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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2I assume you mean $x^2\in H$ for all $x\in G$? – Tobias Kildetoft May 21 '13 at 17:54
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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
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Which part did you not understand? – Tobias Kildetoft May 21 '13 at 17:59
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$H$ 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
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This question has a counter part also, saying, then prove that G/H(the quotient group) is abelian. Can anyone prove this? – Arc Aug 13 '19 at 20:21
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@Arc Note every element of $G/H$ is its own inverse. That $G/H$ is abelian is a consequence of considering $(xy)^2=e$ in $G/H$. This is the same as $xyxy=e$ and so $xy=y^{-1}x^{-1}=yx$. – Jul 09 '24 at 03:17
3 Answers
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$H$ 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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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
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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
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@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
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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
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I already posted a new, hopefully corrected, one in the same spirit, @TobiasKildetoft . – DonAntonio May 21 '13 at 18:19
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@CameronBuie, the poster begins with $,g^{-1}hg,$ without assuming anything (well, only that $,g\in G;,;h\in H,$) , and he reaches the conclusion that this product is in $,H,$ – DonAntonio May 21 '13 at 18:20
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1@Don: I see it now. Awkwardly presented, but a very nice approach. – Cameron Buie May 21 '13 at 18:32
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As I began to correct my former post: Hints
- $G^2:=\langle x^2\;;\;x\in G\rangle\lhd G$
- $G^2\le H$
- The group $G/G^2$ is abelian and thus $G'\le G^2$
Shaun
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DonAntonio
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I am not able to understand this hint. Can you please add more details. I mean what is the idea here. – Shweta Aggrawal Jun 21 '19 at 11:50
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@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...? – DonAntonio Jun 21 '19 at 12:18
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@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 $G$? – hew Sep 26 '19 at 05:21
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Well, from what I wrote $;G'\le H;$, and any subgroup containing the derived subgroup is normal, so... – DonAntonio Oct 16 '19 at 13:37
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This is a proof supporting Don's claim that $G^2 \trianglelefteq G:$
Let $f:G \rightarrow G$ be an automorphism.
Note that $f(g^2)=f(g)^2$ for all $g \in G$.
Thus, $f(G^2) \subseteq G^2$.
This holds for every automorphism, hence for every inner automorphism in particular.
Thus, $G^2 \trianglelefteq G$.
Avi123
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Whoops! I considered $H$ to be the subgroup 'generated' by the squares(call it $G^2$. Well in that case, $G^2$ is contained in $H$ and the proof follows from @DonAntonio's hints. – Avi123 Jul 09 '24 at 08:35