Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$ , $\forall x\in G$ . Prove that $H$ is a normal subgroup of $G$

Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.

I have tried to using the definition but failed. Can someone help me please.

Solution 1:

$H$ is a normal subgroup of $G$ $\iff\forall~h\in H ~\forall~ g\in G:g^{-1}hg \in H$

$g^{-1}hg=g^{-1}g^{-1}ghg=(g^{-1})^2h^{-1}hghg=(g^{-1})^2h^{-1}(hg)^2\in H(hg\in G \to (hg)^2\in H)$ then $$g^{-1}hg \in H$$

Solution 2:

As I began to correct my former post: Hints

$$\begin{align*}\bullet&\;\;\;G^2:=\langle x^2\;;\;x\in G\rangle\lhd G\\ \bullet&\;\;\;G^2\le H\\ \bullet&\;\;\;\text{The group}\;\;G/G^2\;\;\text{is abelian and thus}\;\;G'\le G^2\end{align*}$$