1

[1 mark]

Note: Accept alternative set configurations

     A1

Note:     Accept alternative set configurations.

[1 mark]

LHS:

\(Q\Delta R = \{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}5\} \)     (A1)

\(P \cup (Q\Delta R) = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\} \)     A1

RHS:

\(P \cup Q = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} \) and \(P \cup R = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}5\} \)     (A1)

\((P \cup Q)\Delta (P \cup R) = \{ 4,{\text{ }}5\} \)     A1

hence \(P \cup (Q\Delta R) \ne (P \cup Q)\Delta (P \cup R)\)     AG

[4 marks]

the result shows that union is not distributive over symmetric difference     A1R1

Notes:     Award A1 for “union is not distributive” and R1 for “over symmetric difference”. Condone use of \( \cup \) and \(\Delta \).

[2 marks]

Prove that \(f \circ f\) is the identity function.

Show that \(f\) is injective.

Show that \(f\) is surjective.

METHOD 1

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^{n + {{\left( { - 1} \right)}^n}}}\)     M1A1

\( = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^n} \times {\left( { - 1} \right)^{{{\left( { - 1} \right)}^n}}}\)     (A1)

considering \({\left( { - 1} \right)^n}\) for even and odd \(n\)     M1

if \(n\) is odd, \({\left( { - 1} \right)^n} = - 1\) and if \(n\) is even, \({\left( { - 1} \right)^n} = 1\) and so \({\left( { - 1} \right)^{ \pm 1}} =  - 1\)      A1

\( = n + {\left( { - 1} \right)^n} - {\left( { - 1} \right)^n}\)    A1

= \(n\) and so \(f \circ f\) is the identity function     AG

METHOD 2

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^{n + {{\left( { - 1} \right)}^n}}}\)      M1A1

\( = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^n} \times {\left( { - 1} \right)^{{{\left( { - 1} \right)}^n}}}\)     (A1)

\( = n + {\left( { - 1} \right)^n} \times \left( {1 + {{\left( { - 1} \right)}^{{{\left( { - 1} \right)}^n}}}} \right)\)     M1

\({\left( { - 1} \right)^{ \pm 1}} =  - 1\)     R1

\(1 + {\left( { - 1} \right)^{{{\left( { - 1} \right)}^n}}} = 0\)     A1

\(\left( {f \circ f} \right)\left( n \right) = n\) and so \(f \circ f\) is the identity function     AG

METHOD 3

\(\left( {f \circ f} \right)\left( n \right) = f\left( {n + {{\left( { - 1} \right)}^n}} \right)\)      M1

considering even and odd \(n\)       M1

if \(n\) is even, \(f\left( n \right) = n + 1\) which is odd    A1

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n + 1} \right) = \left( {n + 1} \right) - 1 = n\)    A1

if \(n\) is odd, \(f\left( n \right) = n - 1\) which is even    A1

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n - 1} \right) = \left( {n - 1} \right) + 1 = n\)     A1

\(\left( {f \circ f} \right)\left( n \right) = n\) in both cases

hence \(f \circ f\) is the identity function     AG

[6 marks]

suppose \(f\left( n \right) = f\left( m \right)\)     M1

applying \(f\) to both sides \( \Rightarrow n = m\)     R1

hence \(f\) is injective     AG

[2 marks]

\(m = f\left( n \right)\) has solution \(n = f\left( m \right)\)       R1

hence surjective       AG

[1 mark]

(i)     Write the permutation \(\alpha = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&4&6&2&1&5 \end{array}} \right)\) as a composition of disjoint cycles.

(ii)     State the order of \(\alpha \).

(i)     Write the permutation \(\beta = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 6&4&3&5&1&2 \end{array}} \right)\) as a composition of disjoint cycles.

(ii)     State the order of \(\beta \).

Write the permutation \(\alpha  \circ \beta \) as a composition of disjoint cycles.

Write the permutation \(\beta  \circ \alpha \) as a composition of disjoint cycles.

State the order of \(\{ G,{\text{ }} \circ \} \).

Find the number of permutations in \(\{ G,{\text{ }} \circ \} \) which will result in \(A\), \(B\) and \(A \cap B\) remaining unchanged.

(i)     \((1{\text{ }}3{\text{ }}6{\text{ }}5)(2{\text{ }}4)\)     A1A1

(ii)     4     A1

Note: In (b) (c) and (d) single cycles can be omitted.

[3 marks]

(i)     \((1{\text{ }}6{\text{ }}2{\text{ }}4{\text{ }}5)(3)\)     A1

(ii)     5     A1

[2 marks]

\(\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&2&6&1&3&4 \end{array}} \right) = (1{\text{ }}5{\text{ }}3{\text{ }}6{\text{ }}4)(2)\)    (M1)A1

[2 marks]

\(\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&5&2&4&6&1 \end{array}} \right) = (1{\text{ }}3{\text{ }}2{\text{ }}5{\text{ }}6)(4)\)    (M1)A1

Note: Award A2A0 for (c) and (d) combined, if answers are the wrong way round.

[2 marks]

\(6! = 720\)    A2

[2 marks]

any composition of the cycles (1 2), (3 4) and (5 6)     (M1)

so \({2^3} = 8\)     A1

[2 marks]

\( \odot \) is commutative;

\( * \) is associative;

\( * \) is distributive over \( \odot \) ;

\( \odot \) has an identity element.

\(a \odot b = \sqrt {ab}  = \sqrt {ba}  = b \odot a\)     A1

since \(a \odot b = b \odot a\) it follows that \( \odot \) is commutative     R1

[2 marks]

\(a * (b * c) = a * {b^2}{c^2} = {a^2}{b^4}{c^4}\)     M1A1

\((a * b) * c = {a^2}{b^2} * c = {a^4}{b^4}{c^2}\)     A1

these are different, therefore \( * \) is not associative     R1

Note: Accept numerical counter-example.

[4 marks]

\(a * (b \odot c) = a * \sqrt {bc}  = {a^2}bc\)     M1A1

\((a * b) \odot (a * c) = {a^2}{b^2} \odot {a^2}{c^2} = {a^2}bc\)     A1

these are equal so \( * \) is distributive over \( \odot \)     R1

[4 marks]

the identity e would have to satisfy

\(a \odot e = a\) for all a     M1

now \(a \odot e = \sqrt {ae}  = a \Rightarrow e = a\)     A1

therefore there is no identity element     A1

[3 marks]

Let \(\{ G,{\text{ }} * \} \) be a finite group that contains an element a (that is not the identity element) and \(H = \{ {a^n}|n \in {\mathbb{Z}^ + }\} \), where \({a^2} = a * a,{\text{ }}{a^3} = a * a * a\) etc.

Show that \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \).

since G is closed, H will be a subset of G

closure: \(p,{\text{ }}q \in H \Rightarrow p = {a^r},{\text{ }}q = {a^s},{\text{ }}r,{\text{ }}s \in {\mathbb{Z}^ + }\)     A1

\(p * q = {a^r} * {a^s} = {a^{r + s}}\)     A1

\(r + s \in {\mathbb{Z}^ + } \Rightarrow p * q \in H\) hence H is closed     R1

associativity follows since \( * \) is associative on G     (R1)

EITHER

identity: let the order of a in G be \(m \in {\mathbb{Z}^ + },{\text{ }}m \geqslant 2\)     M1

then \({a^m} = e \in H\)     R1

inverses: \({a^{m - 1}} * a = e \Rightarrow {a^{m - 1}}\) is the inverse of a     A1

\({({a^{m - 1}})^n} * {a^n} = e\), showing that \({a^n}\) has an inverse in H     R1

hence H is a subgroup of G     AG

OR

since \((G,{\text{ }} * )\) is a finite group, and H is a non-empty closed subset of G, then \((H,{\text{ }} * )\) is

a subgroup of \((G,{\text{ }} * )\)     R4

Note: To receive the R4, the candidate must explicitly state the theorem, i.e. the three given conditions, and conclusion.

[8 marks]

This question was generally answered very poorly, if attempted at all. Candidates failed to realize that the property of closure needed to be properly proved. Others used negative indices when the question specifically states that the indices are positive integers.

Write down all the elements of \(A\) and all the elements of \(B\).

Determine the symmetric difference, \(A\Delta B\), of the sets \(A\) and \(B\).

Write down all the elements of \(A \cap B,{\text{ }}A \cap C\) and \(B \cup C\).

Hence by considering \(A \cap (B \cup C)\), verify that in this case the operation \( \cap \) is distributive over the operation \( \cup \).

the elements of \(A\) are: 3, 7, 11, 15, 19     A1

the elements of \(B\) are 2, 3, 5, 7, 11, 13, 17, 19     A1

Note:     Accept \(A = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}15,{\text{ }}19\} \) and \(B = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13,{\text{ }}17,{\text{ }}19\} \)

[2 marks]

attempt to determine \(A\backslash B \cup B\backslash A\) or \((A \cup B) \cap (A \cap B)'\)     (M1)

symmetric difference \( = \{ 2,{\text{ }}5,{\text{ }}13,{\text{ }}15,{\text{ }}17\} \)     A1

Note:     Allow (M1)A1FT.

[2 marks]

the elements of \(A \cap B\) are 3, 7, 11 and 19     A1

the elements of \(A \cap C\) are 7 and19     A1

the elements of \(B \cup C\) are 2, 3, 5, 7, 9, 11, 13, 17 and 19     A1

Note:     Accept \(A \cap B = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}19\} ,{\text{ }}A \cap C = \{ 7,{\text{ }}19\} \) and \(B \cup C = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}9,{\text{ }}11,{\text{ }}13,{\text{ }}17,{\text{ }}19\} \).

[3 marks]

we need to show that

\(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)     (M1)

\(A \cap (B \cup C) = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}19\} \)     A1

\((A \cap B) \cup (A \cap C) = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}19\} \)     A1

hence showing the required result

Note:     Allow (M1)A1FTA1FT.

[3 marks]

Show that R is an equivalence relation.

Determine the equivalence class of R containing the element \((1,{\text{ }}2)\) and illustrate this graphically.

R is an equivalence relation if

R is reflexive, symmetric and transitive     A1

\({x_1}{y_1} = {x_1}{y_1} \Rightarrow ({x_1},{\text{ }}{y_1})R({x_1},{\text{ }}{y_1})\)     A1

so R is reflexive

\(({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2}) \Rightarrow {x_1}{y_1} = {x_2}{y_2} \Rightarrow {x_2}{y_2} = {x_1}{y_1} \Rightarrow ({x_2},{\text{ }}{y_2})R({x_1},{\text{ }}{y_1})\)     A1

so R is symmetric

\(({x_1},{\text{ }}{y_1})R({x_2},{\text{ }}{y_2})\) and \(({x_2},{\text{ }}{y_2})R({x_3},{\text{ }}{y_3}) \Rightarrow {x_1}{y_1} = {x_2}{y_2}\) and \({x_2}{y_2} = {x_3}{y_3}\)     M1

\( \Rightarrow {x_1}{y_1} = {x_3}{y_3} \Rightarrow ({x_1},{\text{ }}{y_1})R({x_3},{\text{ }}{y_3})\)    A1

so R is transitive

R is an equivalence relation     AG

[5 marks]

\((x,{\text{ }}y)R(1,{\text{ }}2)\)     (M1)

the equivalence class is \(\{ (x,{\text{ }}y)|xy = 2\} \)     A1

correct graph     A1

\((1,{\text{ }}2)\) indicated on the graph     A1

Note:     Award last A1 only if plotted on a curve representing the class.

[4 marks]

(i)     Write down the Cayley table for \(\{ G,{\text{ }}{ \times _7}\} \) .

(ii)     Determine whether or not \(\{ G,{\text{ }}{ \times _7}\} \) is cyclic.

(iii)     Find the subgroup of G of order 3, denoting it by H .

(iv)     Identify the element of order 2 in G and find its coset with respect to H .

The group \(\{ K,{\text{ }} \circ \} \) is defined on the six permutations of the integers 1, 2, 3 and \( \circ \) denotes composition of permutations.

(i)     Show that \(\{ K,{\text{ }} \circ \} \) is non-Abelian.

(ii)     Giving a reason, state whether or not \(\{ G,{\text{ }}{ \times _7}\} \) and \(\{ K,{\text{ }} \circ \} \) are isomorphic.

(i)     the Cayley table is

     A3

Note: Deduct 1 mark for each error up to a maximum of 3.

(ii)     by considering powers of elements,     (M1)

it follows that 3 (or 5) is of order 6     A1

so the group is cyclic     A1

(iii)     we see that 2 and 4 are of order 3 so the subgroup of order 3 is {1, 2, 4}     M1A1

(iv)     the element of order 2 is 6     A1

the coset is {3, 5, 6}     A1

[10 marks]

(i)     consider for example

\(\left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&1&3
\end{array}} \right) \circ \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&3&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  1&3&2
\end{array}} \right)\)     M1A1

\(\left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&3&1
\end{array}} \right) \circ \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  2&1&3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  3&2&1
\end{array}} \right)\)     M1A1

Note: Award M1A1M1A0 if both compositions are done in the wrong order.

Note: Award M1A1M0A0 if the two compositions give the same result, if no further attempt is made to find two permutations which are not commutative.

these are different so the group is not Abelian     R1AG

(ii)     they are not isomorphic because \(\{ G,{\text{ }}{ \times _7}\} \) is Abelian and \(\{ K,{\text{ }} \circ \} \) is not     R1

[6 marks]

Find the order of \(\{ G,{\text{ }} \circ \} \).

State the identity element in \(\{ G,{\text{ }} \circ \} \).

Find

(i)     \(p \circ p\);

(ii)     the inverse of \(p \circ p\).

(i)     Find the maximum possible order of an element in \(\{ H,{\text{ }} \circ \} \).

(ii)     Give an example of an element with this order.

the order of \((G,{\text{ }} \circ )\) is \({\text{lcm}}(6,{\text{ }}4)\)     (M1)

\( = 12\)     A1

[2 marks]

\(\left( 1 \right){\rm{ }}\left( 2 \right){\rm{ }}\left( 3 \right){\rm{ }}\left( 4 \right){\rm{ }}\left( 5 \right){\rm{ }}\left( 6 \right){\rm{ }}\left( 7 \right){\rm{ }}\left( 8 \right){\rm{ }}\left( 9 \right){\rm{ }}\left( {10} \right)\)     A1

Note:     Accept ( ) or a word description.

[1 mark]

(i)     \(p \circ p = (1{\text{ }}3{\text{ }}5)(2{\text{ }}4{\text{ }}6)(7{\text{ }}9)(810)\)     (M1)A1

(ii)     its inverse \( = (1{\text{ }}5{\text{ }}3)(2{\text{ }}6{\text{ }}4)(7{\text{ }}9)(810)\)     A1A1

Note:     Award A1 for cycles of 2, A1 for cycles of 3.

[4 marks]

(i)     considering LCM of length of cycles with length \(2\), \(3\) and \(5\)     (M1)

\(30\)     A1

(ii)     eg\(\;\;\;(1{\text{ }}2)(3{\text{ }}4{\text{ }}5)(6{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}10)\)     A1

Note:     allow FT as long as the length of cycles adds to \(10\) and their LCM is consistent with answer to part (i).

Note: Accept alternative notation for each part

[3 marks]

Total [10 marks]

Show that R is an equivalence relation.

Find the equivalence class containing 0.

Denote the equivalence class containing n by C_n .

List the first six elements of \({C_1}\).

Denote the equivalence class containing n by C_n .

Prove that \({C_n} = {C_{n + 7}}\) for all \(n \in \mathbb{N}\).

reflexive: \({a^3} - {a^3} = 0{\text{ }},{\text{ }} \Rightarrow R\) is reflexive     R1

symmetric: if \({a^3} \equiv {b^3}(\bmod 7)\) , then \({b^3} \equiv {a^3}(\bmod 7)\)     M1

\( \Rightarrow R\) is symmetric     R1

transitive: \({a^3} = {b^3} + 7n\) and \({b^3} = {c^3} + 7m\)     M1

then \({a^3} = {c^3} + 7(n + m)\)

\( \Rightarrow {a^3} \equiv {c^3}(\bmod 7)\)     R1

\( \Rightarrow R\) is transitive     A1

and is an equivalence relation     AG 

Note: Allow arguments that use \({a^3} - {b^3} \equiv 0(\bmod 7)\) etc.

