Derived subgroup of a group whose all it's Sylow subgroups are cyclic is abelian. 
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.)

 A: 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. $\Box$ 
A: 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.
