Marks available | 3 |

Reference code | 11N.3srg.hl.TZ0.1 |

## Question

Consider the following Cayley table for the set *G* = {1, 3, 5, 7, 9, 11, 13, 15} under the operation \({ \times _{16}}\), where \({ \times _{16}}\) denotes multiplication modulo 16.

(i) Find the values of *a*, *b*, *c*, *d*, *e*, *f*, *g*, *h*, *i* and *j*.

(ii) Given that \({ \times _{16}}\) is associative, show that the set *G*, together with the operation \({ \times _{16}}\), forms a group.

The Cayley table for the set \(H = \{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{b_1},{\text{ }}{b_2},{\text{ }}{b_3},{\text{ }}{b_4}\} \) under the operation \( * \), is shown below.

(i) Given that \( * \) is associative, show that *H* together with the operation \( * \) forms a group.

(ii) Find two subgroups of order 4.

Show that \(\{ G,{\text{ }}{ \times _{16}}\} \) and \(\{ H,{\text{ }} * \} \) are not isomorphic.

Show that \(\{ H,{\text{ }} * \} \) is not cyclic.

## Markscheme

(i) \(a = 9,{\text{ }}b = 1,{\text{ }}c = 13,{\text{ }}d = 5,{\text{ }}e = 15,{\text{ }}f = 11,{\text{ }}g = 15,{\text{ }}h = 1,{\text{ }}i = 15,{\text{ }}j = 15\) *A3*

**Note:** Award ** A2** for one or two errors,

** A1** for three or four errors,

** A0** for five or more errors.

(ii) since the Cayley table only contains elements of the set *G*, then it is closed *A1*

there is an identity element which is 1 *A1*

{3, 11} and {5, 13} are inverse pairs and all other elements are self inverse *A1*

hence every element has an inverse *R1*

**Note:** Award ** A0R0** if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group *AG*

*[7 marks]*

(i) since the Cayley table only contains elements of the set *H*, then it is closed *A1*

there is an identity element which is *e* *A1*

\(\{ {a_1},{\text{ }}{a_3}\} \) form an inverse pair and all other elements are self inverse *A1*

hence every element has an inverse *R1*

**Note:** Award ** A0R0** if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group *AG*

(ii) any 2 of \(\{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_1},{\text{ }}{b_2}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_3},{\text{ }}{b_4}\} \) *A2A2** *

*[8 marks]*

the groups are not isomorphic because \(\{ H,{\text{ }} * \} \) has one inverse pair whereas \(\{ G,{\text{ }}{ \times _{16}}\} \) has two inverse pairs *A2*

**Note:** Accept any other valid reason:

*e.g.* the fact that \(\{ G,{\text{ }}{ \times _{16}}\} \) is commutative and \(\{ H,{\text{ }} * \} \) is not.

* *

*[2 marks]*

**EITHER**

a group is not cyclic if it has no generators *R1*

for the group to have a generator there must be an element in the group of order eight *A1*

* *

since there is no element of order eight in the group, it is not cyclic *A1*

**OR**

a group is not cyclic if it has no generators *R1*

only possibilities are \({a_1}\), \({a_3}\) since all other elements are self inverse *A1*

this is not possible since it is not possible to generate any of the “*b*” elements from the “*a*” elements – the elements \({a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{a_4}\) form a closed set *A1*

*[3 marks]*

## Examiners report

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Marks available | 4 |

Reference code | 13M.3srg.hl.TZ0.2 |

## Question

Consider the set *S* = {1, 3, 5, 7, 9, 11, 13} under the binary operation multiplication modulo 14 denoted by \({ \times _{14}}\).

Copy and complete the following Cayley table for this binary operation.

Give one reason why \(\{ S,{\text{ }}{ \times _{14}}\} \) is not a group.

Show that a new set *G* can be formed by removing one of the elements of *S* such that \(\{ G,{\text{ }}{ \times _{14}}\} \) is a group.

Determine the order of each element of \(\{ G,{\text{ }}{ \times _{14}}\} \).

Find the proper subgroups of \(\{ G,{\text{ }}{ \times _{14}}\} \).