[6 marks]

\(\{ 0,{\text{ }}7,{\text{ }}14,{\text{ }}21,{\text{ }}...\} \)     A2

[2 marks]

\(\{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}8,{\text{ }}9,{\text{ }}11\} \)     A3 

Note: Deduct 1 mark for each error or omission.

[3 marks]

consider \({(n + 7)^3} = {n^3} + 21{n^2} + 147n + 343 = {n^3} + 7N\)     M1A1

\( \Rightarrow {n^3} \equiv {(n + 7)^3}(\bmod 7) \Rightarrow n\) and \(n + 7\) are in the same equivalence class     R1

[3 marks]

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

Candidates were mostly aware of the conditions required to show an equivalence relation although many seemed unsure as to the degree of detail required to show that the different conditions are met for the example in this question. In part (b) many candidates found the correct set although a number were unable to write down the set correctly, including or excluding elements that were not part of the equivalence class. Part (c) saw candidate being less successful than (b) and relatively few candidates were able to prove the equivalence class in part (d) although there were a number of very good solutions.

Prove that \(f\) is an injection.

(i)     Prove that \(f\) is a surjection.

(ii)     Hence, or otherwise, write down the inverse function \({f^{ - 1}}\).

let \((a,{\text{ }}b)\) and \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)

suppose that \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\)     (M1)

so that \(\sqrt {ab}  = \sqrt {cd} \) and \(\frac{a}{b} = \frac{c}{d}\)     A1

leading to either \({a^2} = {c^2}\) or \({b^2} = {d^2}\) or equivalent     M1

state \(a = c\) and \(b = d\)     A1

this shows that \(f\) is an injection since \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d)\)     R1AG

Note:     Accept final statement seen anywhere for R1.

[5 marks]

(i)     now let \((u,{\text{ }}v) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) and suppose that \(f(x,{\text{ }}y) = (u,{\text{ }}v)\)     (M1)

then, \(u = \sqrt {xy} ,{\text{ }}v = \frac{x}{y}\)     A1

attempt to eliminate \(x\) or \(y\)     M1

\( \Rightarrow x = u{v^{1/2}};{\text{ }}y = u{v^{ - 1/2}}\)    A1A1

this shows that \(f\) is a surjection since, given \((u,{\text{ }}v)\), there exists \((x,{\text{ }}y)\) such that \(f(x,{\text{ }}y) = (u,{\text{ }}v)\)     R1AG

Note: Accept final statement, seen anywhere, for R1.

(ii)     \({f^{ - 1}}(x,{\text{ }}y) = (x{y^{1/2}},{\text{ }}x{y^{ - 1/2}})\)     A1A1

[8 marks]

Those candidates who formulated their responses in terms of the basic mathematical definitions of injectivity and surjectivity were usually successful. Otherwise, verbal attempts such as ‘\(f\) is one-to-one \( \Rightarrow f\) is injective’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit.

(i)     Those candidates who formulated their responses in terms of the basic mathematical definitions of injectivity and surjectivity were usually successful. Otherwise, verbal attempts such as ‘\(f\) is one-to-one \( \Rightarrow f\) is injective’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit.

(ii)     It was surprising to see that some candidates were unable to relate what they had done in part (b)(i) to this part.

Show that \(R\) is an equivalence relation.

Given that \(n = 2\) and \(p = 7\), determine the first four members of each of the four equivalence classes of \(R\).

since \({a^n} - {a^n} = 0\),     A1

it follows that \((aRa)\) and \(R\) is reflexive     R1

if \(aRb\) so that \({b^n} - {a^n} \equiv 0(\bmod p)\)     M1

then, \({a^n} - {b^n} \equiv 0(\bmod p)\) so that \(bRa\) and \(R\) is symmetric     A1

if \(aRb\) and \(bRc\) so that \({b^n} - {a^n} \equiv 0(\bmod p)\) and \({c^n} - {b^n} \equiv 0(\bmod p)\)     M1

adding, (it follows that \({c^n} - {a^n} \equiv 0(\bmod p)\))     M1

so that \(aRc\) and \(R\) is transitive     A1

Note:     Only accept the correct use of the terms “reflexive, symmetric, transitive”.

[7 marks]

we are now given that \(aRb\) if \({b^2} - {a^2} \equiv 0(\bmod 7)\)

attempt to find at least one equivalence class     (M1)

the equivalence classes are

\(\{ 1,{\text{ }}6,{\text{ }}8,{\text{ }}13,{\text{ }} \ldots \} \)    A1

\(\{ 2,{\text{ }}5,{\text{ }}9,{\text{ }}12,{\text{ }} \ldots \} \)    A1

\(\{ 3,{\text{ }}4,{\text{ }}10,{\text{ }}11,{\text{ }} \ldots \} \)    A1

\(\{ 7,{\text{ }}14,{\text{ }}21,{\text{ }}28,{\text{ }} \ldots \} \)    A1

[5 marks]

Most candidates were familiar with the terminology of the required conditions to be satisfied for a relation to be an equivalence relation. The execution of the proofs was variable. It was grating to see such statements as \(R\) is symmetric because \(aRb = bRa\) or \(aRa = {a^n} - {a^n} = 0\), often without mention of \(\bmod p\), and such responses were not fully rewarded.

This was not well answered. Few candidates displayed a strategy to find the equivalence classes.

Simplify \(\frac{c}{2} * \frac{{3c}}{4}\) .

State the identity element for G under \( * \).

For \(x \in G\) find an expression for \({x^{ - 1}}\) (the inverse of x under \( * \)).

Show that the binary operation \( * \) is commutative on G .

Show that the binary operation \( * \) is associative on G .

(i)     If \(x,{\text{ }}y \in G\) explain why \((c - x)(c - y) > 0\) .

(ii)     Hence show that \(x + y < c + \frac{{xy}}{c}\) .

Show that G is closed under \( * \).

Explain why \(\{ G, * \} \) is an Abelian group.

\(\frac{c}{2} * \frac{{3c}}{4} = \frac{{\frac{c}{2} + \frac{{3c}}{4}}}{{1 + \frac{1}{2} \cdot \frac{3}{4}}}\)     M1

\( = \frac{{\frac{{5c}}{4}}}{{\frac{{11}}{8}}} = \frac{{10c}}{{11}}\)     A1

[2 marks]

identity is 0     A1

[1 mark]

inverse is –x     A1

[1 mark]

\(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}},{\text{ }}y * x = \frac{{y + x}}{{1 + \frac{{yx}}{{{c^2}}}}}\)     M1

(since ordinary addition and multiplication are commutative)

\(x * y = y * x{\text{ so }} * \) is commutative     R1

Note: Accept arguments using symmetry.

[2 marks]

\((x * y) * z = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} * z = \frac{{\left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right) + z}}{{1 + \left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right)\frac{z}{{{c^2}}}}}\)     M1

\( = \frac{{\frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}} + \frac{{yz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\)     A1

\(x * (y * z) = x * \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right) = \frac{{x + \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}{{1 + \frac{x}{{{c^2}}}\left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}\)

\( = \frac{{\frac{{\left( {x + \frac{{xyz}}{{{c^2}}} + y + z} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{yz}}{{{c^2}}} + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\)     A1

since both expressions are the same \( * \) is associative     R1

Note: After the initial M1A1, correct arguments using symmetry also gain full marks.

[4 marks]

(i)     \(c > x{\text{ and }}c > y \Rightarrow c - x > 0{\text{ and }}c - y > 0 \Rightarrow (c - x)(c - y) > 0\)     R1AG

(ii)     \({c^2} - cx - cy + xy > 0 \Rightarrow {c^2} + xy > cx + cy \Rightarrow c + \frac{{xy}}{c} > x + y{\text{ (as }}c > 0)\)

so \(x + y < c + \frac{{xy}}{c}\)     M1AG

[2 marks]

if \(x,{\text{ }}y \in G{\text{ then }} - c - \frac{{xy}}{c} < x + y < c + \frac{{xy}}{c}\)

thus \( - c\left( {1 + \frac{{xy}}{{{c^2}}}} \right) < x + y < c\left( {1 + \frac{{xy}}{{{c^2}}}} \right){\text{ and }} - c < \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} < c\)     M1

\(({\text{as }}1 + \frac{{xy}}{{{c^2}}} > 0){\text{ so }} - c < x * y < c\)     A1

proving that G is closed under \( * \)     AG

[2 marks]

as \(\{ G, * \} \) is closed, is associative, has an identity and all elements have an inverse     R1

it is a group     AG

as \( * \) is commutative     R1

it is an Abelian group     AG

[2 marks]

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully - with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Below are the graphs of the two functions \(F:P \to Q{\text{ and }}g:A \to B\) .

Determine, with reference to features of the graphs, whether the functions are injective and/or surjective.

Given two functions \(h:X \to Y{\text{ and }}k:Y \to Z\) . 

Show that

(i)     if both h and k are injective then so is the composite function \(k \circ h\) ;

(ii)     if both h and k are surjective then so is the composite function \(k \circ h\) .

f is surjective because every horizontal line through Q meets the graph somewhere     R1

f is not injective because it is a many-to-one function     R1

g is injective because it always has a positive gradient     R1

(accept horizontal line test reasoning)

g is not surjective because a horizontal line through the negative part of B would not meet the graph at all     R1

[4 marks]

(i)     EITHER

Let \({x_1},{\text{ }}{x_2} \in X{\text{ and }}{y_1} = h({x_1}){\text{ and }}{y_2} = h({x_2})\)     M1

Then

\(k \circ \left( {h({x_1})} \right) = k \circ \left( {h({x_2})} \right)\)

\( \Rightarrow k({y_1}) = k({y_2})\)     A1

\( \Rightarrow {y_1} = {y_2}\,\,\,\,\,{\text{(}}k{\text{ is injective)}}\)     A1

\( \Rightarrow h({x_1}) = h({x_2})\,\,\,\,\,\left( {h({x_1}) = {y_1}{\text{ and }}h({x_2}) = {y_2}} \right)\)     A1

\( \Rightarrow {x_1} \equiv {x_2}\,\,\,\,\,(h{\text{ is injective)}}\)     A1

Hence \(k \circ h\) is injective     AG

OR

\({{\text{x}}_1},{\text{ }}{x_2} \in X,{\text{ }}{x_1} \ne {x_2}\)     M1

since h is an injection \( \Rightarrow h({x_1}) \ne h({x_2})\)     A1

\(h({x_1}),{\text{ }}h({x_2}) \in Y\)     A1

since k is an injection \( \Rightarrow k\left( {h({x_1})} \right) \ne k\left( {h({x_2})} \right)\)     A1

\(k\left( {h({x_1})} \right),{\text{ }}k\left( {h({x_2})} \right) \in \mathbb{Z}\)     A1

so \(k \circ h\) is an injection.     AG

(ii)     h and k are surjections and let \(z \in \mathbb{Z}\)

Since k is surjective there exists \(y \in Y\) such that k(y) = z     R1

Since h is surjective there exists \(x \in X\) such that h(x) = y     R1

Therefore there exists \(x \in X\) such that

\(k \circ h(x) = k\left( {h(x)} \right)\)

\( = k(y)\)     R1

\( = z\)     A1

So \(k \circ h\) is surjective     AG

[9 marks]

‘Using features of the graph’ should have been a fairly open hint but too many candidates contented themselves with describing what injective and surjective meant rather than explaining which graph had which properties. Candidates found considerable difficulty with presenting a convincing argument in part (b).

‘Using features of the graph’ should have been a fairly open hint but too many candidates contented themselves with describing what injective and surjective meant rather than explaining which graph had which properties. Candidates found considerable difficulty with presenting a convincing argument in part (b).

Find the order of elements 5, 7 and 17 in \(\{ G,{\text{ }}{ \times _{18}}\} \).

State whether or not \(\{ G,{\text{ }}{ \times _{18}}\} \) is cyclic, justifying your answer.

Write down the elements in set \(K\).

Find the left cosets of \(K\) in \(\{ G,{\text{ }}{ \times _{18}}\} \).

considering powers of elements     (M1)

5 has order 6     A1

7 has order 3     A1

17 has order 2     A1

[4 marks]

\(G\) is cyclic     A1

because there is an element (are elements) of order 6     R1

Note:     Accept “there is a generator”; allow A1R0.

[3 marks]

\(\{ 1,{\text{ }}17\} \)     A1

[1 mark]

multiplying \(\{ 1,{\text{ }}17\} \) by each element of \(G\)     (M1)

\(\{ 1,{\text{ }}17\} ,{\text{ }}\{ 5,{\text{ }}13\} ,{\text{ }}\{ 7,{\text{ }}11\} \)     A1A1A1

[4 marks]

Prove that the function \(f\) is a homomorphism from the group \(\{ \mathbb{Z},{\text{ }} + \} {\text{ to }}\{ D,{\text{ }}{ \times _3}\} \).

Find the kernel of \(f\).

Prove that \(\{ {\text{Ker}}(f),{\text{ }} + \} \) is a subgroup of \(\{ \mathbb{Z},{\text{ }} + \} \).

consider the cases, \(a\) and \(b\) both even, one is even and one is odd and \(a\) and \(b\) are both odd     (M1)

calculating \(f(a + b)\) and \(f(a){ \times _3}f(b)\) in at least one case     M1

if \(a\) is even and \(b\)  is even, then \(a + b\) is even

so\(\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}1 = 1\)     A1

so\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\)

if one is even and the other is odd, then \(a + b\) is odd

so\(\;\;\;f(a + b) = 2.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}2 = 2\)     A1

so\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\)

if \(a\) is odd and \(b\) is odd, then \(a + b\) is even

so\(\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 2{ \times _3}2 = 1\)     A1

so\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\)

as\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\;\;\;\)in all cases, so\(\;\;\;f:\mathbb{Z} \to D\) is a homomorphism     R1AG

[6 marks]

\(1\) is the identity of \(\{ D,{\text{ }}{ \times _3}\} \)     (M1)(A1)

so\(\;\;\;{\text{Ker}}(f)\) is all even numbers     A1

[3 marks]

METHOD 1

sum of any two even numbers is even so closure applies     A1

associative as it is a subset of \(\{ \mathbb{Z},{\text{ }} + \} \)     A1

identity is \(0\), which is in the kernel     A1

the inverse of any even number is also even     A1

METHOD 2

\({\text{ker}}(f) \ne \emptyset \)

\({b^{ - 1}} \in {\text{ker}}(f)\) for any \(b\)

\(a{b^{ - 1}} \in {\text{ker}}(f)\) for any \(a\) and \(b\)

Note:     Allow a general proof that the Kernel is always a subgroup.

[4 marks]

Total [13 marks]

Associativity and commutativity are two of the five conditions for a set S with the binary operation \( * \) to be an Abelian group; state the other three conditions.

The Cayley table for the binary operation \( \odot \) defined on the set T = {p, q, r, s, t} is given below.

(i)     Show that exactly three of the conditions for {T , \( \odot \)} to be an Abelian group are satisfied, but that neither associativity nor commutativity are satisfied. 

(ii)     Find the proper subsets of T that are groups of order 2, and comment on your result in the context of Lagrange’s theorem. 

(iii)     Find the solutions of the equation \((p \odot x) \odot x = x \odot p\) .

closure, identity, inverse     A2 

Note: Award A1 for two correct properties, A0 otherwise.

[2 marks]

(i)     closure: there are no extra elements in the table     R1

identity: s is a (left and right) identity     R1

inverses: all elements are self-inverse     R1

commutative: no, because the table is not symmetrical about the leading diagonal, or by counterexample     R1

associativity: for example, \((pq)t = rt = p\)     M1A1

not associative because \(p(qt) = pr = t \ne p\)     R1 

Note: Award M1A1 for 1 complete example whether or not it shows non-associativity.

