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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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    Also, notice the difference between $\to$ and $\mapsto$ (\to and \mapsto): you write $f:{\bf R}\to {\bf R}$ but $f:x\mapsto x^2$.2012-11-01

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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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    Oh yeah! Thanks for your help.2012-11-01
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Solution:

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

Let $ac\not= ca$, $c\notin $, then denote maximal group $M: M\not= G, \subset M$, 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 $, then $[a,b]\in Z_G()\subset Z_G(c)$. done