Let {G , \( * \)} be a finite group of order n and let H be a non-empty subset of G .

(a)     Show that any element \(h \in H\) has order smaller than or equal to n .

(b)     If H is closed under \( * \), show that {H , \( * \)} is a subgroup of {G , \( * \)}.

(a)     if \(h \in H\) then \(h \in G\)     R1

hence, (by Lagrange) the order of h exactly divides n

and so the order of h is smaller than or equal to n     R2

[3 marks]

(b)     the associativity in G ensures associativity in H     R1

(closure within H is given)

as H is non-empty there exists an \(h \in H\) , let the order of h be m then \({h^m} = e\) and as H is closed \(e \in H\)     R2

it follows from the earlier result that \(h * {h^{m - 1}} = {h^{m - 1}} * h = e\)     R1

thus, the inverse of h is \({h^{m - 1}}\) which \( \in H\)     R1

the four axioms are satisfied showing that \(\{ H{\text{ , }} * \} \) is a subgroup     R1

[6 marks]

Total [9 marks]

Solutions to this question were extremely disappointing. This property of subgroups is mentioned specifically in the Guide and yet most candidates were unable to make much progress in (b) and even solutions to (a) were often unconvincing.