(ii)     \(\{ s,\,p\} ,{\text{ }}\{ s,\,q\} ,{\text{ }}\{ s,\,r\} ,{\text{ }}\{ s,\,t\} \)     A2 

Note: Award A1 for 2 or 3 correct sets.

as 2 does not divide 5, Lagrange’s theorem would have been contradicted if T had been a group     R1

(iii)     any attempt at trying values     (M1)

the solutions are q, r, s and t     A1A1A1A1 

Note: Deduct A1 if p is included.

[15 marks]

This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.

This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.

Show that \(\{ \mathbb{Z},{\text{ }} * \} \) is an Abelian group.

Show that there is no element of order 2.

Find a proper subgroup of \(\{ \mathbb{Z},{\text{ }} * \} \).

Show that the groups \(\{ \mathbb{Z},{\text{ }} * \} \) and \(\{ \mathbb{Z},{\text{ }} \circ \} \) are isomorphic.

closure: \(\{ \mathbb{Z},{\text{ }} * \} \) is closed because \(a + b - 3 \in \mathbb{Z}\)     R1

identity: \(a * e = a + e - 3 = a\)     (M1)

\(e = 3\)     A1

inverse: \(a * {a^{ - 1}} = a + {a^{ - 1}} - 3 = 3\)     (M1)

\({a^{ - 1}} = 6 - a\)     A1

associative: \(a * (b * c) = a * (b + c - 3) = a + b + c - 6\)     A1

\(\left( {a{\text{ }}*{\text{ }}b} \right){\text{ }}*{\text{ }}c{\text{ }} = \left( {a{\text{ }} + {\text{ }}b{\text{ }} - {\text{ }}3} \right)*{\text{ }}c{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c{\text{ }} - {\text{ }}6\)    A1

associative because \(a * (b * c) = (a * b) * c\)     R1

\(b * a = b + a - 3 = a + b - 3 = a * b\) therefore commutative hence Abelian     R1

hence \(\{ \mathbb{Z},{\text{ }} * \} \) is an Abelian group     AG

[9 marks]

if \(a\) is of order 2 then \(a * a = 2a - 3 = 3\) therefore \(a = 3\)     A1

which is a contradiction

since \(e = 3\) and has order 1     R1

Note:     R1 for recognising that the identity has order 1.

[2 marks]

for example \(S = \{ - 6,{\text{ }} - 3,{\text{ }}0,{\text{ }}3,{\text{ }}6 \ldots \} \) or \(S = \{ \ldots ,{\text{ }} - 1,{\text{ }}1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \} \)     A1R1

Note:     R1 for deducing, justifying or verifying that \(\left\{ {S, * } \right\}\) is indeed a proper subgroup.

[2 marks]

we need to show that \(f(a * b) = f(a) \circ f(b)\)     R1

\(f(a * b) = f(a + b - 3) = a + b - 9\)     A1

\(f(a) \circ f(b) = (a - 6) \circ (b - 6) = a + b - 9\)     A1

hence isomorphic     AG

Note:     R1 for recognising that \(f\) preserves the operation; award R1A0A0 for an attempt to show that \(f(a \circ b) = f(a) * f(b)\).

[3 marks]

Show that \(x * y \in S\) for all \(x,{\text{ }}y \in S\).

Show that the operation \( * \) on the set \(S\) is commutative.

Show that the operation \( * \) on the set \(S\) is associative.

Show that 2 is the identity element.

Show that each element \(a \in S\) has an inverse.

\(x,{\text{ }}y > 1 \Rightarrow (x - 1)(y - 1) > 0\)     M1

\((x - 1)(y - 1) + 1 > 1\)     A1

so \(x * y \in S\) for all \(x,{\text{ }}y \in S\)     AG

[2 marks]

\(x * y = (x - 1)(y - 1) + 1 = (y - 1)(x - 1) + 1 = y * x\)     M1A1

so \( * \) is commutative     AG

[2 marks]

\(x * (y * z) = x * \left( {(y - 1)(z - 1) + 1} \right)\)     M1

\( = (x - 1)\left( {(y - 1)(z - 1) + 1 - 1} \right) + 1\)     (A1)

\( = (x - 1)(y - 1)(z - 1) + 1\)     A1

\((x * y) * z = \left( {(x - 1)(y - 1) + 1} \right) * z\)     M1

\( = \left( {(x - 1)(y - 1) + 1 - 1} \right)(z - 1) + 1\)

\( = (x - 1)(y - 1)(z - 1) + 1\)     A1

so \( * \) is associative     AG

[5 marks]

\(2 * x = (2 - 1)(x - 1) + 1 = x,{\text{ }}x * 2 = (x - 1)(2 - 1) + 1 = x\)     M1

\(2 * x = x * 2 = 2{\text{ }}(2 \in S)\)     R1

Note:     Accept reference to commutativity instead of explicit expressions.

so 2 is the identity element     AG

[2 marks]

\(a * {a^{ - 1}} = 2 \Rightarrow (a - 1)({a^{ - 1}} - 1) + 1 = 2\)     M1

so \({a^{ - 1}} = 1 + \frac{1}{{a - 1}}\)     A1

since \(a - 1 > 0 \Rightarrow {a^{ - 1}} > 1{\text{ }}({a^{ - 1}} * a = a * {a^{ - 1}})\)     R1

Note:     R1 dependent on M1.

so each element, \(a \in S\), has an inverse     AG

[3 marks]

Given that \(R = (P \cap Q')'\) , list the elements of R .

For a set S , let \({S^ * }\) denote the set of all subsets of S ,

(i)     find \({P^ * }\) ;

(ii)     find \(n({R^ * })\) .

\(P = \{ 1,{\text{ }}2,{\text{ }}3\} \)

\(Q' = \{ 1,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}9\} \)

so \(P \cap Q' = \{ 1,{\text{ }}3\} \)     (M1)(A1)

so \((P \cap Q')' = \{ 2,{\text{ }}4,{\text{ }}5,{\text{ }}6,{\text{ }}7,{\text{ }}8,{\text{ }}9,{\text{ }}10\} \)     A1

[3 marks]

(i)     \({P^ * } = \left\{ {\{ 1\} ,{\text{ }}\{ 2\} ,{\text{ }}\{ 3\} ,{\text{ }}\{ 1,{\text{ }}2\} ,{\text{ }}\{ 2,{\text{ }}3\} ,{\text{ }}\{ 3,{\text{ }}1\} ,{\text{ }}\{ 1,{\text{ }}2,{\text{ }}3),{\text{ }}\emptyset } \right\}\)     A2 

Note: Award A1 if one error, A0 for two or more.

(ii)     \({R^ * }\) contains: the empty set (1 element); sets containing one element (8 elements); sets containing two elements     (M1)

\( = \left( {\begin{array}{*{20}{c}}
  8 \\
  0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  8 \\
  1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  8 \\
  2
\end{array}} \right) + ...\left( {\begin{array}{*{20}{c}}
  8 \\
  8
\end{array}} \right)\)     (A1)

\( = {2^8}{\text{ }}( = 256)\)     A1 

Note: FT in (ii) applies if no empty set included in (i) and (ii).

[5 marks]

This was also a well answered question with many candidates obtaining full marks on both parts of the problem. Some candidates attempted to use a factorial rather than a sum of combinations to solve part (b) (ii) and this led to incorrect answers.

This was also a well answered question with many candidates obtaining full marks on both parts of the problem. Some candidates attempted to use a factorial rather than a sum of combinations to solve part (b) (ii) and this led to incorrect answers.

Show that \(R\) is an equivalence relation.

Determine the equivalence classes of \(R\).

Determine the number of equivalence classes of \(S\).

METHOD 1

reflexive: \({4^a} - {4^a} = 0\) which is divisible by 7 (for all \(a \in \mathbb{Z}\))     R1

so \(aRa\) therefore reflexive

symmetric: Let \(aRb\) so that \({4^a} - {4^b}\) is divisible by 7     M1

it follows that \({4^b} - {4^a} = - ({4^a} - {4^b})\) is also divisible by 7     A1

it follows that \(bRa\) therefore symmetric

transitive: let \(aRb\) and \(bRc\) so that \({4^a} - {4^b}\) and \({4^b} - {4^c}\) are divisible by 7     M1

it follows that \({4^a} - {4^b} = 7M\) and \({4^b} - {4^c} = 7N\) so that \(({4^a} - {4^b}) + ({4^b} - {4^c}) = {4^a} - {4^c} = 7(M + N)\)     A1

therefore \(aRb\) and \(bRc \Rightarrow aRc\)     R1

so that \(R\) is transitive

Note:     For transitivity, award A0 if the same variable is used to express the multiples of 7; R1 is dependent on the M mark.

since \(R\) R is reflexive, symmetric and transitive, it is an equivalence relation     AG

METHOD 2

reflexive: \({4^a} - {4^a} \equiv 0\bmod 7\) (for all \(a \in \mathbb{Z}\))     R1

so \(aRa\) therefore reflexive

symmetric: let \(aRb\). Then \({4^a} - {4^b} \equiv 0\bmod 7\)     M1

it follows that \({4^b} - {4^a} \equiv - ({4^a} - {4^b}) \equiv 0\bmod 7\)     A1

it follows that \(bRa\) therefore symmetric

transitive: let \(aRb\) and \(bRc\), ie, \({4^a} - {4^b} \equiv 0\bmod 7\) and \({4^b} - {4^c} \equiv 0\bmod 7\)     M1

so that \({4^a} - {4^c} \equiv ({4^a} - {4^b}) + ({4^b} - {4^c}) \equiv 0\bmod 7\)     A1

therefore \(aRb\) and \(bRc \Rightarrow aRc\)     R1

so \(R\) is transitive

Note:     For transitivity, award A0 if mod 7 is omitted; R1 is dependent on the M mark.

since \(R\) is reflexive, symmetric and transitive, it is an equivalence relation     AG

[6 marks]

attempt to solve \({4^a} \equiv 4\bmod 7\) or \({4^a} \equiv {4^2} \equiv 2\bmod 7\) or \({4^a} \equiv {4^3} \equiv 1\bmod 7\)     (M1)

the equivalence classes are

\(\{ 1,{\text{ }}4,{\text{ }}7,{\text{ }} \ldots \} ,{\text{ \{ }}2,{\text{ }}5,{\text{ }}8,{\text{ }} \ldots \} \) and \(\{ 3,{\text{ }}6,{\text{ }}9,{\text{ }} \ldots \} \)     A2

Note:     Award (M1)A1 for one or two correct equivalence classes.

[3 marks]

starting with 1, we find that 2, 3, 4, … all belong to the same equivalence class or \({4^c} - 4 \equiv 4({4^{c - 1}} - 1) \equiv 4({2^{c - 1}} - 1)({2^{c - 1}} - 1) \equiv 0\bmod 6\) or \({4^c} \equiv 4\bmod 6\)     (M1)

therefore there is one equivalence class     A1

[2 marks]

State the possible orders of an element of \(\{ G,{\text{ }} * \} \) and for each order give an example of an element of that order.

(i)     State a generator for \(\{ H,{\text{ }} * \} \).

(ii)     Write down the elements of \(\{ H,{\text{ }} * \} \).

(iii)     Write down the elements of the coset of \(H\) containing \(a\).

orders are 1 2 3 4 6 12     A2

Note: A1 for four or five correct orders.

Note: For the rest of this question condone absence of xxx and accept equivalent expressions.

\(\begin{array}{*{20}{l}} {{\text{order:}}}&1&{{\text{element:}}}&2&{A1} \\ {}&2&{}&{{a^2}}&{A1} \\ {}&3&{}&{b{\text{ or }}{{\text{b}}^2}}&{A1} \\ {}&4&{}&{a{\text{ or }}{a^3}}&{A1} \\ {}&6&{}&{{a^2} * b{\text{ or }}{a^2} * {b^2}}&{A1} \\ {}&{12}&{}&{a * b{\text{ or }}a * {b^2}{\text{ or }}{a^3} * b{\text{ or }}{a^3} * {b^2}}&{A1} \end{array}\)

[8 marks]

(i)     \(H\) has order 6     (R1)

generator is \({a^2} * b\) or \({a^2} * {b^2}\)     A1

(ii)     \(H = \left\{ {e,{\text{ }}{a^2} * b,{\text{ }}{b^2},{\text{ }}{a^2},{\text{ }}b,{\text{ }}{a^2} * {b^2}} \right\}\)     A3

Note: A2 for 4 or 5 correct. A1 for 2 or 3 correct.

(iii)     required coset is \(Ha\) (or \(aH\))     (R1)

\(Ha = \left\{ {a,{\text{ }}{a^3} * b,{\text{ }}a * {b^2},{\text{ }}{a^3},{\text{ }}a * b,{\text{ }}{a^3} * {b^2}} \right\}\)    A1

[7 marks]

Show that \(R\) is reflexive.

Show that \(R\) is symmetric.

Show, by means of an example, that \(R\) is not transitive.

(for \(x \in \mathbb{R}\)), \(\left| x \right| + \left| x \right| = 2\left| x \right|\)    A1

and \(\left| x \right| + \left| x \right| = \left| {2x} \right| = 2\left| x \right|\)    A1

hence \(xRx\)

so \(R\) is reflexive    AG

Note: Award A1 for correct verification of identity for \(x\) > 0; A1 for correct verification for \(x\) ≤ 0.

[2 marks]

if \(xRy \Rightarrow \left| x \right| + \left| y \right| = \left| {x + y} \right|\)

\(\left| x \right| + \left| y \right| = \left| y \right| + \left| x \right|\)    A1

\(\left| {x + y} \right| = \left| {y + x} \right|\)     A1

hence \(yRx\)

so \(R\) is symmetric      AG

[2 marks]

recognising a condition where transitivity does not hold     (M1)

(eg, \(x\) > 0, \(y\) = 0 and \(z\) < 0)

for example, 1\(R\)0 and 0\(R\)(−1)    A1

however \(\left| 1 \right| + \left| { - 1} \right| \ne \left| {1 +  - 1} \right|\)      A1

so 1\(R\)(−1) (for example) is not true      R1

hence \(R\) is not transitive       AG

[4 marks]

(i)     Show that \(gh{g^{ - 1}}\) has order 2 for all \(g \in G\).

(ii)     Deduce that gh = hg for all \(g \in G\).

(i)     consider \({(gh{g^{ - 1}})^2}\)     M1

\( = gh{g^{ - 1}}gh{g^{ - 1}} = g{h^2}{g^{ - 1}} = g{g^{ - 1}} = e\)     A1

\(gh{g^{ - 1}}\) cannot be order 1 (= e) since h is order 2     R1

so \(gh{g^{ - 1}}\) has order 2     AG

(ii)     but h is the unique element of order 2     R1

hence \(gh{g^{ - 1}} = h \Rightarrow gh = hg\)     A1AG

[5 marks]

This question was by far the problem to be found most challenging by the candidates. Many were able to show that \(gh{g^{ - 1}}\) had order one or two although hardly any candidates also showed that the order was not one thus losing a mark. Part a (ii) was answered correctly by a few candidates who noticed the equality of h and \(gh{g^{ - 1}}\). However, many candidates went into algebraic manipulations that led them nowhere and did not justify any marks. Part (b) (i) was well answered by a small number of students who appreciated the nature of the identity and element h thus forcing the other two elements to have order four. However, (ii) was only occasionally answered correctly and even in these cases not systematically. It is possible that candidates lacked time to fully explore the problem. A small number of candidates “guessed” the correct answer.

Two functions, F and G , are defined on \(A = \mathbb{R}\backslash \{ 0,{\text{ }}1\} \) by

\[F(x) = \frac{1}{x},{\text{ }}G(x) = 1 - x,{\text{ for all }}x \in A.\]

(a)     Show that under the operation of composition of functions each function is its own inverse.

(b)     F and G together with four other functions form a closed set under the operation of composition of functions.

Find these four functions.

(a)     the following two calculations show the required result

\(F \circ F(x) = \frac{1}{{\frac{1}{x}}} = x\)

\(G \circ G(x) = 1 - (1 - x) = x\)     M1A1A1

[3 marks]

(b)     part (a) shows that the identity function defined by I(x) = x belongs to S     A1

the two compositions of F and G are:

\(F \circ G(x) = \frac{1}{{1 - x}};\)     (M1)A1

\(G \circ F(x) = 1 - \frac{1}{x}\left( { = \frac{{x - 1}}{x}} \right)\)     (M1)A1

the final element is

\(G \circ F \circ G(x) = 1 - \frac{1}{{1 - x}}\left( { = \frac{x}{{x - 1}}} \right)\)     (M1)A1

[7 marks]

Total [10 marks]

This question was generally well done. In part(a), the quickest answer involved showing that squaring the function gave the identity. Some candidates went through the more elaborate method of finding the inverse function in each case.

