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Let $G$ be a group such that all of its subgroups are normal, prove that $[G,G]\subset Z_G$ where $[G,G]$ is the subgroup generated by the commutators $$[G,G]=\langle [g,h]: g,h \in G\rangle \;$$ where $[g,h]=ghg^{-1}h^{-1}$ and $Z_G$ is the center of $G$ $$Z_G=\{g \in G : (\forall x \in G)gx=xg\}.$$

I tried fixing $x\in G$ and tried proving that all commutators commutate with $x$. Doing this will show that the set of all commutators is a subset of the center, and thus the group generated by the commutators is a subgroup of the center.

Let us consider $\langle x\rangle$, if it is normal, then $yxy^{-1}=x^{\gamma(y)}$; then the we have that $$\gamma:y\mapsto\gamma(y)$$ is such that $$\gamma(y_1y_2)=\gamma(y_1)\gamma(y_2)$$

Then for any commutator $c=ghg^{-1}h^{-1}$ we have $$cxc^{-1}=ghg^{-1}h^{-1}xhgh^{-1}g^{-1}=x^{\gamma(ghg^{-1}h^{-1})}=x^{\gamma(g)\gamma(h)\gamma(g^{-1})\gamma(h^{-1})}=x^{\gamma(gg^{-1})\gamma(hh^{-1})}=x^{\gamma(e)\gamma(e)}=x$$$$cxc^{-1}=x \Longrightarrow cx=xc$$

Is it correct? Are there any more elegant methods? In my proof the fact that $\gamma:G\rightarrow Z/nZ$ is not an homomorphism is giving me some trouble, although everything seems to work.

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I improved some formatting, spelling and wording. For future posts, notice that $<$ and $\langle$ (and $>$ and $\rangle$) are different symbols. To obtain the latter, use the commands \langle, \rangle. Also, you may want to omit the definitions of most basic objects (the center, commutator subgroup), since they make your posts longer and more tedious to read. –  tomasz Nov 1 '12 at 11:40
    
Also, notice the difference between $\to$ and $\mapsto$ (\to and \mapsto): you write $f:{\bf R}\to {\bf R}$ but $f:x\mapsto x^2$. –  tomasz Nov 1 '12 at 11:43
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2 Answers

up vote 1 down vote accepted

Why do you think $\gamma$ isn't a homomorphism? It seems like a homomorphism into the multiplicative group $({\bf Z}/n{\bf Z})^*$ (or ${\bf Z}^*=\{-1,1\}$ if $x$ is of infinite order); I believe it is enough to show what you have stated: that $\gamma(yy')=\gamma(y)\gamma(y')$ which I presume you can do, then you can show that it is into the group by noticing that $\gamma(e)=\gamma(y)\gamma(y^{-1})=1$, which you have used anyway; you may want to provide a short argument as to why it is well defined.

Though you may want to show these things in more detail if it's a homework or something like that. Otherwise, it seems correct and actually quite elegant.

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How can I be sure that $\gamma$ doesn't output values outside of $(Z/nZ)^*$? –  Temitope.A Nov 1 '12 at 12:44
    
@Temitope.A: by exhibiting the fact that each $\gamma(y)$ is invertible, as I have written. –  tomasz Nov 1 '12 at 12:47
    
Oh yeah! Thanks for your help. –  Temitope.A Nov 1 '12 at 12:49
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Solution:

Easy to see that it's enough to prove for $G=<a,b,c>$ that $[a,b]\subset Z_G(c)$.

Let $ac\not= ca$, $c\notin <a,b>$, then denote maximal group $M: M\not= G, <a,b>\subset M$, so $|G/M|=p$, $p$ is prime, so $G'\subset M$, $c\notin M$, so $c\notin G'$, $aca^{-1}=c^l$, $l\not= 0$, if $p|l$, then $[a,c]\notin M$, but $[a,c]\in G'\subset M$, so $\exists k\in Z: p|kl-2$, so $[a^k,c]=c^{kl-1}=c^{kl-2}*c\notin M$, but $[a^k,c]\in G'\subset M$, and if $ac=ca$, then $[a,b]c=a^ic=ca^i=c[a,b]$, if $c\in <a,b>$, then $[a,b]\in Z_G(<a,b>)\subset Z_G(c)$. done

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