Let $ G $ be a finite group such that all its Sylow subgroups are cyclic. Prove that $ G'$ is abelian. (Here $ G' $ denotes the derived subgroup.)


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    $\begingroup$ This:groupprops.subwiki.org/wiki/… shows that $G$ has a cyclic subgroup $N$ with cyclic quotient (it's metacyclic). Since the derived subgroup must evidently be contained in $N$, it actually follows that $G'$ is cyclic. $\endgroup$ – Alex Youcis Sep 18 '13 at 6:55
  • $\begingroup$ How advanced results are you allowed to use for this? $\endgroup$ – Tobias Kildetoft Sep 18 '13 at 6:57
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    $\begingroup$ @TobiasKildetoft: Besides to your question,I think the title should have been more descriptive. $\endgroup$ – mrs Sep 18 '13 at 7:06

First let's prove the solvability of a group all of whose Sylow subgroups are cyclic - we need this to get an induction argument started.

Let $p$ be the smallest prime dividing the order of $G$ and let $P$ be a Sylow $p$-subgroup.. The choice of $p$ implies that the condition of Burnside's theorem is satisfied, that is $P\subset Z(N_G(P))$ (this is equivalent to $N_G(P)=C_G(P)$, and this follows from the fact that $N_G(P)/C_G(P)$ injects homomorphically in $\operatorname{Aut}(P)$, which is an abelian group of order $\varphi(|P|)$ and the fact that $P \subset C_G(P)$. This last inclusion follows from $P$ being abelian and hence $|N_G(P)/C_G(P)$| is not divisible by $p$.) So $G$ has a normal $p$-complement, say $N$. In other words, $G=NP$ and $N \cap P=1$.

The solvability follows easily now by induction on $|G|$, since $N$ is a proper normal subgroup of $G$ and inherits from $G$ all of its Sylow subgroups being cyclic. It follows by induction that $N$ is solvable. But $G/N \cong P$ is a $p$-group whence solvable, so after all, $G$ itself is solvable.

Now we are almost done, but need a small general observation.

Lemma Let $G$ be a group with both $G'/G''$ and $G''$ cyclic. Then $G'$ must be cyclic.

Proof. (sketch) $G/C_G(G'')$ injects homomorphically in $\operatorname{Aut}(G'')$. This last group is abelian and it follows that $G' \subset C_G(G'')$, thus $G'' \subset Z(G') \subset G'$. Since $G'/G''$ is cyclic, this implies $G'/Z(G')$ is cyclic, and it is well-known that this means $G'$ is abelian. But then $G''=1$, so $G'$ must in fact be cyclic. $\Box$

We are now ready to finish the proof. Since $G$ is solvable, $G'$ must be a proper subgroup and by induction on $|G|$ it follows that $G''$ is abelian. Obviously $G'/G''$ is also an abelian group. But both groups have cyclic Sylow subgroups (inherited from $G$ itself) and hence the condition of the previous lemma is satisfied. This concludes the final argument. $\Box$

  • $\begingroup$ Good luck with your exam! And if you consider my answer as THE answer, please tick it as such. $\endgroup$ – Nicky Hekster Sep 20 '13 at 5:48

This is a non-trivial theorem and even something stronger is true.

Theorem. Suppose that all Sylow subgroups of the finite group $G$ are cyclic. Then both $G'$ and $G/G'$ are cyclic and $\gcd(|G'|,|G/G'|)=1$.

The proof depends heavily on the famous normal $p$-complement theorem of Burnside who showed that if a Sylow $p$-subgroup of a group $G$ is in the center of its normalizer then $G$ has a normal $p$-complement. This implies that if $p$ is the smallest prime dividing the order of a group $G$ and the Sylow $p$-subgroup is cyclic, then $G$ has a normal $p$-complement. It is exactly this result that can be used to prove that a group with only cyclic Sylow subgroups must be solvable. That is a first step.

Another result one needs for the "$\gcd$"-part of the above theorem is the following.

Lemma Let $P$ be a cyclic Sylow $p$-subgroup of $G$. Then $p$ divides at most one of the numbers $|G'|$ and $|G/G'|$.

Details of the proofs can be found for example in Chapter 5 (Transfer) of Marty Isaacs' book Finite Group Theory.

  • $\begingroup$ Thank you very much, please explain more, I quickly need it. $\endgroup$ – mehranian Sep 19 '13 at 11:51
  • $\begingroup$ @mehranian what do you need it for? $\endgroup$ – Tobias Kildetoft Sep 19 '13 at 11:59
  • $\begingroup$ I have an exam for PHD and I guess this question will be given $\endgroup$ – mehranian Sep 19 '13 at 12:51
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    $\begingroup$ @mehranian In that case you really ought to be able to do this yourself given the current amount of information given. $\endgroup$ – Tobias Kildetoft Sep 19 '13 at 18:09
  • $\begingroup$ OK, I will give another answer with the proofs and I will leave out the gcd-part. $\endgroup$ – Nicky Hekster Sep 19 '13 at 20:51

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