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**Defunkt** Part b doesn't mean that infinite groups don't have any subgroups -- it means that the condition, which you have proved sufficient for finite groups, does not work for infinite groups ie. if $\displaystyle (G, \cdot)$ is an infinite group and $\displaystyle H \subseteq G$ is closed under $\displaystyle \cdot$ then $\displaystyle (H, \cdot)$ is not neccesarily a subgroup of $\displaystyle G$.

Now, regarding part a:

Let $\displaystyle h \in H$. $\displaystyle H$ is closed under * and so $\displaystyle h^n \in H \ \forall n \in \mathbb{N}$. $\displaystyle H$ is finite , thus $\displaystyle o(h)$ is finite (this is also why this condition is not sufficient for infinite groups) and both conclusions easily follow (inverse and identity).