If $G$ is abelian, prove that the set of finite-order elements of $G$ is a subgroup of $G$ Let $G$  be abelian group. Prove that $ H = \{g\in G : |g| < ∞ \}$ is a subgroup of $G$. By |g| we denote the order of the element $g \in G$.
Proof: Denote by $e \in G$ the neutral element in $G$ (which trivially lies in $H$). Now let $a,b \in H $ and assume that $|a| = m$ and $|b| = n$. Then because $G$ is abelian we have 
$$(ab)^{mn} = a^{mn}b^{mn} = (a^m)^n(b^n)^m = e*e = e $$
So it follows that $ab \in H$.
Can someone please help me for proving $a^{-1} $ for any element $a \in H$, to conclude $H$ is a subgroup of $G$.
Thank you very much.
 A: Well, let $a \in H$. As in your question, $a$ has finite order, i.e. $\vert a\vert = m$ for some $m \in \mathbb N$. Then it holds that $a^{m-1}*a = a^m = e$ and thus $a^{-1} = a^{m-1}$. Since you have already proven closure of $H$ under multiplication, it (inductively) follows that $a^{m-1} \in H$ and hence $a^{-1} \in H$.
A: Hint The trick used in the statement of the question to prove closure under multiplication suggests a similar tack:

Suppose $a \in H$ has order $n$, and evaluate $(aa^{-1})^n$ in two ways.

A: First critique: Why do you say "suppose $e\in G$"? Isn't $e$ supposed to be the identity element of $G$?
Second critique: You haven't shown that $e\in H,$ yet (though this is simple), so how can you conclude that $ab\in H$?
Hint: One can prove that for any group $G$ and any $a\in G,$ we have $|a|=\left\lvert a^{-1}\right\rvert.$
A: Consider the cyclic subgroup generated by $g$, which has finite order $n$. By a simple application of Lagrange every element of that subgroup has order which divides $n$, hence finite order.
