Show that a^2=1 is a subgroup of H Is my inverse right?
For the inverse, $\displaystyle (a^{-1})^2=1$. So, $\displaystyle a^{-1}a^{-1}=1$ and thus $\displaystyle a^{-1} \in H$.
Assume H is an abelian group. Let K be a subset of H such that $\displaystyle K=\{a \in H | a^2=1\}$. We show that K is a subgroup of H. It suffices to show that whenever x and y are in K, then $\displaystyle xy^{-1}$ is also in K (link).
Since H is an abelian group, we have $\displaystyle {(xy^{-1})}^2=xy^{-1}xy^{-1}=xxy^{-1}y^{-1}=1$. Thus K is a subgroup of H.
Anyhow I don't think K is necessarily a subgroup of H if H is a non-abelian group. Take an example of $\displaystyle S_3$.
Problem: Let $\displaystyle H$ be any group. Define $\displaystyle S=\{a\in H~|~a^2=1\}$. Prove that $\displaystyle S$ is a subgroup of $\displaystyle H$.
Proof (of inverse): Given $\displaystyle a^{-1}\in H$, we have to show that $\displaystyle a^{-1}\in S$ if $\displaystyle a\in S$.
$\displaystyle (a^{-1})^2=a^2(a^{-1})^2=1$ so $\displaystyle a^{-1}\in S$
$\displaystyle \square$