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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.

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For $(d)$, do you mean the subgroup $\{0,4\}=H\leq\Bbb Z/8\Bbb Z$? If so, then it is not a counterexample since $1+1=2\notin H$. Or do you mean something else by your notation? –  Clayton Jan 10 '13 at 3:16
    
@ Clayton Thank you. The question is edited. –  aliakbar Jan 10 '13 at 3:20
    
what does $G'$ denote? –  rondo9 Jan 10 '13 at 3:29
    
The commutator subgroup of $G$ –  Clayton Jan 10 '13 at 3:31
    

3 Answers 3

up vote 8 down vote accepted

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?

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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 –  Babak S. 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. –  Babak S. Jan 10 '13 at 11:42
1  
@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

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.

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$(a)$ is correct. It suffices to prove that $[a,b] \in H$ for all $a,b \in G$. It is an easy job by some commutator relations.

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