# 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.

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.

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 $$x \in Z(P)$$.