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Letting $H$ be a subgroup of $G$ such that for every $x\in G$, $x^{2}\in H$. So, which of the following is true?

a. $H$ is a normal subgroup containing $G'$.

b. $H$ is a normal abelian subgroup.

c. $H=G$.

d. $H$ is a maximal subgroup.

c need not be true. For example, consider $H=2\Bbb Z$ and $G=\Bbb Z$. I want to know which of the a, b or d is true?

Thank you.

Andrew Uzzell
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Aliakbar
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2 Answers2

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Hints:

(1) In any group $\,G\,$ , the subgroup $\,G^n:=\langle\,x^n\;;\;x\in G\,\,,\,n\in\Bbb N\,\rangle\,$ is normal (in fact, it is a fully invariang subgroup)

(2) Any group $\,G\,$ for which $\,x^2=1\,\,\,\forall\,x\in G\,$ is abelian

(3) In your case, $\,H\triangleleft G\,$ and $\,G/H\,$ is abelian.

(4) $\,\forall\,N\triangleleft G\,$ in any group $\,G\,$ , $\,G/N\,$ is abelian iff $\,G'\leq N\,$

Can you take it from here?

DonAntonio
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  • Note that in $\mathbb Q_8$, the subgroup $H={1,-1}$ is as you defined above but it is not maximal at all. Nice suggestions +1 – Mikasa Jan 10 '13 at 10:08
  • @Babak I do not understand that how $H$ is normal – Aliakbar Jan 10 '13 at 11:40
  • @aliakbar: Google the Quaternion subgroups to see that $H$ is a char subgroup so is normal. – Mikasa Jan 10 '13 at 11:42
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    @aliakbar, look: $$\forall,x,g\in G;,;;g^{-1}x^ng=\left(g^{-1}xg\right)^n $$ and this shows that any generator of $,G^n,$ remains a generator of $,G^n,$ after being conjugated (an element of $,G,$ to the $,n-$th power) , so $,G^n,$ is normal as it is generated by a normal set. In the case of $,{-1,1}\leq Q_8,$ , you can try to prove the beautiful lemma: if a group has one unique element $,t,$ of order two, then $,{1,t}\leq Z(G),$ which, in particular, means that this subgroup's normal. – DonAntonio Jan 10 '13 at 12:56
  • @ DonAntonio I still do not understand why $H$ is a normal subgroup. I would be most grateful if you would explain – Aliakbar Jan 23 '13 at 15:19
  • $$\forall,g,x\in G,,,,g^{-1}x^ng=(g^{-1}xg)^n\Longrightarrow$$ every conjugate of a generator of $,G^n,$ is again a generator of $,G^n,$ and thus $,G^n\triangleleft G,$ . For $,n=2,$ we get that $,(xG)^2=x^2G=\overline 1\in G/G^2\Longrightarrow G/G^2,$ is abelian, and since $,G^2\leq H,$ then $,H/G^2\leq G/G^2\Longrightarrow H/G^2\triangleleft G/G^2\Longleftrightarrow H\triangleleft G,$ – DonAntonio Jan 23 '13 at 15:36
  • @DonAntonio Thank you very much – Aliakbar Jan 23 '13 at 16:31
  • @DonAntonio I have another question on http://math.stackexchange.com/questions/285076/which-of-the-following-statements-is-not-true Please help me. – Aliakbar Jan 23 '13 at 16:33
  • What does the notation with the triangle mean: $,H\triangleleft G,$? – MyUserIsThis Jun 22 '13 at 11:27
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I'm not entirely sure about $(a)$ and $(b)$, let me think about it for a while.

For $(d)$, $H$ need not be maximal, for we can consider a group of order $8$, where all elements have order $2$ except for the identity. Then $H=\{e,a\}$ is a subgroup, every element not in $H$ has order $2$, i.e., $g^2=e\in H$, but $J=\{e,a,b,ab\}$ is a subgroup which contains $H$, hence $H$ is not maximal.

Clayton
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