# Is a subgroup $H$ of a group $G$ normal if $g^2 \in H$ for all $g \in G$?

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. – RSh Sep 8 '13 at 5:36
I changed the title to something more useful for searching. – Mikko Korhonen Sep 10 '13 at 13:23

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! – RSh 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. – RSh 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 Gupta Sep 8 '13 at 5:45
@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
@SamiBenRomdhane: I changed my avatar to this one to make the OP accept my ideas. :D)) – Babak S. Sep 8 '13 at 13:06
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
@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