Let $G'$ be the commutator group of the group $G$, and $G''$ the commutator group of $G'$. Show that if $G''$ is cyclic then $G''\subseteq Z(G')$, where $Z(G')$ is the centre of $G'$).

This can't be that difficult - it is half an exercise in Martin Isaac's Algebra (and most exercises have solutions that are only a few lines long). But I have wasted an inordinate amount of time failing to find the solution, so I must be missing something obvious.


Hint: $G/C_G(G$''$)$ can be homomorphically embedded in Aut$(G$''$)$, which is abelian.

  • $\begingroup$ I can see that Aut($G''$) is abelian, and that there is an obvious homomorphism from $N_G(G'')$ with kernel $C_G(G'')$ (namely $x\to x^g$). Is that what you are referring to? If so, I don't see how it helps. $\endgroup$ – almagest Jan 9 '18 at 19:32
  • $\begingroup$ But $N_G(G”)=G$, since $G” \unlhd G$. $\endgroup$ – Nicky Hekster Jan 9 '18 at 19:34
  • 1
    $\begingroup$ Excelent hint...@almagest, perhaps you should google "N/C theorem"...and think a little about the above comment by Nicky. Also, you surely know that $\;G/N\;$ abelian $\,\iff G'\subset N\;$ ...+1 $\endgroup$ – DonAntonio Jan 9 '18 at 19:37
  • $\begingroup$ Why is $G''$ normal? $\endgroup$ – almagest Jan 10 '18 at 8:59
  • $\begingroup$ All the members of the derived series of a group are normal, even characteristic subgroups. A proof of this basically boils down to a property of commutators: $[x,y]^g=[x^g,y^g]$. I think that by now you have all the ammunition to solve your problem, also the remarks of @DonAntonio are very useful. $\endgroup$ – Nicky Hekster Jan 10 '18 at 9:27

For any finite cyclic group $H$ with generator $t$, an automorphism of $H$ is completely determined by its effect on $t$ (which it must take to $t^k$ for some $k$ prime to the order of $t$). Hence Aut($H$) is abelian.

Next we show that if $H$ is any subgroup of $G$, then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of Aut($H$). Define $\theta:N_G(H)\to $Aut($H$) by $\theta(x)(h)=x^{-1}hx$. Evidently it has kernel $C_G(H)$, so the claim follows.

Applying these two observations to $H=G''$, where $G''$ is cyclic, we conclude that $N_G(G'')/C_G(G'')$ is isomorphic to a subgroup of an abelian group and hence abelian.

Finally, repeated application of $(xy)^g=x^gy^g$ shows that $[x,y]^g=[x^g,y^g]$, and repeated application of that shows that if $k\in G''$ and $g\in G$, then $k^g\in G''$ and so $N_G(G'')=G$.

Take any $x,y\in G$. Then $xyC_G(G'')=yxC_G(G'')$ and hence $x^{-1}y^{-1}xyC_G(G'')=C_G(G'')$, so $[x,y]\in C_G(G'')$ and hence $G'\subseteq C_G(G'')$. In other words, any element of $G'$ commutes with any element of $G''$ and so $G''\subseteq Z(G')$


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