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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.

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I assume you mean $x^2\in H$ for all $x\in G$? –  Tobias Kildetoft May 21 '13 at 17:54
    
Assuming that, hint: Note that $H$ contains the subgroup generated by all the elements of the form $x^2$, which is normal. What do you know about the quotient of $G$ with that normal subgroup? –  Tobias Kildetoft May 21 '13 at 17:56
    
Which part did you not understand? –  Tobias Kildetoft May 21 '13 at 17:59
    
$H$ did not become normal. But the subgroup $\left< x^2\mid x\in G\right>$ is normal and contained in $H$. –  Tobias Kildetoft May 21 '13 at 18:05

2 Answers 2

$H$ 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 $$g^{-1}hg \in H$$

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This is very nice.+1 –  DonAntonio May 21 '13 at 18:10
    
That's the elementary solution we were looking for :) Good job. I was halfway there when you completed it. –  rschwieb May 21 '13 at 18:10
    
I though do not understand why the poster deleted his post the first time...perhaps he feared the quick-downvoters... –  DonAntonio May 21 '13 at 18:13
    
@DonAntonio why did you delete your answer? The only part wrong was the index of the subgroups generated by squares (and it was essentially what I was hinting in my comments). –  Tobias Kildetoft May 21 '13 at 18:17
    
It looks like you're saying that $g^{-1}hg\in H$ implies $g^{-1}hg\in H$. Can you clarify what you're doing, here? –  Cameron Buie May 21 '13 at 18:17

As I began to correct my former post: Hints

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

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