Let $G$ be a group and $H$ a normal subgroup of $G$.
Prove: $x^2 \in H$ for every $x \in G$ iff every element of $G/H$ is its own inverse.
Here is my proof. I've only tried proving one way so far, please indicate if I'm on the right path.
If $x^2 \in H, \space \forall x \in G$, then $x^2 = h_1$ for some $h_1 \in H$.
So, $$x=h_1x^{-1}$$ $$x \in Hx^{-1}$$ $$Hx=Hx^{-1}$$
Therefore $G/H = \{Hx:x=x^{-1}\}$
Is this correct?
Also, since $G/H$ is the group $G$ with $H$ factored out, and since $x^2 \in H, \forall x \in G$, does this mean $G/H$ is the group that does not include the elements in $G$ whose square is in $H$? If this is so, I don't understand because if $y \in G/H$ then $y=y^{-1}$ and so $y^2=e$, but $e \in H$. So $G/H$ still has an element whose square is in $H$.
The first part of your proof looks correct to me. The cosets in $G/H$ are on the form $aH$, where $a$ is not the square of some element in $G$.
For the other direction, you want to consider that $xH = x^{-1}H$. This happens, by definition iff $(x^{-1})^{-1}x \in H$, so $x^2 \in H$. This is just the definition of two cosets being equal.
For your intuition about the factor group, if $yH \in G/H$, then $yH = y^{-1}H$, so $G/H$ contains only elements who are their own inverses. What you are doing when you take the factor group is like setting $x^2 = e$ for all $x \in G$, so you are 'forcing' every element to be its own inverse.