On special normal subgroup of a group Let $G$ be a group and $H$ be a subgroup of $G$ such that for any $x\in G$ we have $x^2\in H$.  prove that  $H$ is normal in $G$.
I think this true, but can not prove it. for example this is true for $$D_8 = <a,b \space | a^4=b^2=1, \space ab = ba^{-1}>$$
 A: Let $x\in G$ and $h\in H$. We have that $xhx^{-1}=x^2(x^{-1}h)^2h^{-1}$ which belongs to $H$ since $H$ is a subgroup and the given hypothesis.
A: Zero has already posted a nice and elementary answer, but here is another way to see this, which also indicates how one might "guess" that it is true (without which one would probably not try enough to come up with the expression in his answer).
We will need a few basic results that make for good exercises themselves.


*

*If $S$ is a subset of $G$ which is closed under conjugation, then $\langle S\rangle$ (i.e. the subgroup generated by $S$) is normal in $G$.

*The set of squares in $G$ is closed under conjugation.

*If $S$ is the set of squares in $G$ and $K = \langle S\rangle$ then $G/K$ is abelian.

*Any subgroup containing $K$ is normal.

*If a subgroup contains all squares, then it contains $K$. In particular, $H$ contains $K$ and is thus normal.
A: Following @zero, apparently every element of $G/H$ has order at most $2$. This means $G' \subseteq H$ and $G/H \cong C_2 \times \cdots \times C_2$ (this could be an empty product, since $H$ could equal $G$).
