## Question

Determine, giving reasons, which of the following sets form groups under the operations given below. Where appropriate you may assume that multiplication is associative.

(a) \(\mathbb{Z}\) under subtraction.

(b) The set of complex numbers of modulus 1 under multiplication.

(c) The set {1, 2, 4, 6, 8} under multiplication modulo 10.

(d) The set of rational numbers of the form

\[\frac{{3m + 1}}{{3n + 1}},{\text{ where }}m,{\text{ }}n \in \mathbb{Z}\]

under multiplication.

**Answer/Explanation**

## Markscheme

(a) not a group *A1*

**EITHER**

subtraction is not associative on \(\mathbb{Z}\) (or give counter-example) *R1*

**OR**

there is a right-identity, 0, but it is not a left-identity *R1*

*[2 marks]*

* *

(b) the set forms a group *A1*

the closure is a consequence of the following relation (and the closure of \(\mathbb{C}\) itself):

\(\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|\) *R1*

the set contains the identity 1 *R1*

that inverses exist follows from the relation

\(\left| {{z^{ – 1}}} \right| = {\left| z \right|^{ – 1}}\)

for non-zero complex numbers *R1*

*[4 marks]*

* *

(c) not a group *A1*

for example, only the identity element 1 has an inverse *R1*

*[2 marks]*

* *

(d) the set forms a group *A1*

\(\frac{{2m + 1}}{{3n + 1}} \times \frac{{3s + 1}}{{3t + 1}} = \frac{{9ms + 3s + 3m + 1}}{{9nt + 3n + 3t + 1}} = \frac{{3(3ms + s + m) + 1}}{{3(3nt + n + t) + 1}}\) *M1R1*

shows closure

the identity 1 corresponds to *m* = *n* = 0 *R1*

an inverse corresponds to interchanging the parameters *m* and *n* *R1*

*[5 marks]*

*Total [13 marks]*

## Examiners report

There was a mixed response to this question. Some candidates were completely out of their depth. Stronger candidates provided satisfactory answers to parts (a) and (c). For the other parts there was a general lack of appreciation that, for example, closure and the existence of inverses, requires that products and inverses have to be shown to be members of the set.

## 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.

**Answer/Explanation**

## 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]*

## Examiners report

[N/A]

[N/A]