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.

(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]

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.