Marks available | 17 |

Reference code | 09M.3srg.hl.TZ0.1 |

## Question

(a) Show that {1, −1, i, −i} forms a group of complex numbers *G* under multiplication.

(b) Consider \(S = \{ e,{\text{ }}a,{\text{ }}b,{\text{ }}a * b\} \) under an associative operation \( * \) where *e* is the identity element. If \(a * a = b * b = e\) and \(a * b = b * a\) , show that

(i) \(a * b * a = b\) ,

(ii) \(a * b * a * b = e\) .

(c) (i) Write down the Cayley table for \(H = \{ S{\text{ , }} * \} \).

(ii) Show that *H* is a group.

(iii) Show that *H* is an Abelian group.

(d) For the above groups, *G* and *H* , show that one is cyclic and write down why the other is not. Write down all the generators of the cyclic group.

(e) Give a reason why *G* and *H* are not isomorphic.

## Markscheme

(a)

see the Cayley table, (since there are no new elements) the set is closed *A1*

1 is the identity element *A1*

1 and –1 are self inverses and i and -i form an inverse pair, hence every element has an inverse *A1*

multiplication is associative *A1*

hence {1, –1, i, –i} form a group *G* under the operation of multiplication *AG*

*[4 marks]*

(b) (i) *aba* = *aab*

= *eb* *A1*

= *b* *AG*

* *

(ii) *abab* = *aabb*

= *ee* *A1*

= *e* *AG*

*[2 marks]*

(c) (i)

* A2*

**Note:** Award ** A1** for 1 or 2 errors,

**for more than 2.**

*A0*

(ii) see the Cayley table, (since there are no new elements) the set is closed *A1*

*H* has an identity element *e* *A1*

all elements are self inverses, hence every element has an inverse *A1*

the operation is associative as stated in the question

hence {*e *, *a *, *b *, *ab*} forms a group *G* under the operation \( * \) *AG*

* *

(iii) since there is symmetry across the leading diagonal of the group table, the group is Abelian *A1*

*[6 marks]*

(d) consider the element i from the group *G* *(M1)*

\({{\text{i}}^2} = – 1\)

\({{\text{i}}^3} = – {\text{i}}\)

\({{\text{i}}^4} = 1\)

thus i is a generator for *G* and hence *G* is a cyclic group *A1*

–i is the other generator for *G* *A1*

for the group *H* there is no generator as all the elements are self inverses *R1*

*[4 marks]*

(e) since one group is cyclic and the other group is not, they are not isomorphic *R1*

*[1 mark]*

*Total [17 marks]*

## Examiners report

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. A number of candidates did not understand the term “Abelian”. Many candidates understood the conditions for a group to be cyclic. Many candidates did not realise that the answer to part (e) was actually found in part (d), hence the reason for this part only being worth 1 mark. Overall, a number of fully correct solutions to this question were seen.

Marks available | 20 |

Reference code | 10N.3srg.hl.TZ0.4 |

## Question

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 \(\{ S,{\text{ }} \circ \} \).

## Markscheme

(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,

**for one error,**

*A3***for two errors,**

*A2***for three errors and**

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

*A0*

*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]*

## Examiners report

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.

Marks available | 2 |

Reference code | 12M.3srg.hl.TZ0.1 |

## Question

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 \((p \odot x) \odot x = x \odot p\)* *.

## Markscheme

closure, identity, inverse *A2*** **

**Note: **Award ** A1 **for two correct properties,

**otherwise.**

*A0*

*[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]*

## Examiners report

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.

Marks available | 15 |

Reference code | 12M.3srg.hl.TZ0.1 |

## Question

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 \((p \odot x) \odot x = x \odot p\)* *.

## Markscheme

closure, identity, inverse *A2*** **

**Note: **Award ** A1 **for two correct properties,

**otherwise.**

*A0*

*[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]*

## Examiners report

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.

Marks available | 7 |

Reference code | 15N.3srg.hl.TZ0.4 |

## 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 \(H\).

## 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

Marks available | 5 |

Reference code | 16M.3srg.hl.TZ0.1 |

## 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 \(2 * x * 4 * x * 4 = 2\).

## 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.

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

Marks available | 9 |

Reference code | 17M.3srg.hl.TZ0.4 |

## Question

The binary operation \( * \) is defined by

\(a * b = a + b – 3\) for \(a,{\text{ }}b \in \mathbb{Z}\).

The binary operation \( \circ \) is defined by

\(a \circ b = a + b + 3\) for \(a,{\text{ }}b \in \mathbb{Z}\).

Consider the group \(\{ \mathbb{Z},{\text{ }} \circ {\text{\} }}\) and the bijection \(f:\mathbb{Z} \to \mathbb{Z}\) given by \(f(a) = a – 6\).

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.

## Markscheme

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

**for an attempt to show that \(f(a \circ b) = f(a) * f(b)\).**

*R1A0A0**[3 marks]*

## Examiners report

Marks available | 4 |

Reference code | 18M.3srg.hl.TZ0.1 |

## 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.

## 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

Marks available | 1 |

Reference code | 18M.3srg.hl.TZ0.1 |

## 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.

## 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]**