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) $\forall x \in H:x^{-1} \in H.$
 A: Unless you know $H$ is a nonempty subset of $G$, we need also to show that the identity $e \in G$ is also in $H$:
(o) Clearly, the identity $e \in G$ is its own inverse, hence $e = e^{-1} \in H$.
$(ii)\;$ $\forall x \in H$, $x\in H \implies x = x^{-1}  \in H$. Hence we have closure under inverses.
$(i)$ $x\circ y \in H$?
$$\; x, y \in H, \implies x = x^{-1}, y = y^{-1}$$ $$ x \circ y = x^{-1}\circ y^{-1} = y^{-1}\circ x^{-1} \quad\quad\quad\quad\tag{G is abelian}$$
$$y^{-1}\circ x^{-1} = (x\circ y)^{-1} \implies x\circ y \in H$$
Hence we have closure under the group operation.
Therefore, $H \leq G$. That is, $H$ is a subgroup of $G$.

Added: Note that this problem is equivalent to the task of proving that if $G$ is an abelian group, and $H$ is a subset of $G$ such that $H = \{x \in G\mid x^2 = e\}$, then $H$ is a subgroup of $G$. Why? For $x \in G$ such that $x^2 = e$, note that $$x^2 = e \iff x^{-1}\circ x^2 = x^{-1} \circ e  \iff x^{-1}\circ (x\circ x) = x^{-1}$$ $$\iff (x^{-1}\circ x) \circ x = x^{-1} \iff e\circ x = x^{-1} \iff x = x^{-1}$$
A: Consider the map $\varphi:G\to G$ given by $\varphi(x)=x^2$. Since $G$ is abelian, then $\varphi$ is a homomorphism, and it's readily seen that $H$ is precisely the kernel of $\varphi$. Thus, $H$ is a subgroup of $G$.
A: 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!
A: You know what you need to do, so proceed and do it. For example, to show that if $x\in H$ then $y=x^{-1}\in H$, all you have to do is note that it is given that $x=x^{-1}$, and thus $y=x$, and thus clearly in $H$. 
Now, all that is left for you is to start by saying: suppose $x,y\in H$, then I need to show that $xy\in H$. 
Now write down what you know: $x^{-1}=?$ and $y^{-1}=?$. And write what you want to show: $(xy)^{-1}=?$. Fill in the question marks, and complete the proof. 
A: $x,y \in H \implies (xy)^{-1} = y^{-1} x^{-1} = yx = xy$
A: It is clear that $e\in H$ as $e=e^{-1}$. So $H\neq $ the empty set. 
Pick $x,y\in H$, then consider $xy$. Since $G$ is ablian and elements of $H$ are their own inverse we can see that $(xy)^{-1}=x^{-1}y^{-1}=xy$ so $xy\in H$. So $H$ is closed.
Then pick $x\in H$ since $x\in H$, $x=x^{-1}$ so $xx=x^2=e$ so inverses are in $H$
Therefore $H$ is nonempty, closed subset of $G$ with inverses, so $H$ is a subgroup of $G$.
