Show that $M=\{g \in G \mid g^m=e\}$ is a subgroup of $G$. Let $G$ be an abelian group and $m$ a positive integer. Show that $M=\{g \in G \mid g^m=e\}$ is a subgroup of $G$.
What I am thinking about is that our set is the set of all groups generators, since $m$ appears to be the order of $G$ which therefore must be a cyclic group. So, and since each group generated by a single element in a subgroup of $G$, the union of all of these must also be a subgroup.
But...I have not used commutativity yet...of course know that since $G$ is abelian, $g^m h^m = h^m g^m$ for all $h,g \in G$. 
How can I do this?
 A: 
Prove that $M=\{g \in G \mid g^m=e\}$ is a subgroup of $G$.

Proof: $e\in G$ and $e^m=e$ and so, by the definition of the set $M$, we get $e\in M$ which shows that $M\neq \emptyset$.
Let $x,y\in M$. Then $x,y\in G$ such that $x^m=e$ and $y^m=e$. Note that $y^{-1}\in G$ and so, $xy^{-1}\in G$. Moreover,
$$\overbrace{(xy^{-1})^m=x^m(y^{-1})^m}^{\text{since $G$ is abelian. }}=x^m(y^m)^{-1}=ee^{-1}=e.$$ 
This implies that $xy^{-1}\in M$. Using the Subgroup Criterion, the result follows.
A: Hint: The map $x \mapsto x^m$ is a homomorphism $G \to G$. Find its kernel.
A: You wrote "$m$ appears to be the order of $G$". That is wrong. The number $m$ is the order of one element of $g$. For example, if $G$ is the Klein group, the only non-cyclic abelian group of order $4$, then there are three elements of $G$ of order $2$, and $2$ is not the order of $G$. Likewise if $G$ is the cyclic group of order $4$, then $G$ has one element of order $2$, and again $2$ is not the order of $G$.
The essential point is not that $g^m h^m = h^m g^m,$ but rather that $(gh)^m = g^m h^m.$ For example:
$$
(gh)^6 = (gh)(gh)(gh)(gh)(gh)(gh) = (gggggg)(hhhhhh).
$$
If $g^6=e$ and $h^6=e$, then this proves $(gh)^6=e$, so $gh$ is a member of that same set, and so that set is closed under multiplication. You also need to show it's closed under inversion.
A: Let $h \in H$, then $h \in G$, so $H \subseteq G$. Then we just need to check that $H$ satisfies the group axioms. 
For $h_1$ and $h_2$ in $H$, and assuming that $m$ is fixed in your definition, we have
$$(h_1 h_2 )^m = h_1 h_2 \cdots h_1 h_2 \text{ ($m$ times)},$$
then you can use the fact $h_1$ and $h_2$ commute to get $h_1^m h_2^m = e$, hence $H$ is closed. Now check the other axioms.
