A group G in which every subgroup is normal includes [G,G] in the center 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.
 A: 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.
A: 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
