if $G$ is a finite group and every sylow subgroup of it is cyclic,prove that $G$ is super-solvable.

Question

if $G$ is a finite group and every sylow subgroup of it is cyclic,prove that $G$ is super-solvable.

we have that $G$ is solvable,I want to show that all factors of derived series are cyclic. but no achievement,so it will be great to help me with this,thanks.

Answer 1

Start with the smallest prime divisor $p_1$ of $|G|$. Let $P_1 \in Syl_{p_1}(G)$, $|P_1| = p_1^{e_1}$. We'll show that $G$ has a normal $p_1$-complement.

Now $N_G(P_1) / C_G(P_1) \cong K \leq Aut(P_1)$ by $N/C$-theorem. As $P_1$ is cyclic, $|Aut(P_1)| = \phi(p_1^{e_1}) = p_1^{e_1-1}(p_1-1)$, hence $|N_G(P_1) / C_G(P_1)|$ has no greater prime divisors than $p_1$. As $P_1$ is cyclic, $P_1 \leq C_G(P_1)$, hence $|N_G(P_1) / C_G(P_1)|$ is prime to $p_1$. As $p_1$ is the smallest prime divisor of $|G|$, we conclude that $N_G(P_1) = C_G(P_1)$. Thus by Burnside's transfer theorem, $G$ has a normal $p_1$-complement $N$, and $G/N \cong P_1$ is cyclic.

Now take the second smallest prime divisor $p_2$ of $|G|$, which is the smallest prime divisor of $|N|$ and continue inductively.