A subgroup that contains all squares is normal Let $G$ is a finite group. $H\leq G$. $g^2 \in H$ for $\forall g (\in G)$ $\Rightarrow$ $H$ is normal subgroup of $G$
I tried to prove it by the different way using cosets. Let me show my proof.
Given the assumption I could get the cosets of $G$, $G = \{H, Hg\}$ (Here $g(\in G\setminus H)$ is the fixed element). Another $\forall x(\neq g)$ with $x \in G\setminus H$, I could also get $G=\{H,Hx\}$. Therefore My conclusion is $[G;H]=2$ since $Hg=Hx$ and $H$ is a subgroup of $G$, Finally $H$ is normal subgroup.
I can't find my errors in my proof. Is my proof right? If not please tell me which point I was wrong.
Best regards.

Maybe this is simpler - take $g \in G$. Then $g^2 = h$, for some $h \in H$. In words,
$g=hg^{-1}$. Multiply both sides by $g$ to get $g^2= ghg^{-1}$. But this is $h$ so it is in $H$.