Find the element \(e\) such that \(e * y = y\), for all \(y \in S\).

(i)     Find the least solution of \(x * x = e\).

(ii)     Deduce that \((S,{\text{ }} * )\) is not a group.

Determine whether or not \(e\) is an identity element.

attempt to solve \({e^3}y - ey \equiv y\bmod 7\)     (M1)

the only solution is \(e = 5\)     A1

[2 marks]

(i)     attempt to solve \({x^4} - {x^2} \equiv 5\bmod 7\)     (M1)

least solution is \(x = 2\)     A1

(ii)     suppose \((S,{\text{ }} * )\) is a group with order 7     A1

\(2\) has order \(2\)     A1

since \(2\) does not divide \(7\), Lagrange’s Theorem is contradicted     R1

hence, \((S,{\text{ }} * )\) is not a group     AG

[5 marks]

(\(5\) is a left-identity), so need to test if it is a right-identity:

ie, is \(y * 5 = y\)?     M1

\(1 * 5 = 0 \ne 1\)     A1

so \(5\) is not an identity     A1

[3 marks]

Total [10 marks]

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange's theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange's theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange's theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow x + y > 7\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow - 2 < x - y < 2\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow xy > 0\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

Decide, giving a proof or a counter-example, whether \(xRy \Leftrightarrow \frac{x}{y} \in \mathbb{Z}\) is

(i)     reflexive;

(ii)     symmetric;

(iii)     transitive.

One of the relations from parts (a), (b), (c) and (d) is an equivalence relation.

For this relation, state what the equivalence classes are.

(i)     not reflexive e.g. 1 + 1 = 2     R1

(ii)     symmetric since x + y = y + x     R1

(iii)     e.g. 1 + 11 > 7, 11 + 2 > 7 but 1 + 2 = 3, so not transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

(i)     reflexive since \(x - x = 0\)     R1

(ii)     symmetric since \(\left| {x - y} \right| = \left| {y - x} \right|\)     R1

(iii)     e.g. 1R2, 2R3 but 1 − 3 = −2 , so not transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

(i)     reflexive since \({x^2} > 0\)     R1

(ii)     symmetric since \(xy = yx\)     R1

(iii)     \(xy > 0{\text{ and }}yz > 0 \Rightarrow x{y^2}z > 0 \Rightarrow xz > 0{\text{ since }}{y^2} > 0\), so transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

(i)     reflexive since \(\frac{x}{x} = 1\)     R1

(ii)     not symmetric e.g. \(\frac{2}{1} = 2{\text{ but }}\frac{1}{2} = 0.5\)     R1

(iii)     \(\frac{x}{y} \in \mathbb{Z}{\text{ and }}\frac{y}{z} \in \mathbb{Z} \Rightarrow \frac{{xy}}{{yz}} = \frac{x}{z} \in \mathbb{Z}\), so transitive     M1A1

Note: For each R1 mark the correct decision and a valid reason must be given.

[4 marks]

only (c) is an equivalence relation     (A1)

the equivalence classes are

{1, 2, 3,…} and {−1,−2,−3,…}     A1A1

[3 marks]

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

Generally this question was well answered, with students showing a sound knowledge of relations. There were a few candidates who mixed reflexive and symmetric qualities and marks were also lost because reasoning was either unclear or absent. Most students were able to offer counterexamples for transitivity in parts (a) and (b) but a number lost marks in failing to give adequate working to show transitivity in parts (c) and (d). That said, there were a pleasing number of good solutions here showing all the required rigour. Whilst most students were able to identify part (c) as an equivalence relation, surprisingly few gave the correct equivalence classes.

Write down all four subsets of A .

Construct the Cayley table for \(P(A)\) under \(\Delta \) .

Prove that \(\left\{ {P(A),{\text{ }}\Delta } \right\}\) is a group. You are allowed to assume that \(\Delta \) is associative.

Is \(\{ P(A){\text{, }}\Delta \} \) isomorphic to \(\{ {\mathbb{Z}_4},{\text{ }}{ + _4}\} \) ? Justify your answer.

(i)     State the identity element for \(\{ P(S){\text{, }}\Delta \} \).

(ii)     Write down \({X^{ - 1}}\) for \(X \in P(S)\) .

(iii)     Hence prove that \(\{ P(S){\text{, }}\Delta \} \) is a group.

Explain why \(\{ P(S){\text{, }} \cup \} \) is not a group.

Explain why \(\{ P(S){\text{, }} \cap \} \) is not a group.

\(\emptyset {\text{, \{ a\} , \{ b\} , \{ a, b\} }}\)     A1

[1 mark]

 A3

Note: Award A2 for one error, A1 for two errors, A0 for more than two errors.

[3 marks]

closure is seen from the table above     A1

\(\emptyset \) is the identity     A1

each element is self-inverse     A1

Note: Showing each element has an inverse is sufficient.

associativity is assumed so we have a group     AG

[3 marks]

not isomorphic as in the above group all elements are self-inverse whereas in \(({\mathbb{Z}_4},{\text{ }}{ + _4})\) there is an element of order 4 (e.g. 1)     R2

[2 marks]

(i)     \(\emptyset \) is the identity     A1

(ii)     \({X^{ - 1}} = X\)     A1

(iii)     if X and Y are subsets of S then \(X\Delta Y\) (the set of elements that belong to X or Y but not both) is also a subset of S, hence closure is proved     R1

\(\{ P(S){\text{, }}\Delta \} \) is a group because it is closed, has an identity, all elements have inverses (and \(\Delta \) is associative)     R1AG

[4 marks]

not a group because although the identity is \(\emptyset {\text{, if }}X \ne \emptyset \) it is impossible to find a set Y such that \(X \cup Y = \emptyset \), so there are elements without an inverse     R1AG

[1 mark]

not a group because although the identity is S, if \(X \ne S\) is impossible to find a set Y such that \(X \cap Y = S\), so there are elements without an inverse     R1AG

[1 mark]

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

Copy and complete the following Cayley table for \(\{ T,{\text{ }} * \} \).

Prove that \(\{ T,{\text{ }} * \} \) forms an Abelian group.

Find the order of each element in \(T\).

Given that \(\{ H,{\text{ }} * \} \) is the subgroup of \(\{ T,{\text{ }} * \} \) of order \(2\), partition \(T\) into the left cosets with respect to \(H\).

Cayley table is

     A4

award A4 for all 16 correct, A3 for up to 2 errors, A2 for up to 4 errors, A1 for up to 6 errors

[4 marks]

closed as no other element appears in the Cayley table     A1

symmetrical about the leading diagonal so commutative     R1

hence it is Abelian

\(0\) is the identity

as \(x * 0( = 0 * x) = x + 0 - 0 = x\)     A1

\(0\) and \(2\) are self inverse, \(3\) and \(5\) is an inverse pair, \(4\) and \(6\) is an inverse pair     A1

Note:     Accept “Every row and every column has a \(0\) so each element has an inverse”.

\((a * b) * c = (a + b - ab) * c = a + b - ab + c - (a + b - ab)c\)     M1

\( = a + b + c - ab - ac - bc + abc\)     A1

\(a * (b * c) = a * (b + c - bc) = a + b + c - bc - a(b + c - bc)\)     A1

\( = a + b + c - ab - ac - bc + abc\)

so \((a * b) * c = a * (b * c)\) and \( * \) is associative

Note:     Inclusion of mod 7 may be included at any stage.

[7 marks]

\(0\) has order \(1\) and \(2\) has order \(2\)    A1

\({3^2} = 4,{\text{ }}{3^3} = 2,{\text{ }}{3^4} = 6,{\text{ }}{3^5} = 5,{\text{ }}{3^6} = 0\) so \(3\) has order \(6\)     A1

\({4^2} = 6,{\text{ }}{4^3} = 0\) so \(4\) has order \(3\)     A1

\(5\) has order \(6\) and \(6\) has order \(3\)     A1

[4 marks]

\(H = \{ 0,{\text{ }}2\} \)     A1

\(0 * \{ 0,{\text{ }}2\}  = \{ 0,{\text{ }}2\} ,{\text{ }}2 * \{ 0,{\text{ }}2\}  = \{ 2,{\text{ }}0\} ,{\text{ }}3 * \{ 0,{\text{ }}2\}  = \{ 3,{\text{ }}6\} ,{\text{ }}4 * \{ 0,{\text{ }}2\}  = \{ 4,{\text{ }}5\} ,\)

\(5 * \{ 0,{\text{ }}2\}  = \{ 5,{\text{ }}4\} ,{\text{ }}6 * \{ 0,{\text{ }}2\}  = \{ 6,{\text{ }}3\} \)     M1

Note:     Award the M1 if sufficient examples are used to find at least two of the cosets.

so the left cosets are \(\{ 0,{\text{ }}2\} ,{\text{ }}\{ 3,{\text{ }}6\} ,{\text{ }}\{ 4,{\text{ }}5\} \)     A1

[3 marks]

Total [18 marks]

Show that \(f\) is a bijection.

Hence write down the inverse function \({f^{ - 1}}(x,{\text{ }}y)\).

for \(f\) to be a bijection it must be both an injection and a surjection     R1

Note:     Award this R1 for stating this anywhere.

suppose that \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\)     (M1)

it follows that

\(2{a^3} + {b^3} = 2{c^3} + {d^3}\) and \({a^3} + 2{b^3} = {c^3} + 2{d^3}\)     A1

attempting to solve the two equations     M1

to obtain \(3{a^3} = 3{c^3}\)

Note:     Award M1 only if a good attempt is made to solve the system.

\( \Rightarrow a = c\) and therefore \(b = d\)     A1

\(f\) is an injection because \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d)\)     R1

Note:     Award this R1 for stating this anywhere providing that an attempt is made to prove injectivity.

let \((p,{\text{ }}q) \in \mathbb{R} \times \mathbb{R}\) and let \(f(r,{\text{ }}s) = (p,{\text{ }}q)\)     (M1)

then, \(p = 2{r^3} + {s^3}\) and \(q = {r^3} + 2{s^3}\)     A1

attempting to solve the two equations     M1

\(r = \sqrt[3]{{\frac{{2p - q}}{3}}}\) and \(s = \sqrt[3]{{\frac{{2q - p}}{3}}}\)     A1A1

\(f\) is a surjection because given \((p,{\text{ }}q) \in \mathbb{R} \times \mathbb{R}\), there exists \((r,{\text{ }}s) \in \mathbb{R} \times \mathbb{R}\) such that \(f(r,{\text{ }}s) = (p,{\text{ }}q)\)     R1

Note:   Award this R1 for stating this anywhere providing that an attempt is made to prove surjectivity.

[12 marks]

\(\left( {{f^{ - 1}}(x,{\text{ }}y) = } \right)\,\,\,\left( {\sqrt[3]{{\frac{{2x - y}}{3},}}{\text{ }}\sqrt[3]{{\frac{{2y - x}}{3}}}} \right)\)     A1

Note:     A1 for correct expressions in \(x\) and \(y\), allow FT only if the expression is deduced in part (a).

[1 mark]

(i)     Sketch the graph of \(y = f(x)\) and hence justify whether or not \(f\) is a bijection.

(ii)     Show that \(A\) is a group under the binary operation of multiplication.

(iii)     Give a reason why \(B\) is not a group under the binary operation of multiplication.

(iv)     Find an example to show that \(f(a \times b) = f(a) \times f(b)\) is not satisfied for all \(a,{\text{ }}b \in A\).

(i)     Sketch the graph of \(y = g(x)\) and hence justify whether or not \(g\) is a bijection.

(ii)     Show that \(g(a + b) = g(a) \times g(b)\) for all \(a,{\text{ }}b \in \mathbb{R}\).

(iii)     Given that \(\{ \mathbb{R},{\text{ }} + \} \) and \(\{ D,{\text{ }} \times \} \) are both groups, explain whether or not they are isomorphic.

(i)          A1

Notes: Award A1 for general shape, labelled asymptotes, and showing that \(x \ne 0\).

graph shows that it is injective since it is increasing or by the horizontal line test     R1

graph shows that it is surjective by the horizontal line test     R1

Note: Allow any convincing reasoning.

so \(f\) is a bijection     A1

(ii)     closed since non-zero real times non-zero real equals non-zero real     A1R1

we know multiplication is associative     R1

identity is 1     A1

inverse of \(x\) is \(\frac{1}{x}(x \ne 0)\)     A1

hence it is a group     AG

(iii)     \(B\) does not have an identity     A2

hence it is not a group     AG

(iv)     \(f(1 \times 1) = f(1) = \frac{1}{2}\) whereas \(f(1) \times f(1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) is one counterexample     A2

hence statement is not satisfied     AG

[13 marks]

award A1 for general shape going through (0, 1) and with domain \(\mathbb{R}\)     A1

graph shows that it is injective since it is increasing or by the horizontal line test and graph shows that it is surjective by the horizontal line test     R1

Note: Allow any convincing reasoning.

so \(g\)  is a bijection     A1

(ii)     \(g(a + b) = {{\text{e}}^{a + b}}\) and \(g(a) \times g(b) = {{\text{e}}^a} \times {{\text{e}}^b} = {{\text{e}}^{a + b}}\)     M1A1

hence \(g(a + b) = g(a) \times g(b)\)     AG

(iii)     since \(g\) is a bijection and the homomorphism rule is obeyed     R1R1

the two groups are isomorphic     A1

[8 marks]

(a)     Show that \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) defined by \(f(x,{\text{ }}y) = (2x + y,{\text{ }}x - y)\) is a bijection.

(b)     Find the inverse of f .

(a)     we need to show that the function is both injective and surjective to be a bijection     R1

suppose \(f(x,{\text{ }}y) = f(u,{\text{ }}v)\)     M1

\((2x + y,{\text{ }}x - y) = (2u + v,{\text{ }}u - v)\)

forming a pair of simultaneous equations     M1

\(2x + y = 2u + v\)     (i)

\(x - y = u - v\)     (ii)

\((i) + (ii) \Rightarrow 3x = 3u \Rightarrow x = u\)     A1

\((i) - 2(ii) \Rightarrow 3y = 3v \Rightarrow y = v\)     A1

hence function is injective     R1

let \(2x + y = s\) and \(x - y = t\)     M1

\( \Rightarrow 3x = s + t\)

\( \Rightarrow x = \frac{{s + t}}{3}\)     A1

also \(3y = s - 2t\)

\( \Rightarrow y = \frac{{s - 2t}}{3}\)     A1

for any \((s,{\text{ }}t) \in \mathbb{R} \times \mathbb{R}\) there exists \((x,{\text{ }}y) \in \mathbb{R} \times \mathbb{R}\) and the function is surjective     R1

[10 marks]

(b)     the inverse is \({f^{ - 1}}(x,{\text{ }}y) = \left( {\frac{{x + y}}{3},{\text{ }}\frac{{x - 2y}}{3}} \right)\)     A1

[1 mark]

Total [11 marks]

Many students were able to show that the expression was injective, but found more difficulty in showing it was subjective. As with question 1 part (e), a number of candidates did not realise that the answer to part (b) came directly from part (a), hence the reason for it being worth only one mark.

(i)     Show that \( * \) is commutative.

(ii)     Find the identity element.

(iii)     Find the inverse of the element a .

The binary operation \( \cdot \) is defined on \(\mathbb{R}\) as follows. For any elements a , \(b \in \mathbb{R}\)

\(a \cdot b = 3ab\) . The set S is the set of all ordered pairs \((x,{\text{ }}y)\) of real numbers and the binary operation \( \odot \) is defined on the set S as

\(({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2}) = ({x_1} * {x_2},{\text{ }}{y_1} \cdot {y_2}){\text{ }}.\)

Determine whether or not \( \odot \) is associative.

