7
$\begingroup$

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 $1$?

$\endgroup$
  • $\begingroup$ 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 $1$). $\endgroup$ – Lord Shark the Unknown May 25 '17 at 4:58
  • $\begingroup$ @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. $\endgroup$ – Guacho Perez May 25 '17 at 5:03
  • $\begingroup$ 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 $G'''=1$. $\endgroup$ – Derek Holt May 25 '17 at 8:33
5
$\begingroup$

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 $G''=G'''=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.