## Question

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 2*k*.

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

**Answer/Explanation**

## Markscheme

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 2*k*. *AG*

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

*[12 marks]*

## Examiners report

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

## Question

Let {*G* , \( * \)} be a finite group of order *n* and let *H* be a non-empty subset of *G* .

(a) Show that any element \(h \in H\) has order smaller than or equal to *n* .

(b) If *H* is closed under \( * \), show that {*H* , \( * \)} is a subgroup of {*G* , \( * \)}.

**Answer/Explanation**

## Markscheme

(a) if \(h \in H\) then \(h \in G\) *R1*

hence, (by Lagrange) the order of *h* exactly divides *n*

and so the order of *h* is smaller than or equal to *n* *R2*

*[3 marks]*

* *

(b) the associativity in *G* ensures associativity in *H* *R1*

(closure within *H* is given)

as *H* is non-empty there exists an \(h \in H\) , let the order of *h* be *m* then \({h^m} = e\) and as *H* is closed \(e \in H\) *R2*

it follows from the earlier result that \(h * {h^{m – 1}} = {h^{m – 1}} * h = e\) *R1*

thus, the inverse of *h* is \({h^{m – 1}}\) which \( \in H\) *R1*

the four axioms are satisfied showing that \(\{ H{\text{ , }} * \} \) is a subgroup *R1*

*[6 marks]*

*Total [9 marks]*

## Examiners report

Solutions to this question were extremely disappointing. This property of subgroups is mentioned specifically in the Guide and yet most candidates were unable to make much progress in (b) and even solutions to (a) were often unconvincing.

## Question

The binary operator multiplication modulo 14, denoted by \( * \), is defined on the set *S* = {2, 4, 6, 8, 10, 12}.

Copy and complete the following operation table.

(i) Show that {*S* , \( * \)} is a group.

(ii) Find the order of each element of {*S* , \( * \)}.

(iii) Hence show that {*S* , \( * \)} is cyclic and find all the generators.

The set *T* is defined by \(\{ x * x:x \in S\} \). Show that {*T* , \( * \)} is a subgroup of {*S* , \( * \)}.

**Answer/Explanation**

## Markscheme

*A4 *

**Note:** Award ** A4** for all correct,

**for one error,**

*A3***for two errors,**

*A2***for three errors and**

*A1***for four or more errors.**

*A0*

*[4 marks]*

(i) closure: there are no new elements in the table *A1*

identity: 8 is the identity element *A1*

inverse: every element has an inverse because there is an 8 in every row and column *A1*

associativity: (modulo) multiplication is associative *A1*

therefore {*S *, \( * \)} is a group *AG*

* *

(ii) the orders of the elements are as follows

*A4*

**Note:** Award ** A4** for all correct,

**for one error,**

*A3***for two errors,**

*A2***for three errors and**

*A1***for four or more errors.**

*A0*

(iii) **EITHER**

the group is cyclic because there are elements of order 6 *R1*

**OR**

the group is cyclic because there are generators *R1*

**THEN**

10 and 12 are the generators *A1A1** *

*[11 marks]*

looking at the Cayley table, we see that

*T* = {2, 4, 8} *A1*

this is a subgroup because it contains the identity element 8, no new elements are formed and 2 and 4 form an inverse pair *R2*

**Note:** Award ** R1** for any two conditions

*[3 marks]*

## Examiners report

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of *T*. This approach was invariably unsuccessful.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of *T*. This approach was invariably unsuccessful.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of *T*. This approach was invariably unsuccessful.

## Question

The group *G *has a unique element, *h *, of order 2.

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

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

**Answer/Explanation**

## Markscheme

(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]*

## Examiners report

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.

## Question

A group with the binary operation of multiplication modulo 15 is shown in the following Cayley table.

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.

**Answer/Explanation**

## Markscheme

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

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

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

**Note: **Award ** A3 **for 9 correct answers,

**for 6 or more, and**

*A2***for 3 or more.**

*A1***[3 marks]**

*A3*

**Note: **Award ** A3 **for 8 correct answers,

**for 6 or more, and**

*A2***for 4 or more.**

*A1***[3 marks]**

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

**Note: **Award ** A1 **for 1 correct answer and

**for all 3 (and no extras).**

*A2***[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]*

## Examiners report

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.

## Question

Consider the set \({S_3} = \{ {\text{ }}p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\} \) of permutations of the elements of the set \(\{ 1,{\text{ }}2,{\text{ }}3\} \), defined by

\(p = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&2&3 \end{array}} \right),{\text{ }}q = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&3&2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&2&1 \end{array}} \right),{\text{ }}s = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&1&3 \end{array}} \right),{\text{ }}t = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&1 \end{array}} \right),{\text{ }}u = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&1&2 \end{array}} \right).\)

Let \( \circ \) denote composition of permutations, so \(a \circ b\) means \(b\) followed by \(a\). You may assume that \(({S_3},{\text{ }} \circ )\) forms a group.