(i)     if \( * \) is commutative \(a * b = b * a\)

since \(a + b + 1 = b + a + 1\) , \( * \) is commutative     R1

(ii)     let e be the identity element

\(a * e = a + e + 1 = a\)     M1

\( \Rightarrow e = - 1\)     A1

(iii)     let a have an inverse, \({a^{ - 1}}\)

\(a * {a^{ - 1}} = a + {a^{ - 1}} + 1 = - 1\)     M1

\( \Rightarrow {a^{ - 1}} = - 2 - a\)     A1

[5 marks]

\(({x_1},{\text{ }}{y_1}) \odot \left( {({x_2},{\text{ }}{y_2}) \odot ({x_3},{\text{ }}{y_3})} \right) = ({x_1},{\text{ }}{y_1}) \odot ({x_2} + {x_3} + 1,{\text{ }}3{y_2}{y_3})\)     M1

\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\)     A1A1

\(\left( {({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2})} \right) \odot ({x_3},{\text{ }}{y_3}) = ({x_1} + {x_2} + 1,{\text{ }}3{y_1}{y_2}) \odot ({x_3},{\text{ }}{y_3})\)     M1

\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\)     A1

hence \( \odot \) is associative     R1

[6 marks]

Part (a) of this question was the most accessible on the paper and was completed correctly by the majority of candidates.

Part (b) was completed by many candidates, but a significant number either did not understand what was meant by associative, confused associative with commutative, or were unable to complete the algebra.

(a)     Draw the Cayley table for the set of integers G = {0, 1, 2, 3, 4, 5} under addition modulo 6, \({ + _6}\).

(b)     Show that \(\{ G,{\text{ }}{ + _6}\} \) is a group.

(c)     Find the order of each element.

(d)     Show that \(\{ G,{\text{ }}{ + _6}\} \) is cyclic and state its generators.

(e)     Find a subgroup with three elements. 

(f)     Find the other proper subgroups of \(\{ G,{\text{ }}{ + _6}\} \).

(a)         A3

Note: Award A2 for 1 error, A1 for 2 errors and A0 for more than 2 errors.

[3 marks]

(b)     The table is closed     A1

Identity element is 0     A1

Each element has a unique inverse (0 appears exactly once in each row and column)     A1

Addition mod 6 is associative     A1

Hence \(\{ G,{\text{ }}{ + _6}\} \) forms a group     AG

[4 marks]

(c)     0 has order 1 (0 = 0),

1 has order 6 (1 + 1 + 1 + 1 + 1 + 1 = 0),

2 has order 3 (2 + 2 + 2 = 0),

3 has order 2 (3 + 3 = 0),

4 has order 3 (4 + 4 + 4 = 0),

5 has order 6 (5 + 5 + 5 + 5 + 5 + 5 = 0).     A3

Note: Award A2 for 1 error, A1 for 2 errors and A0 for more than 2 errors.

[3 marks]

(d)     Since 1 and 5 are of order 6 (the same as the order of the group) every element can be written as sums of either 1 or 5. Hence the group is cyclic.     R1

The generators are 1 and 5.     A1

[2 marks]

(e)     A subgroup of order 3 is \(\left( {\{ 0,{\text{ }}2,{\text{ }}4\} ,{\text{ }}{ + _6}} \right)\)     A2

Note: Award A1 if only {0, 2, 4} is seen.

[2 marks]

(f)     Other proper subgroups are \(\left( {\{ 0\} { + _6}} \right),{\text{ }}\left( {\{ 0,{\text{ }}3\} { + _6}} \right)\)     A1A1

Note: Award A1 if only {0}, {0, 3} is seen.

[2 marks]

Total [16 marks]

The table was well done as was showing its group properties. The order of the elements in (b) was done well except for the order of 0 which was often not given. Finding the generators did not seem difficult but correctly stating the subgroups was not often done. The notion of a ‘proper’ subgroup is not well known.

The function \(f:[0,{\text{ }}\infty [ \to [0,{\text{ }}\infty [\) is defined by \(f(x) = 2{{\text{e}}^x} + {{\text{e}}^{ - x}} - 3\) .

(a)     Find \(f'(x)\) .

(b)     Show that f is a bijection.

(c)     Find an expression for \({f^{ - 1}}(x)\) .

(a)     \(f'(x) = 2{{\text{e}}^x} - {{\text{e}}^{ - x}}\)     A1

[1 mark]

(b)     f is an injection because \(f'(x) > 0\) for \(x \in [0,{\text{ }}\infty [\)     R2

(accept GDC solution backed up by a correct graph)

since \(f(0) = 0\) and \(f(x) \to \infty \) as \(x \to \infty \) , (and f is continuous) it is a surjection     R1

hence it is a bijection     AG

[3 marks]

(c)     let \(y = 2{{\text{e}}^x} + {{\text{e}}^{ - x}} - 3\)     M1

so \(2{{\text{e}}^{2x}} - (y + 3){{\text{e}}^x} + 1 = 0\)     A1

\({{\text{e}}^x} = \frac{{y + 3 \pm \sqrt {{{(y + 3)}^2} - 8} }}{4}\)     A1

\(x = \ln \left( {\frac{{y + 3 \pm \sqrt {{{(y + 3)}^2} - 8} }}{4}} \right)\)     A1

since \(x \geqslant 0\) we must take the positive square root     (R1)

\({f^{ - 1}}(x) = \ln \left( {\frac{{x + 3 + \sqrt {{{(x + 3)}^2} - 8} }}{4}} \right)\)     A1

[6 marks]

Total [10 marks]

In many cases the attempts at showing that f is a bijection were unconvincing. The candidates were guided towards showing that f is an injection by noting that \(f'(x) > 0\) for all x, but some candidates attempted to show that \(f(x) = f(y) \Rightarrow x = y\) which is much more difficult. Solutions to (c) were often disappointing, with the algebra defeating many candidates.

A \ B ;

\(A\Delta B\) .

A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}     (A1)

B = {3, 8, 13, 18, 23, 28}     (A1)

Note: FT on their A and B

A \ B = {elements in A that are not in B}     (M1)

= {2, 5, 7, 11, 17, 19, 29}     A1

[4 marks]

\(B\backslash A\) = {8, 18, 28}     (A1)

\(A\Delta B = (A\backslash B) \cup (B\backslash A)\)     (M1)

= {2, 5, 7, 8, 11, 17, 18, 19, 28, 29}     A1

[3 marks]

It was disappointing to find that many candidates wrote the elements of A and B incorrectly. The most common errors were the inclusion of 1 as a prime number and the exclusion of 3 in B. It has been suggested that some candidates use N to denote the positive integers. If this is the case, then it is important to emphasise that the IB notation is that N denotes the positive integers and zero and IB candidates should all be aware of that. Most candidates solved the remaining parts of the question correctly and follow through ensured that those candidates with incorrect A and/or B were not penalised any further.

It was disappointing to find that many candidates wrote the elements of A and B incorrectly. The most common errors were the inclusion of 1 as a prime number and the exclusion of 3 in B. It has been suggested that some candidates use N to denote the positive integers. If this is the case, then it is important to emphasise that the IB notation is that N denotes the positive integers and zero and IB candidates should all be aware of that. Most candidates solved the remaining parts of the question correctly and follow through ensured that those candidates with incorrect A and/or B were not penalised any further.

is closed;

is commutative;

is associative;

has an identity element.

\( * \) is closed     A1

because \(1 + ab \in \mathbb{N}\) (when \(a,b \in \mathbb{N}\))     R1

[2 marks]

consider

\(a * b = 1 + ab = 1 + ba = b * a\)     M1A1

therefore \( * \) is commutative

[2 marks]

EITHER

\(a * (b * c) = a * (1 + bc) = 1 + a(1 + bc){\text{ }}( = 1 + a + abc)\)     A1

\((a * b) * c = (1 + ab) * c = 1 + c(1 + ab){\text{ }}( = 1 + c + abc)\)     A1

(these two expressions are unequal when \(a \ne c\)) so \( * \) is not associative     R1

OR

proof by counter example, for example

\(1 * (2 * 3) = 1 * 7 = 8\)     A1

\((1 * 2) * 3 = 3 * 3 = 10\)     A1

(these two numbers are unequal) so \( * \) is not associative     R1

[3 marks]

let e denote the identity element; so that

\(a * e = 1 + ae = a\) gives \(e = \frac{{a - 1}}{a}\) (where \(a \ne 0\))     M1

then any valid statement such as: \(\frac{{a - 1}}{a} \notin \mathbb{N}\) or e is not unique     R1

there is therefore no identity element     A1

Note: Award the final A1 only if the previous R1 is awarded.

[3 marks]

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

Find the values represented by each of the letters in the table.

Find the order of each of the elements of the group.

Write down the three sets that form subgroups of order 2.

Find the three sets that form subgroups of order 4.

\(a = 1\;\;\;b = 8\;\;\;c = 4\)

\(d = 8\;\;\;e = 4\;\;\;f = 2\)

\(g = 4\;\;\;h = 2\;\;\;i = 1\)     A3

Note:     Award A3 for 9 correct answers, A2 for 6 or more, and A1 for 3 or more.

[3 marks]

    A3

Note:     Award A3 for 8 correct answers, A2 for 6 or more, and A1 for 4 or more.

[3 marks]

\(\{ 1,{\text{ }}4\} ,{\text{ }}\{ 1,{\text{ }}11\} ,{\text{ }}\{ 1,{\text{ }}14\} \)     A1A1

Note:     Award A1 for 1 correct answer and A2 for all 3 (and no extras).

[2 marks]

\(\{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}8\} ,{\text{ }}\{ 1,{\text{ }}4,{\text{ }}7,{\text{ }}13\} ,\)     A1A1

\(\{ 1,{\text{ }}4,{\text{ }}11,{\text{ }}14\} \)     A2

[4 marks]

Total [12 marks]

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

Given that p , q and r are elements of a group, prove the left-cancellation rule, i.e. \(pq = pr \Rightarrow q = r\) .

Your solution should indicate which group axiom is used at each stage of the proof.

Consider the group G , of order 4, which has distinct elements a , b and c and the identity element e .

(i)     Giving a reason in each case, explain why ab cannot equal a or b .

(ii)     Given that c is self inverse, determine the two possible Cayley tables for G .

(iii)     Determine which one of the groups defined by your two Cayley tables is isomorphic to the group defined by the set {1, −1, i, −i} under multiplication of complex numbers. Your solution should include a correspondence between a, b, c, e and 1, −1, i, −i .

\(pq = pr\)

\({p^{ - 1}}(pq) = {p^{ - 1}}(pr)\) , every element has an inverse     A1

\(({p^{ - 1}}p)q = ({p^{ - 1}}p)r\) , Associativity     A1

Note: Brackets in lines 2 and 3 must be seen.

\(eq = er\), \({p^{ - 1}}p = e\), the identity     A1

\(q = r\), \(ea = a\) for all elements a of the group     A1

[4 marks]

(i)     let ab = a so b = e be which is a contradiction     R1

let ab = b so a = e which is a contradiction     R1

therefore ab cannot equal either a or b     AG

(ii)     the two possible Cayley tables are

table 1

     A2

table 2

     A2

(iii)     the group defined by table 1 is isomorphic to the given group     R1

because

EITHER

both contain one self-inverse element (other than the identity)     R1

OR

both contain an inverse pair     R1

OR

both are cyclic     R1

THEN

the correspondence is \(e \to 1\), \(c \to - 1\), \(a \to i\), \(b \to - i\)

(or vice versa for the last two)     A2

Note: Award the final A2 only if the correct group table has been identified.

[10 marks]

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see \(pq = pr\)

\({p^{ - 1}}pq = {p^{ - 1}}pr\)

\(q = r\)

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as ab and bc without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see \(pq = pr\)

\({p^{ - 1}}pq = {p^{ - 1}}pr\)

\(q = r\)

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as ab and bc without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

A binary operation is defined on {−1, 0, 1} by

\[A \odot B = \left\{ {\begin{array}{*{20}{c}}
  { - 1,}&{{\text{if }}\left| A \right| < \left| B \right|} \\
  {0,}&{{\text{if }}\left| A \right| = \left| B \right|} \\
  {1,}&{{\text{if }}\left| A \right| > \left| B \right|{\text{.}}}
\end{array}} \right.\]

(a)     Construct the Cayley table for this operation.

(b)     Giving reasons, determine whether the operation is

(i)     closed;

(ii)     commutative;

(iii)     associative.

(a)     the Cayley table is

\(\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&{ - 1}&0&1
\end{array} \\
  \begin{array}{*{20}{c}}
  { - 1} \\
  0 \\
  1
\end{array}\left( {\begin{array}{*{20}{c}}
  0&1&0 \\
  { - 1}&0&{ - 1} \\
  0&1&0
\end{array}} \right) \\
\end{gathered} \)     M1A2

Notes: Award M1 for setting up a Cayley table with labels.

Deduct A1 for each error or omission.

[3 marks]

(b)     (i)     closed     A1

because all entries in table belong to {–1, 0, 1}     R1

(ii)     not commutative     A1

because the Cayley table is not symmetric, or counter-example given     R1

(iii)     not associative     A1

for example because     M1

\(0 \odot ( - 1 \odot 0) = 0 \odot 1 = - 1\)

but

\((0 \odot  - 1) \odot 0 = - 1 \odot 0 = 1\)     A1

or alternative counter-example

[7 marks]

Total [10 marks]

This question was generally well done, with the exception of part(b)(iii), showing that the operation is non-associative.

(i)     Sketch the set \(X \times Y\) in the Cartesian plane.

(ii)     Sketch the set \(Y \times X\) in the Cartesian plane.

(iii)     State \((X \times Y) \cap (Y \times X)\).

Consider the function \(f:X \times Y \to \mathbb{R}\) defined by \(f(x,{\text{ }}y) = x + y\) and the function \(g:X \times Y \to \mathbb{R}\) defined by \(g(x,{\text{ }}y) = xy\).

(i)     Find the range of the function f.

(ii)     Find the range of the function g.

(iii)     Show that \(f\) is an injection.

(iv)     Find \({f^{ - 1}}(\pi )\), expressing your answer in exact form.

(v)     Find all solutions to \(g(x,{\text{ }}y) = \frac{1}{2}\).

(i)     

correct horizontal lines     A1

correctly labelled axes     A1

clear indication that the endpoints are not included     A1

(ii)     

fully correct diagram     A1

Note:     Do not penalize the inclusion of endpoints twice.

(iii)     the intersection is empty     A1

[5 marks]

(i)     range \((f) = \left] {0,{\text{ 1}}} \right[ \cup \left] {1,{\text{ 2}}} \right[ \cup \rm{L} \cup \left] {5,{\text{ 6}}} \right[\)     A1A1

Note:     A1 for six intervals and A1 for fully correct notation.

     Accept \(0 < x < 6,{\text{ }}x \ne 0{\text{, 1, 2, 3, 4, 5, 6}}\).

(ii)     range \((g) = \left[ {0,{\text{ 5}}} \right[\)     A1

(iii)     Attempt at solving

\(f({x_1},{\text{ }}{y_1}) = f({x_2},{\text{ }}{y_2})\)     M1

\(f(x,{\text{ }}y) \in \left] {y,{\text{ }}y + 1} \right[ \Rightarrow {y_1} = {y_2}\)     M1

and then \({x_1} = {x_2}\)     A1

so \(f\) is injective     AG

(iv)     \({f^{ - 1}}(\pi ) = (\pi  - 3,{\text{ }}3)\)     A1A1

(v)     solutions: (0.5, 1), (0.25, 2), \(\left( {\frac{1}{6},{\text{ 3}}} \right)\), (0.125, 4), (0.1, 5)     A2

Note:     A2 for all correct, A1 for 2 correct.

[10 marks]

Prove that \(f({e_G}) = {e_H}\), where \({e_G}\) is the identity element in \(G\) and \({e_H}\) is the identity

element in \(H\).

(i)     Prove that the kernel of \(f,{\text{ }}K = {\text{Ker}}(f)\), is closed under the group operation.

(ii)     Deduce that \(K\) is a subgroup of \(G\).

(i)     Prove that \(gk{g^{ - 1}} \in K\) for all \(g \in G,{\text{ }}k \in K\).

