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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 $G$?

Martin Sleziak
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RSh
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2 Answers2

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

Martin Sleziak
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Mikasa
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    Lovely argument! – Vishal Gupta Sep 08 '13 at 05:45
  • Thanks; So tricky! Is there any intuition behind this problem and your answer(I can't find any)? – RSh Sep 08 '13 at 06:04
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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. – Mikasa Sep 08 '13 at 06:09
  • @BabakS. : It seems to be a big terrible (at least for me!)! You say that Abstract Algebra somewhat has no intuition? – RSh Sep 08 '13 at 06:13
  • @I'mtoo: Nope! I didn't say that :-). The magic point which helps us here is to show $h^g\in H$. One can do it differently, but the condition given to us should be applied here, so I think how to manage elements in $G$ till that $h^g\in H$. This leads us to think about the forms (...)^2. – Mikasa Sep 08 '13 at 06:17
  • Thank you again for your patience about me @BabakS. I confess that Algebra (abstract kid!) is too complicated. The origination of my question really was about Algebraic-Thinking! – RSh Sep 08 '13 at 07:02
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    @SamiBenRomdhane: I changed my avatar to this one to make the OP accept my ideas. :D)) – Mikasa Sep 08 '13 at 13:06
  • Short and sweet! +1 – amWhy Sep 08 '13 at 14:14
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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 09 '13 at 08:41
  • @MikkoKorhonen: Very straightforward way dear friend. I think in this easy problem one should feel what is going between subgroup and the group. Yes, you are right, however the OP needs to know more about the structure of $G/N$ and its subgroup. Why don't u make it as an answer? +1 for it. (-: – Mikasa Sep 09 '13 at 08:46
  • @MikkoKorhonen: Does your last claim come from The Correspondence Theorem? – RSh Sep 10 '13 at 12:15
  • @BabakS. : A good mathematician is not one whom sees the problems easy! – RSh Sep 10 '13 at 12:17
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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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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 $G$.

Mikko Korhonen
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  • 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 04:36
  • @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 04:49