Let G be a finite group and let H be a non-empty subset of G that is closed under multiplication.Show that H is closed under inverse and hence prove that H is a subgroup of G.
Clearly since and we have that . Let be fixed and define by . Clearly this is injective since and since is a group we have cancellation and thus . But, every injection from a finite set to itself is a bijection. Thus, since we have that for some . It follows that and since was arbitrary the conclusion follows.