(ii)     Deduce that each left coset of K in G is also a right coset.

\(f(g) = f({e_G}g) = f({e_G})f(g)\) for \(g \in G\)     M1A1

\( \Rightarrow f({e_G}) = {e_H}\)     AG

[2 marks]

(i)     closure: let \({k_1}\) and \({k_2} \in K\), then \(f({k_1}{k_2}) = f({k_1})f({k_2})\)     M1A1

          \( = {e_H}{e_H} = {e_H}\)     A1

          hence \({k_1}{k_2} \in K\)     R1

(ii)     K is non-empty because \({e_G}\) belongs to K     R1

          a closed non-empty subset of a finite group is a subgroup     R1AG

[6 marks]

(i)     \(f(gk{g^{ - 1}}) = f(g)f(k)f({g^{ - 1}})\)     M1

\( = f(g){e_H}f({g^{ - 1}}) = f(g{g^{ - 1}})\)     A1

\( = f({e_G}) = {e_H}\)     A1

\( \Rightarrow gk{g^{ - 1}} \in K\)     AG

(ii)     clear definition of both left and right cosets, seen somewhere.     A1

use of part (i) to show \(gK \subseteq Kg\)     M1

similarly \(Kg \subseteq gK\)     A1

hence \(gK = Kg\)     AG

[6 marks]

Prove that: 

(i)     \(f\) is an injection,

(ii)     \(g\) is a surjection.

Given that \(X = {\mathbb{R}^ + } \cup \{ 0\} \) and \(Y = \mathbb{R}\), choose a suitable pair of functions \(f\) and \(g\) to show that \(g\) is not necessarily a bijection.

(i)     to test injectivity, suppose \(f({x_1}) = f({x_2})\)     M1

apply \(g\) to both sides \(g\left( {f({x_1})} \right) = g\left( {f({x_2})} \right)\)     M1

\( \Rightarrow {x_1} = {x_2}\)     A1

so \(f\) is injective     AG

Note:     Do not accept arguments based on “\(f\) has an inverse”.

(ii)     to test surjectivity, suppose \(x \in X\)     M1

define \(y = f(x)\)     M1

then \(g(y) = g\left( {f(x)} \right) = x\)     A1

so \(g\) is surjective     AG

[6 marks]

choose, for example, \(f(x) = \sqrt x \) and \(g(y) = {y^2}\)     A1

then \(g \circ f(x) = {\left( {\sqrt x } \right)^2} = x\)     A1

the function \(g\) is not injective as \(g(x) = g( - x)\)     R1

[3 marks]

Total [9 marks]

Those candidates who formulated the questions in terms of the basic definitions of injectivity and surjectivity were usually sucessful. Otherwise, verbal attempts such as '\(f{\text{ is one - to - one }} \Rightarrow f{\text{ is injective}}\)' or '\(g\) is surjective because its range equals its codomain', received no credit. Some candidates made the false assumption that \(f\) and \(g\) were mutual inverses.

Few candidates gave completely satisfactory answers. Some gave functions satisfying the mutual identity but either not defined on the given sets or for which \(g\) was actually a bijection.

Let \((H,{\text{ }} * {\text{)}}\) be a subgroup of the group \((G,{\text{ }} * {\text{)}}\).

Consider the relation \(R\) defined in \(G\) by \(xRy\) if and only if \({y^{ - 1}} * x \in H\).

(a)     Show that \(R\) is an equivalence relation on \(G\).

(b)     Determine the equivalence class containing the identity element.

(a)     \(R\) is reflexive as \({x^{ - 1}} * x = e \in H \Rightarrow xRx\) for any \(x \in G\)     A1

if \(xRy\) then \({y^{ - 1}} * x = h \in H\)

but \(h \in H \Rightarrow {h^{ - 1}} \in H\), ie, \(\underbrace {{{({y^{ - 1}} * x)}^{ - 1}}}_{{x^{ - 1}}{\text{*}}y} \in H\)     M1

therefore \(yRx\)

\(R\) is symmetric     A1

if \(xRy\) then \({y^{ - 1}} * x = h \in H\) and if \(yRz\) then \({z^{ - 1}} * y = k \in H\)     M1

\(k * h \in H\), ie, \(\underbrace {({z^{ - 1}} * y) * ({y^{ - 1}} * x)}_{{z^{ - 1}} * x} \in H\)     A1

therefore \(xRz\)

\(R\) is transitive     A1

so \(R\) is an equivalence relation on \(G\)     AG

[6 marks]

(b)     \(xRe \Leftrightarrow {e^{ - 1}} * x \in H\)     M1

\(x \in H\)     A1

\([e] = H\)     A1     N0

[3 marks]

Part (a) was fairly well answered by many candidates. They knew how to apply the equivalence relations axioms in this particular example. Part (b) however proved to be very challenging and hardly any correct answers were seen.

Two members of \(A\) are given by \(p = (1{\text{ }}2{\text{ }}5)\) and \(q = (1{\text{ }}3)(2{\text{ }}5)\).

Find the single permutation which is equivalent to \(q \circ p\).

State a permutation belonging to \(A\) of order

(i)     \(4\);

(ii)     \(6\).

Let \(P = \) {all permutations in \(A\) where exactly two integers change position},

and \(Q = \) {all permutations in \(A\) where the integer \(1\) changes position}.

(i)     List all the elements in \(P \cap Q\).

(ii)     Find \(n(P \cap Q')\).

\(q \circ p = (1{\text{ }}3)(2{\text{ }}5)(1{\text{ }}2{\text{ }}5)\)     (M1)

\( = (1{\text{ }}5{\text{ }}3)\)     M1A1A1

Note:     M1 for an answer consisting of disjoint cycles, A1 for \((1{\text{ }}5{\text{ }}3)\),

A1 for either \((2)\) or \((2)\) omitted.

Note:     Allow \(\left( {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&2&1&4&3 \end{array}} \right)\)

If done in the wrong order and obtained \((1{\text{ }}3{\text{ }}2)\), award A2.

[4 marks]

(i)     any cycle with length \(4\) eg (\(1234\))     A1

(ii)     any permutation with \(2\) disjoint cycles one of length \(2\) and one of length \(3\) eg (\(1\) \(2\)) (\(3\) \(4\) \(5\))     M1A1

Note:     Award M1A0 for any permutation with \(2\) non-disjoint cycles one of length \(2\) and one of length \(3\).

Accept non cycle notation.

[3 marks]

(i)     (\(1\), \(2\)), (\(1\), \(3\)), (\(1\), \(4\)), (\(1\), \(5\))     M1A1

(ii)     (\(2\) \(3\)), (\(2\) \(4\)), (\(2\) \(5\)), (\(3\) \(4\)), (\(3\) \(5\)), (\(4\) \(5\))     (M1)

6     A1

Note:     Award M1 for at least one correct cycle.

[4 marks]

Total [11 marks]

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

Given the sets \(A\) and \(B\), use the properties of sets to prove that \(A \cup (B' \cup A)' = A \cup B\), justifying each step of the proof.

\(A \cup (B' \cup A)' = A \cup (B \cap A')\)     De Morgan     M1A1

\( = (A \cup B) \cap (A \cup A')\)     Distributive property     M1A1

\( = (A \cup B) \cap U\)     (Union of set and its complement)     A1

\( = A \cup B\)     (Intersection with the universal set)     AG

Note:     Do not accept proofs using Venn diagrams unless the properties are clearly stated.

Note:     Accept double inclusion proofs: M1A1 for each inclusion, final A1 for conclusion of equality of sets.

[5 marks]

(a)     Write down why the table below is a Latin square.

\[\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&d&e&b&a&c
\end{array} \\
  \begin{array}{*{20}{c}}
  d \\
  e \\
  b \\
  a \\
  c
\end{array}\left[ {\begin{array}{*{20}{c}}
  c&d&e&b&a \\
  d&e&b&a&c \\
  a&b&d&c&e \\
  b&a&c&e&d \\
  e&c&a&d&b
\end{array}} \right] \\
\end{gathered} \]

(b)     Use Lagrange’s theorem to show that the table is not a group table.

(a)     Each row and column contains all the elements of the set.     A1A1

[2 marks]

(b)     There are 5 elements therefore any subgroup must be of an order that is a factor of 5     R2

But there is a subgroup \(\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&e&a
\end{array} \\
  \begin{array}{*{20}{c}}
  e \\
  a
\end{array}\left( {\begin{array}{*{20}{c}}
  e&a \\
  a&e
\end{array}} \right) \\
\end{gathered} \) of order 2 so the table is not a group table     R2

Note: Award R0R2 for “a is an element of order 2 which does not divide the order of the group”.

[4 marks]

Total [6 marks]

Part (a) presented no problem but finding the order two subgroups (Lagrange’s theorem was often quoted correctly) was beyond some candidates. Possibly presenting the set in non-alphabetical order was the problem.

Let \(p = {2^k} + 1,{\text{ }}k \in {\mathbb{Z}^ + }\) be a prime number and let G be the group of integers 1, 2, ..., p − 1 under multiplication defined modulo p.

By first considering the elements \({2^1},{\text{ }}{2^2},{\text{ ..., }}{2^k}\) and then the elements \({2^{k + 1}},{\text{ }}{2^{k + 2}},{\text{ …,}}\) show that the order of the element 2 is 2k.

Deduce that \(k = {2^n}{\text{ for }}n \in \mathbb{N}\) .

The identity is 1.     (R1)

Consider

\({2^1},{\text{ }}{2^2},{\text{ }}{2^3},{\text{ ..., }}{2^k}\)

\({2^k} = p - 1\)     R1

Therefore all the above powers of two are different     R1

Now consider

\({2^{k + 1}} \equiv 2p - 2(\bmod p) = p - 2\)     M1A1

\({2^{k + 2}} \equiv 2p - 4(\bmod p) = p - 4\)     A1

\({2^{k + 3}} = p - 8\)

etc.

\({2^{2k - 1}} = p - {2^{k - 1}}\)

\({2^{2k}} = p - {2^k}\)     A1

\( = 1\)     A1

and this is the first power of 2 equal to 1.     R2

The order of 2 is therefore 2k.     AG

Using Lagrange’s Theorem, it follows that 2k is a factor of \({2^k}\) , the order of the group, in which case k must be as given.     R2

[12 marks]

Few solutions were seen to this question with many candidates unable even to start.

Prove that \((A \cap B)\backslash (A \cap C) = A \cap (B\backslash C)\) where A, B and C are three subsets of the universal set U.

\((A \cap B)\backslash (A \cap C) = (A \cap B) \cap (A \cap C)'\)     M1

\( = (A \cap B) \cap (A' \cup C')\)     A1

\( = (A \cap B \cap A') \cup (A \cap B \cap C')\)     A1

\( = (A \cap A' \cap B) \cup (A \cap B \cap C')\)     A1

\( = (\emptyset  \cap B) \cup (A \cap B \cap C')\)     (A1)

\( = \emptyset  \cup (A \cap B \cap C')\)

\( = \left( {A \cap (B \cap C')} \right)\)     A1

\( = A \cap (B\backslash C)\)     AG

Note: Do not accept proofs by Venn diagram.

[6 marks]

Venn diagram ‘proof’ are not acceptable. Those who used de Morgan’s laws usually were successful in this question.

Let \(\{ G,{\text{ }} * \} \) be a finite group and let H be a non-empty subset of G . Prove that \(\{ H,{\text{ }} * \} \) is a group if H is closed under \( * \).

the associativity property carries over from G     R1

closure is given     R1

let \(h \in H\) and let n denote the order of h, (this is finite because G is finite)     M1

it follows that \({h^n} = e\), the identity element     R1

and since H is closed, \(e \in H\)     R1

since \(h * {h^{n - 1}} = e\)     M1

it follows that \({h^{n - 1}}\) is the inverse, \({h^{ - 1}}\), of h     R1

and since H is closed, \({h^{ - 1}} \in H\) so each element of H has an inverse element     R1

the four requirements for H to be a group are therefore satisfied     AG

[8 marks]

Prove that for all \(a \in G,{\text{ }}f({a^{ - 1}}) = {\left( {f(a)} \right)^{ - 1}}\).

Let \(\{ H,{\text{ }} \circ \} \) be the cyclic group of order seven, and let \(p\) be a generator.

Let \(x \in G\) such that \(f(x) = {p^{\text{2}}}\).

Find \(f({x^{ - 1}})\).

Given that \(f(x * y) = p\), find \(f(y)\).

\(f({e_G}) = {e_H} \Rightarrow f(a * {a^{ - 1}}) = {e_H}\)     M1

\(f\) is a homomorphism so \(f(a * {a^{ - 1}}) = f(a) \circ f({a^{ - 1}}) = {e_H}\)     M1A1

by definition \(f(a) \circ {\left( {f(a)} \right)^{ - 1}} = {e_H}\) so \(f({a^{ - 1}}) = {\left( {f(a)} \right)^{ - 1}}\) (by the left-cancellation law)     R1

[4 marks]

from (a) \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\)

hence \(f({x^{ - 1}}) = {({p^2})^{ - 1}} = {p^5}\)     M1A1

[2 marks]

\(f(x * y) = f(x) \circ f(y)\;\;\;\)(homomorphism)     (M1)

\({p^2} \circ f(y) = p\)     A1

\(f(y) = {p^5} \circ p\)     (M1)

\( = {p^6}\)     A1

[4 marks]

Total [10 marks]

Part (a) was well answered by those who understood what a homomorphism is. However many candidates simply did not have this knowledge and consequently could not get into the question.

Part (b) was well answered, even by those who could not do (a). However, there were many who having not understood what a homomorphism is, made no attempt on this easy question part. Understandably many lost a mark through not simplifying \({p^{ - 2}}\) to \({p^5}\).

Those who knew what a homomorphism is generally obtained good marks in part (c).

H and K are subgroups of a group G. By considering the four group axioms, prove that \(H \cap K\) is also a subgroup of G.

closure: let \(a,{\text{ }}b \in H \cap K\), so that \(a,{\text{ }}b \in H\) and \(a,{\text{ }}b \in K\)     M1

therefore \(ab \in H\) and \(ab \in K\) so that \(ab \in H \cap K\)     A1

associativity: this carries over from G     R1

identity: the identity \(e \in H\) and \(e \in K\)     M1

therefore \(e \in H \cap K\)     A1

inverse:

\(a \in H \cap K\) implies \(a \in H\) and \(a \in K\)     M1

it follows that \({a^{ - 1}} \in H\) and \({a^{ - 1}} \in K\)     A1

and therefore that \({a^{ - 1}} \in H \cap K\)     A1

the four group axioms are therefore satisfied     AG

[8 marks]

This question presented the most difficulty for students. Overall the candidates showed a lack of ability to present a formal proof. Some gained points for the proof of the identity element in the intersection and the statement that the associative property carries over from the group. However, the vast majority gained no points for the proof of closure or the inverse axioms.

Prove that set difference is not associative.

we are trying to prove \((A\backslash B)\backslash C \ne A\backslash (B\backslash C)\)     M1(A1)

\({\text{LHS}} = (A \cap B')\backslash C\)     (A1)

\( = (A \cap B') \cap C'\)     A1

\({\text{RHS}} = A\backslash (B \cap C')\)

\( = A \cap (B \cap C')'\)     (A1)

\( = A \cap (B' \cup C)\)     A1

as LHS does not contain any element of C and RHS does, \({\text{LHS}} \ne {\text{RHS}}\)     R1

hence set difference is not associative     AG

Note: Accept answers which use a proof containing a counter example.

Total [7 marks]

This question was found difficult by a large number of candidates, but a number of correct solutions were seen. A number of candidates who understood what was required failed to gain the final reasoning mark. Many candidates seemed to be ill-prepared to deal with this style of question.

Prove that \(f\) is an injection.

Prove that \(f\) is not a surjection.

