Let $P$ be a Sylow $p$-subgroup of $G$ and suppose $x\in P$ has order $p$ and $C_G(x)$ has cyclic Sylow $p$-subgroups. Prove $P$ is cyclic.

Question

I can't seem to make any progress in the following question.

Let $P$ be a Sylow $p$-subgroup of a group $G$, and suppose that $x\in P$ has order $p$ and $C_G(x)$ has cyclic Sylow $p$-subgroups. Prove that $P$ is cyclic.

I was given the following hint: ($Z(P)\subseteq C_G(x)$, and that $P$ is cyclic if and only if $x\in Z(P)$)

My current thought process:

I suppose that I am suppose to show that $x\in Z(P)$. But I dont how to even begin doing this. Perhaps $Z(P)=C_G(x)$? Also, I can't seem to see why $x\in Z(P)$ implies that $P$ is cyclic.

Any help provided will be appreciated.

Answer 1

If $x \in Z(P)$ then $x$ commutes with everything in $P$. Hence $P \subset C_G(x)$. Then $P$ is a Sylow $p$-subgroup of $C_G(x)$ hence cyclic. So it is enough to show that $x \in Z(P)$.

Since $Z(P)$ is a $p$-group and $Z(P) \subset C_G(x)$ we have that $Z(P)$ will be contained in a Sylow $p$-subgroup of $C_G(x)$ which is cyclic. Hence $Z(P)$ is itself cyclic. Let $H = \langle x, Z(P)\rangle$, the subgroup generated by $x$ and $Z(P)$. Then $H \subset P$ hence $H$ is a $p$-group. Also $H \subset C_G(x)$, and as before this implies that $H$ is cyclic. A cyclic $p$-group has a unique subgroup of order $p$. So both $H$ and $Z(P)$ contain unique subgroups of order $p$. These two subgroups must be identical since $Z(P) \subset H$. Furthermore the unique subgroup of order $p$ of $H$ is the subgroup generated by $x$. As this is also the unique subgroup of order $p$ in $Z(P)$ we have shown $x \in Z(P)$.