TITLE: Prove that $H = \{x \in G \mid  x=x^{-1}\}$ is a subgroup.
QUESTION [8 upvotes]: Let $G$ be an abelian group. Prove that $H = \{x \in G \mid x=x^{-1}\}$ is a subgroup.   ($x^{-1}$ is $x$'s inverse)
to check if the identity is an element of the subgroup: $ex=x$; $(e^{-1})ex=(e^{-1})x$; $ex=(e^{-1})x$; $e=e^{-1}$.
To check closure: this is where I'm not sure I know what I am doing. To check closure, I need to show that... $(ab)x=(ab)^{-1}(x^{-1})$? Let $a,b\in H$, $(ab)(ab)= e$ if $(ab)=(ab)^{-1}$. From that we can get $(aa)(bb)=e$ because we have commutativity and associativity from our group. We know $a=a^{-1}$, $b=b^{-1}$, so $(aa^{-1})(bb^{-1})=e$, $ee=e$; $e=e$. So... $(ab)=ab^{-1}$. If I plug that into $(ab)x=(ab)^{-1}(x^{-1})$, and say that $(ab)x=(ab)(x^{-1})$.  If I left multiply with a $(ab)^{-1}$, I get $x=x^{-1}$.  Which is what I started with so... I'm done?
To check inverses: the group includes inverses by definition of the group.
H is closed, includes the identity, and inverses, so it is a subgroup of $G$.

REPLY [3 votes]: If you know about homomorphisms, then $H=\ker \phi$, where $\phi(x)=x^2$, which is a homomorphism because $G$ is abelian. (Actually, iff $\phi$ is a homomorphism $G$ is abelian.)