METHOD 1

\(f(x) = f(y) \Rightarrow \frac{{4x + 1}}{{2x - 1}} = \frac{{4y + 1}}{{2y - 1}}\)     M1A1

for attempting to cross multiply and simplify     M1

\((4x + 1)(2y - 1) = (2x - 1)(4y + 1)\)

\( \Rightarrow 8xy + 2y - 4x - 1 = 8xy + 2x - 4y - 1 \Rightarrow 6y = 6x\)

\( \Rightarrow x = y\)     A1

hence an injection     AG

METHOD 2

\(f'(x) = \frac{{4(2x - 1) - 2(4x + 1)}}{{{{(2x - 1)}^2}}} = \frac{{ - 6}}{{{{(2x - 1)}^2}}}\)     M1A1

\( < 0\;\;\;{\text{(for all }}x \ne 0.5{\text{)}}\)     R1

therefore the function is decreasing on either side of the discontinuity

and \(f(x) < 2\) and \(x < 0.5\) for \(f(x) > 0.5\)     R1

hence an injection     AG

Note:     If a correct graph of the function is shown, and the candidate states this is decreasing in each part (or horizontal line test) and hence an injection, award M1A1R1.

[4 marks]

METHOD 1

attempt to solve \(y = \frac{{4x + 1}}{{2x - 1}}\)     M1

\(y(2x - 1) = 4x + 1 \Rightarrow 2xy - y = 4x + 1\)     A1

\(2xy - 4x = 1 + y \Rightarrow x = \frac{{1 + y}}{{2y - 4}}\)     A1

no value for \(y = 2\)     R1

hence not a surjection     AG

METHOD 2

consider \(y = 2\)     A1

attempt to solve \(2 = \frac{{4x + 1}}{{2x - 1}}\)     M1

\(4x - 2 = 4x + 1\)     A1

which has no solution     R1

hence not a surjection     AG

Note:     If a correct graph of the function is shown, and the candidate states that because there is a horizontal asymptote at \(y = 2\) then the function is not a surjection, award M1R1.

[4 marks]

Total [8 marks]

Most students indicated an understanding of the concepts of Injection and Surjection, but many did not give rigorous proofs. Even where graphs were used, it was very common for a sketch to be so imprecise with no asymptotes marked that it was difficult to award even partial credit. Some candidates mistakenly stated that the function was not surjective because 0.5 was not in the domain.

Most students indicated an understanding of the concepts of Injection and Surjection, but many did not give rigorous proofs. Even where graphs were used, it was very common for a sketch to be so imprecise with no asymptotes marked that it was difficult to award even partial credit. Some candidates mistakenly stated that the function was not surjective because 0.5 was not in the domain.

Show that \((G,{\text{ }} + )\) forms a group where \( + \) denotes addition on \(\mathbb{Q}\). Associativity may be assumed.

Assuming that \((H,{\text{ }} + )\) forms a group, show that it is a proper subgroup of \((G,{\text{ }} + )\).

The mapping \(\phi :G \to G\) is given by \(\phi (g) = g + g\), for \(g \in G\).

Prove that \(\phi \) is an isomorphism.

closure: \(\frac{{{n_1}}}{{{6^{{i_1}}}}} + \frac{{{n_2}}}{{{6^{{i_2}}}}} = \frac{{{6^{{i_2}}}{n_1} + {6^{{i_1}}}{n_2}}}{{{6^{{i_1} + {i_2}}}}} \in G\)     A1R1

Note:     Award A1 for RHS of equation. R1 is for the use of two different, but not necessarily most general elements, and the result \( \in G\) or equivalent.

identity: \(0\)     A1

inverse: \(\frac{{ - n}}{{{6^i}}}\)     A1

since associativity is given, \((G,{\text{ }} + )\) forms a group     R1AG

Note:     The R1 is for considering closure, the identity, inverses and associativity.

[5 marks]

it is required to show that \(H\) is a proper subset of \(G\)     (M1)

let \(\frac{n}{{{3^i}}} \in H\)     M1

then \(\frac{n}{{{3^i}}} = \frac{{{2^i}n}}{{{6^i}}} \in G\) hence \(H\) is a subgroup of \(G\)     A1

\(H \ne G\) since \(\frac{1}{6} \in G\) but \(\frac{1}{6} \notin H\)     A1

Note:     The final A1 is only dependent on the first M1.

hence, \(H\) is a proper subgroup of \(G\)     AG

[4 marks]

consider \(\phi ({g_1} + {g_2}) = ({g_1} + {g_2}) + ({g_1} + {g_2})\)     M1

\( = ({g_1} + {g_1}) + ({g_2} + {g_2}) = \phi ({g_1}) + \phi ({g_2})\)     A1

(hence \(\phi \) is a homomorphism)

injectivity: let \(\phi ({g_1}) = \phi ({g_2})\)     M1

working within \(\mathbb{Q}\) we have \(2{g_1} = 2{g_2} \Rightarrow {g_1} = {g_2}\)     A1

surjectivity: considering even and odd numerators     M1

\(\phi \left( {\frac{n}{{{6^i}}}} \right) = \frac{{2n}}{{{6^i}}}\) and \(\phi \left( {\frac{{3(2n + 1)}}{{{6^{i + 1}}}}} \right) = \frac{{2n + 1}}{{{6^i}}}\)     A1A1

hence \(\phi \) is an isomorphism     AG

[7 marks]

Total [16 marks]

This part was generally well done. Where marks were lost, it was usually because a candidate failed to choose two different elements in the proof of closure.

Only a few candidates realised that they did not have to prove that \(H\) is a group - that was stated in the question. Some candidates tried to invoke Lagrange's theorem, even though \(G\) is an infinite group.

Many candidates showed that the mapping is injective. Most attempts at proving surjectivity were unconvincing. Those candidates who attempted to establish the homomorphism property sometimes failed to use two different elements.

Consider the following functions

     \(f:\left] {1,{\text{ }} + \infty } \right[ \to {\mathbb{R}^ + }\) where \(f(x) = (x - 1)(x + 2)\)

     \(g:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) where \(g(x,{\text{ }}y) = \left( {\sin (x + y),{\text{ }}x + y} \right)\)

     \(h:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) where \(h(x,{\text{ }}y) = (x + 3y,{\text{ }}2x + y)\)

(a)     Show that \(f\) is bijective.

(b)     Determine, with reasons, whether

(i)     \(g\) is injective;

(ii)     \(g\) is surjective.

(c)     Find an expression for \({h^{ - 1}}(x,{\text{ }}y)\) and hence justify that \(h\) has an inverse function.

(a)     Method 1

sketch of the graph of \(f\)     (M1)

range of \(f = \) co-domain, therefore \(f\) is surjective     R1

graph of \(f\) passes the horizontal line test, therefore \(f\) is injective     R1

therefore \(f\) is bijective     AG

Note: Other explanations may be given (eg use of derivative or description of parabola).

Method 2

Injective: \(f(a) = f(b) \Rightarrow a = b\)     M1

\((a - 1)(a + 2) = (b - 1)(b + 2)\)

\({a^2} + a = {b^2} + b\)

solving for \(a\) by completing the square, or the quadratic formula,     A1

\(a = b\)

surjective: for all \(y \in {\mathbb{R}^ + }\) there exists \(x \in \left] {1,{\text{ }}\infty } \right[\) such that \(f(x) = y\)

solving \(y = {x^2} + x - 2\) for x, \(x = \frac{{\sqrt {4y + 9}  - 1}}{2}\). For all positive real \(y\), the minimum value for \(\sqrt {4y + 9} \) is \(3\). Hence, \(x \geqslant 1\)     R1

since \(f\) is both injective and surjective, \(f\) is bijective.     AG

Method 3

\(f\) is bijective if and only if \(f\) has an inverse     (M1)

solving \(y = {x^2} + x - 2\) for \(x,{\text{ }}x = \frac{{\sqrt {4y + 9}  - 1}}{2}\). For all positive real \(y\), the minimum value for \(\sqrt {4y + 9} \) is \(3\). Hence, \(x \geqslant 1\)     R1

\({f^{ - 1}}(x) = \frac{{\sqrt {4x + 9}  - 1}}{2}\)     R1

\(f\) has an inverse, hence \(f\) is bijective     AG

[3 marks]

(b)     (i)     attempt to find counterexample     (M1)

eg \(g(x,{\text{ }}y) = g(y,{\text{ }}x),{\text{ }}x \ne y\)     A1

g is not injective     R1

(ii)     \( - 1 \leqslant \sin (x + y) \leqslant 1\)     (M1)

range of \(g\) is \(\left[ { - 1,{\text{ 1}}} \right] \times \mathbb{R} \ne \mathbb{R} \times \mathbb{R}\)     A1

\(g\) is not surjective     R1

[6 marks]

(c)     let \(h(x,{\text{ }}y) = (u,{\text{ }}v)\)

then \(u = x + 3y\)

\(v = 2x + y\)     (M1)

solving simultaneous equations     (M1)

eg \(\left( \begin{array}{l}x\\y\end{array} \right) = {\left( {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right)^{ - 1}}\left( \begin{array}{l}u\\v\end{array} \right) \Rightarrow \left( \begin{array}{l}x\\y\end{array} \right) =  - \frac{1}{5}\left( {\begin{array}{*{20}{c}}1&{ - 3}\\{ - 2}&1\end{array}} \right)\left( \begin{array}{l}u\\v\end{array} \right)\)

\(x = \frac{{ - u + 3v}}{5}y = \frac{{2u - v}}{5}\)     A1

hence \({h^{ - 1}}(x,{\text{ }}y) = \left( {\frac{{ - x + 3y}}{5},{\text{ }}\frac{{2x - y}}{5}} \right)\)     A1

as this expression is defined for any values of \((x,{\text{ }}y) \in \mathbb{R} \times \mathbb{R}\)     R1

the inverse of \(h\) exists     AG

[5 marks] 

For part (a), given the command term ‘show that’ and the number of marks for this part, the best approach is a graphical one, i.e., an informal approach. Many candidates chose an algebraic approach and generally made correct statements for injective and surjective. However, they often did not follow through with the necessary algebraic manipulation to make a valid conclusion. In part (b), many candidates were not able to provide valid counter-examples. In part (c) It was obvious that quite a few candidates had not seen this type of function before. Those that were able to find the inverse generally did not justify their result, and hence could not earn the final R mark.

Let \(f:\mathbb{Z} \times \mathbb{R} \to \mathbb{R},{\text{ }}f(m,{\text{ }}x) = {( - 1)^m}x\). Determine whether f is

(i)     surjective;

(ii)     injective.

P is the set of all polynomials such that \(P = \left\{ {\sum\limits_{i = 0}^n {{a_i}{x^i}|n \in \mathbb{N}} } \right\}\).

Let \(g:P \to P,{\text{ }}g(p) = xp\). Determine whether g is

(i)     surjective;

(ii)     injective.

Let \(h:\mathbb{Z} \to {\mathbb{Z}^ + }\), \(h(x) = \left\{ {\begin{array}{*{20}{c}}
  {2x,}&{x > 0} \\
  {1 - 2x,}&{x \leqslant 0}
\end{array}} \right\}\). Determine whether h is

(i)     surjective;

(ii)     injective.

(i)     let \(x \in \mathbb{R}\)

for example, \(f(0,{\text{ }}x) = x\),     M1

hence f is surjective     A1

(ii)     for example, \(f(2,{\text{ }}3) = f(4,{\text{ }}3) = 3,{\text{ but }}(2,{\text{ }}3) \ne (4,{\text{ }}3)\)     M1

hence f is not injective     A1

[4 marks]

(i)     there is no element of P such that \(g(p) = 7\), for example     R1

hence g is not surjective     A1

(ii)     \(g(p) = g(q) \Rightarrow xp = xq \Rightarrow p = q\), hence g is injective     M1A1

[4 marks]

(i)     for \(x > 0,{\text{ }}h(x) = 2,{\text{ }}4,{\text{ }}6,{\text{ }}8 \ldots \)     A1

for \(x \leqslant 0,{\text{ }}h(x) = 1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \)     A1

therefore h is surjective     A1

(ii)     for \(h(x) = h(y)\), since an odd number cannot equal an even number, there are only two possibilities:     R1

\(x,{\text{ }}y > 0,{\text{ }}2x = 2y \Rightarrow x = y;\)     A1

\(x,{\text{ }}y \leqslant 0,{\text{ }}1 - 2x = 1 - 2y \Rightarrow x = y\)     A1

therefore h is injective     A1

Note: This can be demonstrated in a variety of ways.

[7 marks]

This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples. 

a) Some candidates failed to show convincingly that the function was surjective, and not injective.

This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples. 

b) Some candidates had trouble interpreting the notation used in the question, hence could not answer the question successfully.

This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples. 

c) Many candidates failed to appreciate that the function is discrete, and hence erroneously attempted to differentiate the function to show that it is monotonic increasing, hence injective. Others who provided a graph again showed a continuous rather than discrete function.

Prove that \(f\) is

(i)     not injective;

(ii)     not surjective.

The relation \(R\) is defined for \(a,{\text{ }}b \in \mathbb{R}\) so that \(aRb\) if and only if \(f(a) \times f(b) = 1\).

Show that \(R\) is an equivalence relation.

The relation \(R\) is defined for \(a,{\text{ }}b \in \mathbb{R}\) so that \(aRb\) if and only if \(f(a) \times f(b) = 1\).

State the equivalence classes of \(R\).

(i)     eg\(\;\;\;f(2) = f(3)\)     M1

hence \(f(a) = f(b) \Rightarrow a = b\)     R1

so not injective     AG

(ii)     eg\(\;\;\;\)Codomain is \(\mathbb{R}\) and range is \(\{  - 1,{\text{ }}1\} \)     M1

these not the same so not surjective     R1AG

Note:     if counter example is given it must be stated it is not in the range to obtain the R1. Eg \(f(x) = 2\) has no solution as \(f(x) \in \{  - 1,{\text{ }}1\} \forall x\).

[4 marks]

if \(a \ge 0\) then \(f(a) \times f(a) = 1 \times 1 = 1\)     A1

if \(a < 0\) then \(f(a) \times f(a) =  - 1 \times  - 1 = 1\)     A1

in either case \(aRa\) so \(R\) is reflexive     R1

\(aRb \Rightarrow f(a) \times f(b) = 1 \Rightarrow f(b) \times f(a) = 1 \Rightarrow bRa\)     A1

so \(R\) is symmetric     R1

if \(aRb\) then either \(a \ge 0\) and \(b \ge 0\) or \(a < 0\) and \(b < 0\)

if \(a \ge 0\) and \(b \ge 0\) and \(bRc\) then \(c \ge 0\) so \(f(a) \times f(c) = 1 \times 1 = 1\) and \(aRc\)     A1

if \(a < 0\) and \(b < 0\) and \(bRc\) then \(c < 0\) so \(f(a) \times f(c) =  - 1 \times  - 1 = 1\) and \(aRc\)     A1

in either case \(aRb\) and \(bRc \Rightarrow aRc\) so \(R\) is transitive     R1

Note:     Accept

\(f(a) \times f(b) \times f(b) \times f(c) = 1 \times 1 = 1 \Rightarrow f(a) \times 1 \times {\text{ }}f(c) = 1 \Rightarrow {\text{ }}f(a) \times f(c) = 1\)

Note:     for each property just award R1 if at least one of the A marks is awarded.

as \(R\) is reflexive, symmetric and transitive it is an equivalence relation     AG

[8 marks]

equivalence classes are \([0,{\text{ }}\infty [\) and \(] - \infty ,{\text{ }}0[\)     A1A1

Note:     Award A1A0 for both intervals open.

[2 marks]

Total [14 marks]

The function \(f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) is defined by \(f(x,{\text{ }}y) = \left( {x{y^2},\frac{x}{y}} \right)\).

