# If $G'/G''$ and $G''/G'''$ are cyclic, then $G''=1$

Here, $G'=[G,G]$, $G''=[G',G']$ and $G'''=[G'',G'']$. I had no idea where to start so I read the hint given by the book, which started with "You may assume $G'''=1$". I am uncomfortable with this assumption, since I have been unable to prove that $G''$ is abelian. I tried arguing that $G'$ consisted of elements of the form $x^k[y,k]$ where $x\in G''^c$ and $y,z \in G'$ but after computing the commutator of two such elements, I found no reason for which two such commutators should commute.
If I blindly assumed that was true, then $G''$ would be cyclic, but other than that I do not see how exactly it is necessarily the trivial group.
Why is $G''$ abelian? And why is it necessarily equal to $1$?

• Are there any other hypotheses on $G$? As it stands, this is false. Take $G$ to be non-Abelian simple, then $G'/G''$ etc. are all cyclic (of order $1$). – Lord Shark the Unknown May 25 '17 at 4:58
• @LordSharktheUnknown there are no other hypotheses. This comes from Dummit and Foote's Abstract Algebra, and the groups in question are in the derived or commutator series. – Guacho Perez May 25 '17 at 5:03
• The hypotheses do imply that $G'' = G'''$; that is $G''/G'''=1$. To prove that it would be sufficient to prove it under the assumption $G'''=1$. – Derek Holt May 25 '17 at 8:33

As pointed out by Lord Shark the Unknown, this is not true without any extra conditions on the group $G$. It is true when $G$ is solvable.
Now, put $X=G/G'''$. Then $X' \cong G'/G'''$ and $X'' \cong G''/G'''$. Moreover, $X'''=1$.
So, by the isomorphism theorems, $X'/X'' \cong G'/G''$ is cyclic and $X''$ is cyclic. Observe that $X/C_X(X'')$ embeds homomorphically into Aut$(X'')$ (consider the conjugation action of $X$ on $X''$ and do not forget that $X'' \unlhd X$), and this last automorphism group is abelian, since $X''$ is cyclic. Hence $X' \subseteq C_X(X'')$ and this implies that $X'' \subseteq Z(X')$. And of course $Z(X') \subseteq X'$. This yields $X'/Z(X') \cong (X'/X'') / (Z(X')/X'')$, being the quotient of a cyclic group, is cyclic and a well-known fact then says that $X'$ must be abelian! So, $X''=1$ and writing this back to $G$, this means $G''=G'''$.
If $G$ would be solvable, this would mean $G''=G'''=1$.