Let $G$ be a group such that $|G|=p^{\alpha}m$, and let $P$ be a Sylow $p$-subgroup of $G$. Suppose that $x\in P$ is such that $x$ has order $p$ and that $C_{G}(x)$ has cyclic Sylow $p$-subgroups. Then $P$ is cyclic.
My attempt (edited):
If $x\not\in Z(P)$, choose $y\in Z(P)$ of order $p$. Such $y$ exists since $Z(P)\supsetneq \{1\}$. Then $yxy^{-1}x^{-1}=1$, so that $yx=xy$. Since $\langle x\rangle\cap \langle y\rangle\leq\langle y\rangle$, then $\langle x\rangle\cap\langle y\rangle$ is either trivial or $\langle y\rangle$. It cannot be the latter for otherwise $x\in\langle y\rangle\subseteq Z(P)$, a contradiction. Hence $\langle x,y\rangle\cong\langle x\rangle\times\langle y\rangle$ is a $p$-subgroup of $C_{G}(x)$ and so is cyclic by hypothesis.
Comments (edited):
- I have a feeling that I should conclude that $\langle x,y\rangle$ is not cyclic, but I am not sure how to.
- With this $\langle x,y\rangle$ not being cyclic, we get a contradiction so that $x\in Z(P)$. How then would we be able to conclude that $P\subseteq C_{G}(x)$ as I feel that it should go in this direction.
I would appreciate any hints to point me in the right direction for this problem.