Show that f is a bijection.

for f to be a bijection it must be both an injection and a surjection     R1

Note: Award this R1 for stating this anywhere.

injection:

let \(f(a{\text{, }}b) = f(c,{\text{ }}d)\) so that     (M1)

\(a{b^2} = c{d^2}\) and \(\frac{a}{b} = \frac{c}{d}\)     A1

dividing the equations,

\({b^3} = {d^3}\) so \(b = d\)     A1

substituting,

a = c     A1

it follows that f is an injection because \(f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)\)     R1

surjection:

let \(f(a{\text{, }}b) = (c,{\text{ }}d)\) where \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)     (M1)

then \(c = a{b^2}\) and \(d = \frac{a}{b}\)     A1

dividing,

\({b^3} = \frac{c}{d}\) so \(b = \sqrt[3]{{\frac{c}{d}}}\)     A1

substituting,

\(a = d \times \sqrt[3]{{\frac{c}{d}}}\)     A1

it follows that f is a surjection because

given \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) , there exists \((a,{\text{ }}b) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(f(a{\text{, }}b) = (c,{\text{ }}d)\)     R1

therefore f is a bijection     AG

[11 marks]

Candidates who knew that they were required to give a rigorous demonstration that f was injective and surjective were generally successful, although the formality that is needed in this style of demonstration was often lacking. Some candidates, however, tried unsuccessfully to give a verbal explanation or even a 2-D version of the horizontal line test. In 2-D, the only reliable method for showing that a function f is injective is to show that \(f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)\).

Let G be a finite cyclic group.

(a)     Prove that G is Abelian.

(b)     Given that a is a generator of G, show that \({a^{ - 1}}\) is also a generator.

(c)     Show that if the order of G is five, then all elements of G, apart from the identity, are generators of G.

(a)     let a be a generator and consider the (general) elements \(b = {a^m},{\text{ }}c = {a^n}\)     M1

then

\(bc = {a^m}{a^n}\)     A1

\( = {a^n}{a^m}\) (using associativity)     R1

\( = cb\)     A1

therefore G is Abelian     AG

[4 marks]

(b)     let G be of order p and let \(m \in \{ 1,.......,{\text{ }}p\} \), let a be a generator

consider \(a{a^{ - 1}} = e \Rightarrow {a^m}{({a^{ - 1}})^m} = e\)     M1R1

this shows that \({({a^{ - 1}})^m}\) is the inverse of \({a^m}\)     R1

as m increases from 1 to p, \({a^m}\) takes p different values and it generates G     R1

it follows from the uniqueness of the inverse that \({({a^{ - 1}})^m}\) takes p different values and is a generator     R1

[5 marks]

(c)     EITHER

by Lagrange, the order of any element divides the order of the group, i.e. 5     R1

the only numbers dividing 5 are 1 and 5     R1

the identity element is the only element of order 1     R1

all the other elements must be of order 5     R1

so they all generate G     AG

OR

let a be a generator.

successive powers of a and therefore the elements of G are

\(a,{\text{ }}{a^2},{\text{ }}{a^3},{\text{ }}{a^4}{\text{ and }}{a^5} = e\)     A1

successive powers of \({a^2}\) are \({a^2},{\text{ }}{a^4},{\text{ }}a,{\text{ }}{a^3},{\text{ }}{a^5} = e\)     A1

successive powers of \({a^3}\) are \({a^3},{\text{ }}a,{\text{ }}{a^4},{\text{ }}{a^2},{\text{ }}{a^5} = e\)     A1

successive powers of \({a^4}\) are \({a^4},{\text{ }}{a^3},{\text{ }}{a^2},{\text{ }}a,{\text{ }}{a^5} = e\)     A1

this shows that \({a^2},{\text{ }}{a^3},{\text{ }}{a^4}\)  are also generators in addition to a     AG

[4 marks]

Total [13 marks]

Solutions to (a) were often disappointing with some solutions even stating that a cyclic group is, by definition, commutative and therefore Abelian. Explanations in (b) were often poor and it was difficult in some cases to distinguish between correct and incorrect solutions. In (c), candidates who realised that Lagrange’s Theorem could be used were generally the most successful. Solutions again confirmed that, in general, candidates find theoretical questions on this topic difficult.

The function \(f:\mathbb{R} \to \mathbb{R}\) is defined by

\[f(x) = 2{{\text{e}}^x} - {{\text{e}}^{ - x}}.\]

(a)     Show that f is a bijection.

(b)     Find an expression for \({f^{ - 1}}(x)\).

(a)     EITHER

consider

\(f'(x) = 2{{\text{e}}^x} - {{\text{e}}^{ - x}} > 0\) for all x     M1A1

so f is an injection     A1

OR

let \(2{{\text{e}}^x} - {{\text{e}}^{ - x}} = 2{{\text{e}}^y} - {{\text{e}}^{ - y}}\)     M1

\(2({{\text{e}}^x} - {{\text{e}}^y}) + {{\text{e}}^{ - y}} - {{\text{e}}^{ - x}} = 0\)

\(2({{\text{e}}^x} - {{\text{e}}^y}) + {{\text{e}}^{ - (x + y)}}({{\text{e}}^x} - {{\text{e}}^y}) = 0\)

\(\left( {2 + {{\text{e}}^{ - (x + y)}}} \right)({{\text{e}}^x} - {{\text{e}}^y}) = 0\)

\({{\text{e}}^x} = {{\text{e}}^y}\)

\(x = y\)     A1

Note: Sufficient working must be shown to gain the above A1.

so f is an injection     A1

Note: Accept a graphical justification i.e. horizontal line test.

THEN

it is also a surjection (accept any justification including graphical)     R1

therefore it is a bijection     AG

[4 marks]

(b)     let \(y = 2{{\text{e}}^x} - {{\text{e}}^{ - x}}\)     M1

\(2{{\text{e}}^{2x}} - y{{\text{e}}^x} - 1 = 0\)     A1

\({{\text{e}}^x} = \frac{{y \pm \sqrt {{y^2} + 8} }}{4}\)     M1A1

since \({{\text{e}}^x}\) is never negative, we take the + sign     R1

\({f^{ - 1}}(x) = \ln \left( {\frac{{x + \sqrt {{x^2} + 8} }}{4}} \right)\)     A1

[6 marks]

Total [10 marks]

Solutions to (a) were often disappointing. Many candidates tried to use the result that, for an injection, \(f(a) = f(b) \Rightarrow a = b\) – although this is the definition, it is often much easier to proceed by showing that the derivative is everywhere positive or everywhere negative or even to use a horizontal line test. Although (b) is based on core material, solutions were often disappointing with some very poor use of algebra seen.

Determine the order of S₄.

Find the proper subgroup H of order 6 containing \({p_1}\), \({p_2}\) and their compositions. Express each element of H in cycle form.

Let \(f{\text{:}}\,{S_4} \to {S_4}\) be defined by \(f\left( p \right) = p \circ p\) for \(p \in {S_4}\).

Using \({p_1}\) and \({p_2}\), explain why \(f\) is not a homomorphism.

number of possible permutations is 4 × 3 × 2 × 1       (M1)

= 24(= 4!)      A1

[2 marks]

attempting to find one of \({p_1} \circ {p_1}\), \({p_1} \circ {p_2}\) or \({p_2} \circ {p_1}\)     M1

\({p_1} \circ {p_1} = \left( {132} \right)\) or equivalent (eg, \({p_1}^{ - 1} = \left( {132} \right)\))    A1

\( {p_1} \circ {p_2} = \left( {13} \right)\) or equivalent (eg, \({p_2} \circ {p_1} \circ {p_1} = \left( {13} \right)\))    A1

\({p_2} \circ {p_1} = \left( {23} \right)\) or equivalent (eg, \({p_1} \circ {p_1} \circ {p_2} = \left( {23} \right)\))    A1

Note: Award A1A0A0 for one correct permutation in any form; A1A1A0 for two correct permutations in any form.

\(e = \left( 1 \right)\), \({p_1} = \left( {123} \right)\) and \({p_2} = \left( {12} \right)\)     A1

Note: Condone omission of identity in cycle form as long as it is clear it is considered one of the elements of H.

[5 marks]

METHOD 1

if \(f\) is a homomorphism \(f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)\)

attempting to express one of \(f\left( {{p_1} \circ {p_2}} \right)\) or \(f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)\) in terms of \({p_1}\) and \({p_2}\)      M1

\(f\left( {{p_1} \circ {p_2}} \right) = {p_1} \circ {p_2} \circ {p_1} \circ {p_2}\)     A1

\(f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = {p_1} \circ {p_1} \circ {p_2} \circ {p_2}\)     A1

\( \Rightarrow {p_2} \circ {p_1} = {p_1} \circ {p_2}\)     A1

but \({p_1} \circ {p_2} \ne {p_2} \circ {p_1}\)     R1

so \(f\) is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if \({p_1}\) and \({p_2}\) are used; cycle form is not required.

METHOD 2

if \(f\) is a homomorphism \(f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)\)

attempting to find one of \(f\left( {{p_1} \circ {p_2}} \right)\) or \(f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)\)      M1

\(f\left( {{p_1} \circ {p_2}} \right) = e\)     A1

\(f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = \left( {132} \right)\)     (M1)A1

so \(f\left( {{p_1} \circ {p_2}} \right) \ne f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)\)     R1

so \(f\) is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if \({p_1}\) and \({p_2}\) are used; cycle form is not required.

[5 marks]

The relation aRb is defined on {1, 2, 3, 4, 5, 6, 7, 8, 9} if and only if ab is the square of a positive integer. 

(i)     Show that R is an equivalence relation. 

(ii)     Find the equivalence classes of R that contain more than one element.

Given the group \((G,{\text{ }} * )\), a subgroup \((H,{\text{ }} * )\) and \(a,{\text{ }}b \in G\), we define \(a \sim b\) if and only if \(a{b^{ - 1}} \in H\). Show that \( \sim \) is an equivalence relation.

(i)     \(aRa \Rightarrow a \cdot a = {a^2}\) so R is reflexive     A1

\(aRb = {m^2} \Rightarrow bRa\) so R is symmetric     A1

\(aRb = ab = {m^2}{\text{ and }}bRc = bc = {n^2}\)     M1A1

so \(a = \frac{{{m^2}}}{b}{\text{ and }}c = \frac{{{n^2}}}{b}\)

\(ac = \frac{{{m^2}{n^2}}}{{{b^2}}} = {\left( {\frac{{mn}}{b}} \right)^2},\)     A1

ac is an integer hence \({\left( {\frac{{mn}}{b}} \right)^2}\) is an integer     R1

so aRc, hence R is transitive     R1

R is therefore an equivalence relation     AG

(ii)     1R4 and 4R9 or 2R8     M1

so {1, 4, 9} is an equivalence class     A1

and {2, 8} is an equivalence class     A1

[10 marks]

\(a \sim a{\text{ since }}a{a^{ - 1}} = e \in H\), the identity must be in H since it is a subgroup.     M1

Hence reflexivity.     R1

\(a \sim b \Leftrightarrow a{b^{ - 1}} \in H\) but H is a subgroup so it must contain \({(a{b^{ - 1}})^{ - 1}} = b{a^{ - 1}}\)     M1R1

i.e. \(b{a^{ - 1}} \in H{\text{ so }} \sim \) is symmetric     A1

\(a \sim b{\text{ and }}b \sim c \Rightarrow a{b^{ - 1}} \in H{\text{ and }}b{c^{ - 1}} \in H\)     M1

But H is closed, so

\((a{b^{ - 1}})(b{c^{ - 1}}) \in H{\text{ or }}a({b^{ - 1}}b){c^{ - 1}} \in H\)     R1

\(a{c^{ - 1}} \in H \Rightarrow a \sim c\)     A1

Hence \( \sim \) is transitive and is thus an equivalence relation     R1AG

[9 marks]

Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.

Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.

Set \(S = \{ {x_0},{\text{ }}{x_1},{\text{ }}{x_2},{\text{ }}{x_3},{\text{ }}{x_4},{\text{ }}{x_5}\} \) and a binary operation \( \circ \) on S is defined as \({x_i} \circ {x_j} = {x_k}\), where \(i + j \equiv k(\bmod 6)\).

(a)     (i)     Construct the Cayley table for \(\{ S,{\text{ }} \circ \} \) and hence show that it is a group.

  (ii)     Show that \(\{ S,{\text{ }} \circ \} \) is cyclic.

(b)     Let \(\{ G,{\text{ }} * \} \) be an Abelian group of order 6. The element \(a \in {\text{G}}\) has order 2 and the element \(b \in {\text{G}}\) has order 3.

  (i)     Write down the six elements of \(\{ G,{\text{ }} * \} \).

  (ii)     Find the order of \({\text{a}} * b\) and hence show that \(\{ G,{\text{ }} * \} \) is isomorphic to \(\{ S,{\text{ }} \circ \} \).

(a)     (i)     Cayley table for \(\{ S,{\text{ }} \circ \} \)

\(\begin{array}{*{20}{c|cccccc}}
   \circ &{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\
\hline
  {{x_0}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\
  {{x_1}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}} \\
  {{x_2}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}} \\
  {{x_3}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}} \\
  {{x_4}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}} \\
  {{x_5}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}
\end{array}\)     A4

Note: Award A4 for no errors, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

S is closed under \( \circ \)     A1

\({x_0}\) is the identity     A1

\({x_0}\) and \({x_3}\) are self-inverses,     A1

\({x_2}\) and \({x_4}\) are mutual inverses and so are \({x_1}\) and \({x_5}\)     A1

modular addition is associative     A1

hence, \(\{ S,{\text{ }} \circ \} \) is a group     AG

(ii)     the order of \({x_1}\) (or \({x_5}\)) is 6, hence there exists a generator, and \(\{ S,{\text{ }} \circ \} \) is a cyclic group     A1R1

[11 marks]

(b)     (i)     e, a, b, ab     A1

and \({b^2},{\text{ }}a{b^2}\)     A1A1

Note: Accept \(ba\) and \({b^2}a\).

(ii)     \({(ab)^2} = {b^2}\)     M1A1

\({(ab)^3} = a\)     A1

\({(ab)^4} = b\)     A1

hence order is 6     A1

groups G and S have the same orders and both are cyclic     R1

hence isomorphic     AG

[9 marks]

Total [20 marks]

a) Most candidates had the correct Cayley table and were able to show successfully that the group axioms were satisfied. Some candidates, however, simply stated that an inverse exists for each element without stating the elements and their inverses. Most candidates were able to find a generator and hence show that the group is cyclic.

b) This part was answered less successfully by many candidates. Some failed to find all the elements. Some stated that the order of ab is 6 without showing any working.

(a)     Find the range of f .

(b)     Prove that f is an injection.

(c)     Taking the codomain of f to be equal to the range of f , find an expression for \({f^{ - 1}}(x)\) .

(a)     \(\left] { - 1,1} \right[\)     A1A1

Note: Award A1 for the values –1, 1 and A1 for the open interval.

[2 marks]

(b)     EITHER

Let \(\frac{{1 - {{\text{e}}^{ - x}}}}{{1 + {{\text{e}}^{ - x}}}} = \frac{{1 - {{\text{e}}^{ - y}}}}{{1 + {{\text{e}}^{ - y}}}}\)     M1

\(1 - {{\text{e}}^{ - x}} + {{\text{e}}^{ - y}} - {{\text{e}}^{ - (x + y)}} = 1 + {{\text{e}}^{ - x}} - {{\text{e}}^{ - y}} - {{\text{e}}^{ - (x + y)}}\)     A1

\({{\text{e}}^{ - x}} = {{\text{e}}^{ - y}}\)

\(x = y\)     A1

Therefore f is an injection     AG

OR

Consider

\(f'(x) = \frac{{{{\text{e}}^{ - x}}(1 + {{\text{e}}^{ - x}}) + {{\text{e}}^{ - x}}(1 - {{\text{e}}^{ - x}})}}{{{{(1 + {{\text{e}}^{ - x}})}^2}}}\)     M1

\( = \frac{{2{{\text{e}}^{ - x}}}}{{{{(1 + {{\text{e}}^{ - x}})}^2}}}\)     A1

\( > 0\) for all x.     A1

Therefore f is an injection.     AG

Note: Award M1A1A0 for a graphical solution.

[3 marks]

(c)     Let \(y = \frac{{1 - {{\text{e}}^{ - x}}}}{{1 + {{\text{e}}^{ - x}}}}\)     M1

\(y(1 + {{\text{e}}^{ - x}}) = 1 - {{\text{e}}^{ - x}}\)     A1

\({{\text{e}}^{ - x}}(1 + y) = 1 - y\)     A1

\({{\text{e}}^{ - x}} = \frac{{1 - y}}{{1 + y}}\)

\(x = \ln \left( {\frac{{1 + y}}{{1 - y}}} \right)\)     A1