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 $\displaystyle |G|<\infty$ and $\displaystyle H\leqslant G$ we have that $\displaystyle |H|<\infty$. Let $\displaystyle h_0\in H$ be fixed and define $\displaystyle \theta_{h_0}:H\to H$ by $\displaystyle h\overset{\theta_{h_0}}{\longmapsto}h_0h$. Clearly this is injective since $\displaystyle \theta_{h_0}(h)=\theta_{h_0}(h')\implies h_0h=h_0h'$ and since $\displaystyle G$ is a group we have cancellation and thus $\displaystyle h=h'$. But, every injection from a finite set to itself is a bijection. Thus, since $\displaystyle e\in H$ we have that $\displaystyle \theta_{h_0}(h)=h_0h=e$ for some $\displaystyle h\in H$. It follows that $\displaystyle h=h_0^{-1}$ and since $\displaystyle h_0$ was arbitrary the conclusion follows.