Complete the following Cayley table

**[5 marks]**

(i) State the inverse of each element.

(ii) Determine the order of each element.

Write down the subgroups containing

(i) \(r\),

(ii) \(u\).

**Answer/Explanation**

## Markscheme

**(M1)A4**

**Note:** Award ** M1** for use of Latin square property and/or attempted multiplication,

**for the first row or column,**

*A1***for the squares of \(q\), \(r\) and \(s\), then**

*A1***for all correct.**

*A2*(i) \({p^{ – 1}} = p,{\text{ }}{q^{ – 1}} = q,{\text{ }}{r^{ – 1}} = r,{\text{ }}{s^{ – 1}} = s\) *A1*

\({t^{ – 1}} = u,{\text{ }}{u^{ – 1}} = t\) *A1*

**Note:** Allow FT from part (a) unless the working becomes simpler.

(ii) using the table or direct multiplication *(M1)*

the orders of \(\{ p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\} \) are \(\{ 1,{\text{ }}2,{\text{ }}2,{\text{ }}2,{\text{ }}3,{\text{ }}3\} \) *A3*

**Note:** Award ** A1** for two, three or four correct,

**for five correct.**

*A2***[6 marks]**

(i) \(\{ p,{\text{ }}r\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)\) *A1*

(ii) \(\{ p,{\text{ }}u,{\text{ }}t\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)\) *A1*

**Note:** Award ** A0A1** if the identity has been omitted.

Award ** A0** in (i) or (ii) if an extra incorrect “subgroup” has been included.

**[2 marks]**

**Total [13 marks]**

## Examiners report

The majority of candidates were able to complete the Cayley table correctly. Unfortunately, many wasted time and space, laboriously working out the missing entries in the table – the identity is \(p\) and the elements \(q\), \(r\) and \(s\) are clearly of order two, so 14 entries can be filled in without any calculation. A few candidates thought \(t\) and \(u\) had order two.

Generally well done. A few candidates were unaware of the definition of the order of an element.

Often well done. A few candidates stated extra, and therefore incorrect subgroups.

## Question

The set of all permutations of the elements \(1,{\text{ }}2,{\text{ }} \ldots 10\) is denoted by \(H\) and the binary operation \( \circ \) represents the composition of permutations.

The permutation \(p = (1{\text{ }}2{\text{ }}3{\text{ }}4{\text{ }}5{\text{ }}6)(7{\text{ }}8{\text{ }}9{\text{ }}10)\) generates the subgroup \(\{ G,{\text{ }} \circ \} \) of the group \(\{ H,{\text{ }} \circ \} \).

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.

**Answer/Explanation**

## Markscheme

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,

**for cycles of 3.**

*A1***[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]**

## Examiners report

[N/A]

[N/A]

[N/A]

[N/A]

## Question

The binary operation \( * \) is defined on the set \(T = \{ 0,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6\} \) by \(a * b = (a + b – ab)(\bmod 7),{\text{ }}a,{\text{ }}b \in T\).

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 \(H\).

**Answer/Explanation**

## Markscheme

Cayley table is

*A4*

award ** A4 **for all 16 correct,

**for up to 2 errors,**

*A3***for up to 4 errors,**

*A2***for up to 6 errors**

*A1***[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]*

## Examiners report

[N/A]

[N/A]

[N/A]

## Question

The following Cayley table for the binary operation multiplication modulo 9, denoted by \( * \), is defined on the set \(S = \{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}5,{\text{ }}7,{\text{ }}8\} \).

Copy and complete the table.

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

Determine the orders of all the elements of \(\{ S,{\text{ }} * \} \).

(i) Find the two proper subgroups of \(\{ S,{\text{ }} * \} \).

(ii) Find the coset of each of these subgroups with respect to the element 5.

Solve the equation \(2 * x * 4 * x * 4 = 2\).

