The set $S$ contains the eighth roots of unity given by $\left\{ {{\text{cis}}\left( {\frac{{n\pi }}{4}} \right),{\text{ }}n \in \mathbb{N},{\text{ }}0 \leqslant n \leqslant 7} \right\}$.

(i)     Show that $\{ S,{\text{ }} \times \}$ is a group where $\times$ denotes multiplication of complex numbers.

(ii)     Giving a reason, state whether or not $\{ S,{\text{ }} \times \}$ is cyclic.

(i)     closure: let ${a_1} = {\text{cis}}\left( {\frac{{{n_1}\pi }}{4}} \right)$ and ${a_2} = {\text{cis}}\left( {\frac{{{n_2}\pi }}{4}} \right) \in S$     M1

then ${a_1} \times {a_2} = {\text{cis}}\left( {\frac{{({n_1} + {n_2})\pi }}{4}} \right)$ (which $\in S$ because the addition is carried out modulo 8)     A1

identity: the identity is 1 (and corresponds to $n = 0$)     A1

inverse: the inverse of ${\text{cis}}\left( {\frac{{n\pi }}{4}} \right)$ is ${\text{cis}}\left( {\frac{{(8 - n)\pi }}{4}} \right) \in S$     A1

associatively: multiplication of complex numbers is associative     A1

the four group axioms are satisfied so $S$ is a group     AG

(ii)     $S$ is cyclic     A1

because ${\text{cis}}\left( {\frac{\pi }{4}} \right)$, for example, is a generator     R1

[7 marks]