How to prove that if group $G$ is abelian, $H = \{x \in G : x = x^{-1}\}$ is a subgroup? $G$ is abelian, $H = \{x \in G : x = x^{-1}\}$ is a subgroup?
I know to prove that a subset $H$ of group $G$ be a subgroup, one needs to (i) prove $\forall x,y \in H:x \circ y \in H$ and (ii) $\forall x \in H:x^{-1} \in H.$

Straightforward insertion helps: 
(i) We always have $(xy)^{-1}=y^{-1}x^{-1}$.If we know $x=x^{-1}$ and $y=y^{-1}$, then this becomes $yx$. As $G$ is abelian, ideed $(xy)^{-1}=xy$, so $H$ is closed under multiplication.
(ii) If $x=x^{-1}$ then $(x^{-1})^{-1}=x^{-1}$, so also $x^{-1}\in H$.
(iii) Something is missing. There are subsets that are closed under multiplication and taking inverses, but are not subgroups!
