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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    $\begingroup$ I assume you mean $x^2\in H$ for all $x\in G$? $\endgroup$ – Tobias Kildetoft May 21 '13 at 17:54
  • $\begingroup$ 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? $\endgroup$ – Tobias Kildetoft May 21 '13 at 17:56
  • $\begingroup$ Which part did you not understand? $\endgroup$ – Tobias Kildetoft May 21 '13 at 17:59
  • $\begingroup$ $H$ did not become normal. But the subgroup $\left< x^2\mid x\in G\right>$ is normal and contained in $H$. $\endgroup$ – Tobias Kildetoft May 21 '13 at 18:05
  • $\begingroup$ This question has a counter part also, saying, then prove that G/H(the quotient group) is abelian. Can anyone prove this? $\endgroup$ – Arc Aug 13 '19 at 20:21

$H$ 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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  • $\begingroup$ This is very nice.+1 $\endgroup$ – DonAntonio May 21 '13 at 18:10
  • $\begingroup$ That's the elementary solution we were looking for :) Good job. I was halfway there when you completed it. $\endgroup$ – rschwieb May 21 '13 at 18:10
  • $\begingroup$ I though do not understand why the poster deleted his post the first time...perhaps he feared the quick-downvoters... $\endgroup$ – DonAntonio May 21 '13 at 18:13
  • $\begingroup$ @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). $\endgroup$ – Tobias Kildetoft May 21 '13 at 18:17
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    $\begingroup$ @Don: I see it now. Awkwardly presented, but a very nice approach. $\endgroup$ – Cameron Buie May 21 '13 at 18:32

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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  • $\begingroup$ I am not able to understand this hint. Can you please add more details. I mean what is the idea here. $\endgroup$ – StammeringMathematician Jun 21 '19 at 11:50
  • $\begingroup$ @StammeringMathematician What part you don't understand? Is it clear what $\;G^2\;$ is and why it is a normal subgroup of $\;G\;$ ? Is it clear that this normal subgroup of $\;G\;$ is in fact contained in $\;H\;$ ? Maybe the last point...? $\endgroup$ – DonAntonio Jun 21 '19 at 12:18
  • $\begingroup$ @DonAntonio , May i ask you something? I see now the derived set satisfies $G' \leq G^2$. Then what's next step?? How can i show $H$ is normal of $G$? $\endgroup$ – hew Sep 26 '19 at 5:21
  • $\begingroup$ Well, from what I wrote $\;G'\le H\;$, and any subgroup containing the derived subgroup is normal, so... $\endgroup$ – DonAntonio Oct 16 '19 at 13:37

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