**Answer/Explanation**

## Markscheme

*A3*

**Note: **Award ** A3 **for correct table,

**for one or two errors,**

*A2***for three or four errors and**

*A1***otherwise.**

*A0**[3 marks]*

the table contains only elements of \(S\), showing closure *R1*

the identity is 1 *A1*

every element has an inverse since 1 appears in every row and column, or a complete list of elements and their correct inverses *A1*

multiplication of numbers is associative *A1*

the four axioms are satisfied therefore \(\{ S,{\text{ }} * \} \) is a group

the group is Abelian because the table is symmetric (about the leading diagonal) *A1*

*[5 marks]*

*A3*

**Note: **Award ** A3 **for all correct values,

**for 5 correct,**

*A2***for 4 correct and**

*A1***otherwise.**

*A0**[3 marks]*

(i) the subgroups are \(\{ 1,{\text{ }}8\} \); \(\{ 1,{\text{ }}4,{\text{ }}7\} \) *A1A1*

(ii) the cosets are \(\{ 4,{\text{ }}5\} \); \(\{ 2,{\text{ }}5,{\text{ }}8\} \) *A1A1*

*[4 marks]*

**METHOD 1**

use of algebraic manipulations *M1*

and at least one result from the table, used correctly *A1*

\(x = 2\) *A1*

\(x = 7\) *A1*

**METHOD 2**

testing at least one value in the equation *M1*

obtain \(x = 2\) *A1*

obtain \(x = 7\) *A1*

explicit rejection of all other values *A1*

*[4 marks]*

## Examiners report

The majority of candidates were able to complete the Cayley table correctly.

Generally well done. However, it is not good enough for a candidate to say something along the lines of ‘the operation is closed or that inverses exist by looking at the Cayley table’. A few candidates thought they only had to prove commutativity.

Often well done. A few candidates stated extra, and therefore incorrect subgroups.

[N/A]

The majority found only one solution, usually the obvious \(x = 2\), but sometimes only the less obvious \(x = 7\).

## Question

An Abelian group, \(\{ G,{\text{ }} * \} \), has 12 different elements which are of the form \({a^i} * {b^j}\) where \(i \in \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} \) and \(j \in \{ 1,{\text{ }}2,{\text{ }}3\} \). The elements \(a\) and \(b\) satisfy \({a^4} = e\) and \({b^3} = e\) where \(e\) is the identity.

Let \(\{ H,{\text{ }} * \} \) be the proper subgroup of \(\{ G,{\text{ }} * \} \) having the maximum possible order.

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 \(a\).

**Answer/Explanation**

## Markscheme

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.

**for 2 or 3 correct.**

*A1*(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]*

## Examiners report

[N/A]

[N/A]

## Question

The binary operation multiplication modulo 10, denoted by ×_{10}, is defined on the set *T* = {2 , 4 , 6 , 8} and represented in the following Cayley table.

Show that {*T*, ×_{10}} is a group. (You may assume associativity.)

By making reference to the Cayley table, explain why* T* is Abelian.

Find the order of each element of {*T*, ×_{10}}.

Hence show that {*T*, ×_{10}} is cyclic and write down all its generators.

The binary operation multiplication modulo 10, denoted by ×_{10} , is defined on the set *V* = {1, 3 ,5 ,7 ,9}.

Show that {*V*, ×_{10}} is not a group.

**Answer/Explanation**

## Markscheme

closure: there are no new elements in the table **A1**

identity: 6 is the identity element **A1**

inverse: every element has an inverse because there is a 6 in every row and column (2^{−1} = 8, 4^{−1} = 4, 6^{−1} = 6, 8^{−1} = 2) **A1**

we are given that (modulo) multiplication is associative **R1**

so {*T*, ×_{10}} is a group **AG**

**[4 marks]**

the Cayley table is symmetric (about the main diagonal) **R1**

so* T* is Abelian ** AG**

**[1 mark]**

considering powers of elements **(M1)**

**A2**

**Note:** Award * A2 *for all correct and

*for one error.*

**A1****[3 marks]**

**EITHER**

{*T*, ×_{10}} is cyclic because there is an element of order 4 **R1**

**Note:** Accept “there are elements of order 4”.

**OR**

{*T*, ×_{10}} is cyclic because there is generator **R1**

**Note:** Accept “because there are generators”.

**THEN**

2 and 8 are generators **A1A1**

**[3 marks]**

**EITHER**

considering singular elements **(M1)**

5 has no inverse (5 ×_{10} a = 1, a∈*V* has no solution) **R1**

**OR**

considering Cayley table for {*V*, ×_{10}}

**M1**

the Cayley table is not a Latin square (or equivalent) **R1**

**OR**

considering cancellation law

*eg*, 5 ×_{10}_{ }9 = 5 ×_{10} 1 = 5 **M1**

if {*V*, ×_{10}} is a group the cancellation law gives 9 = 1 **R1**

**OR**

considering order of subgroups

*eg*, {1, 9} is a subgroup **M1**

it is not possible to have a subgroup of order 2 for a group of order 5 (Lagrange’s theorem) **R1**

**THEN**

so {*V*, ×_{10}} is not a group ** AG**

**[2 marks]**

## Examiners report

[N/A]

[N/A]

[N/A]

[N/A]

[N/A]