Subgroup of an abelian Group I think I have the proof correct, but my group theory is not that strong yet. If there is anything I am missing I would appreciate you pointing it out.
Let $G$ be an abelian group (s.t. $gh = hg$  $\forall g,h\in G$). Show that $H = \{g\in G:g^2=e_G\}$ is a subgroup of $G$ where $e_G$ is the identity element of $G$.
Proof: 
$1.$ Take $g \in H$. We know $g^2 = e_G$. So, we know $e_G^2 = e_G*e_G = e_G \in H$.
$2.$ Let $g,h \in H \Rightarrow g^2=e_G, h^2=e_G \Rightarrow (gh)^2=g^2h^2=e_G*e_G=e_G$
$\Rightarrow gh\in H$. So we have closure in H.
$3.$ Take $g \in H. g^2=e_G \Rightarrow g*g=e_G \Rightarrow g*(g*g^{-1})=e_G*g^{-1} \Rightarrow g=g^{-1} \Rightarrow g^{-1}\in H.$ So $H$ is a subgroup of $G.$ $\Box$

Looks very good, but it would be best to explictly establish, in proving closure, that $$(gh)^2 = \underbrace{(gh)(gh) = g^2h^2}_{\large G \;\text{is abelian}}$$ 
I'm not sure if that's what you meant but left that detail out of your proof(?), or if you erroneously made an immediate move by distributing the exponent: $(gh)^2 = g^2 h^2$, which is NOT true, in general.
