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Suppose $G$ is a group, $H\leq G$, and for all $g\in G$ we have $g^2\in H$. Is $H$ a normal subgroup of $G$?

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I saw a positive answer to this question, but it's argument was not correct. –  I'mtoo Sep 8 '13 at 5:36
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I changed the title to something more useful for searching. –  Mikko Korhonen Sep 10 '13 at 13:23

2 Answers 2

up vote 8 down vote accepted

Here is a different solution.

Let $N$ be the subgroup generated by the elements $g^2$, where $g \in G$. Then $N$ is a normal subgroup, and $G/N$ is abelian since $x^2 = 1$ for each $x \in G/N$. So if $g^2 \in H$ for all $g \in G$, it follows that $H$ contains $N$. Because $G/N$ is abelian, $H/N$ is a normal subgroup of $G/N$ and thus $H$ is a normal subgroup of $G$.

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+1 Nice approach, indeed. –  Babak S. Sep 10 '13 at 19:16
    
Now I am faithful that no part of mathematics (also Abstract algebra) is not vacuous which to be a sequence of just playing around! –  I'mtoo Sep 12 '13 at 4:36
    
@Mikko Korhonen : Your answer is perfect and has more Algebraic intuition behind itself. I changed "the Acceptance" of answers. Thank you Mikko. –  I'mtoo Sep 12 '13 at 4:49

Just think about this: $$g^{-2}h^{-1}(hg)^2\in H$$ wherein $g\in G$ is any elemnt and $h\in H$.

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Lovely argument! –  Vishal Sep 8 '13 at 5:45
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@I'mtoo: Just playing with elements regarding the given property $x^2\in H$. However, doing like these before, would help us to get the final point easily. That's it. –  Babak S. Sep 8 '13 at 6:09
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@SamiBenRomdhane: I changed my avatar to this one to make the OP accept my ideas. :D)) –  Babak S. Sep 8 '13 at 13:06
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Perhaps this would be more intuitive. Let $N$ be the subgroup generated by the elements $g^2$, where $g \in G$. Then $N$ is a normal subgroup, and $G/N$ is abelian since the identity $x^2 = 1$ holds in $G/N$. Now if $H$ contains every $g^2$, then $H$ contains $N$. Because $G/N$ is abelian, $H/N$ is normal in $G/N$ and thus $H$ is normal in $G$. –  Mikko Korhonen Sep 9 '13 at 8:41
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@I'mtoo: yes, this is a part of the correspondence theorem. It is true that $H/N$ is normal in $G/N$ if and only if $H$ is normal in $G$ –  Mikko Korhonen Sep 10 '13 at 13:15

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