## Markscheme

*A4*

**Note:** Award ** A3** for one error,

**for two errors,**

*A2***for three errors,**

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

*A0**[4 marks]*

any valid reason, for example *R1*

not a Latin square

7 has no inverse

*[1 mark]*

delete 7 (so that *G* = {1, 3, 5, 9, 11, 13}) *A1*

closure – evident from the table *A1*

associative because multiplication is associative *A1*

the identity is 1 *A1*

13 is self-inverse, 3 and 5 form an inverse

pair and 9 and 11 form an inverse pair *A1*

the four conditions are satisfied so that \(\{ G,{\text{ }}{ \times _{14}}\} \) is a group *AG*

*[5 marks]*

*A4*

**Note:** Award ** A3** for one error,

**for two errors,**

*A2***for three errors,**

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

*A0*

*[4 marks]*

{1}

{1, 13}\(\,\,\,\,\,\){1, 9, 11} *A1A1*

*[2 marks]*

## Examiners report

There were no problems with parts (a), (b) and (d).

There were no problems with parts (a), (b) and (d).

There were no problems with parts (a), (b) and (d) but in part (c) candidates often failed to state that the set was associative under the operation because multiplication is associative. Likewise they often failed to list the inverses of each element simply stating that the identity was present in each row and column of the Cayley table.

The majority of candidates did not answer part (d) correctly and often simply listed all subsets of order 2 and 3 as subgroups.

Marks available | 9 |

Reference code | 13N.3srg.hl.TZ0.2 |

## Question

Let \(G\) be a group of order 12 with identity element *e*.

Let \(a \in G\) such that \({a^6} \ne e\) and \({a^4} \ne e\).

(i) Prove that \(G\) is cyclic and state two of its generators.

(ii) Let \(H\) be the subgroup generated by \({a^4}\). Construct a Cayley table for \(H\).

State, with a reason, whether or not it is necessary that a group is cyclic given that all its proper subgroups are cyclic.

## Markscheme

(i) the order of \(a\) is a divisor of the order of \(G\) *(M1)*

since the order of \(G\) is 12, the order of \(a\)* *must be 1, 2, 3, 4, 6 or 12 *A1*

the order cannot be 1, 2, 3 or 6, since \({a^6} \ne e\) *R1*

the order cannot be 4, since \({a^4} \ne e\) *R1*

so the order of \(a\)* *must be 12

therefore, \(a\)* *is a generator of \(G\), which must therefore be cyclic *R1*

another generator is *eg* \({a^{ – 1}},{\text{ }}{a^5},{\text{ }} \ldots \) *A1*

*[6 marks]*

* *

(ii) \(H = \{ e,{\text{ }}{a^4},{\text{ }}{a^8}\} \) *(A1)*

*M1A1*

* *

*[3 marks]*

no *A1*

*eg *the group of symmetries of a triangle \({S_3}\)* *is not cyclic but all its (proper) subgroups are cyclic

*eg *the Klein four-group is not cyclic but all its (proper) subgroups are cyclic *R1*

*[2 marks]*

## Examiners report

In part (a), many candidates could not provide a logical sequence of steps to show that \(G\) is cyclic. In particular, although they correctly quoted Lagrange’s theorem, they did not always consider all the orders of a, i.e., all the factors of 12, omitting in particular 1 as a factor. Some candidates did not state the second generator, in particular \({a^{ – 1}}\). Very few candidates were successful in finding the required subgroup, although they were obviously familiar with setting up a Cayley table.

Marks available | 2 |

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

## 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.

## 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

Marks available | 3 |

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 | 8 |

Reference code | 16N.3srg.hl.TZ0.3 |

## 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 \(a\).

## 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

Marks available | 7 |

Reference code | 16N.3srg.hl.TZ0.3 |

## 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 \(a\).

## 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

Marks available | 3 |

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 | 3 |

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 | 2 |

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 | 10 |

Reference code | SPNone.3srg.hl.TZ0.3 |

## Question

The group \(\{ G,{\text{ }}{ \times _7}\} \) is defined on the set {1, 2, 3, 4, 5, 6} where \({ \times _7}\) denotes multiplication modulo 7.

(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.

## Markscheme

(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]*