# $p$-nilpotent finite group $G$ is a $p$-normal group?

Is it true that a $p$-nilpotent finite group $G$ is necessarily a $p$-normal group?

The definitions are here: p-nilpotent and p-normal (I like the 2nd definition)

Let $p$ be a prime and $P,Q$ be $p$-Sylow subgroups of $G$ such that $Z(Q)\subseteq P$.

We have to show $Z(Q)=Z(P)$. Writing $Q=P^{g}$ we can write this as $Z(P)^g=Z(P)$.

The group $G$ is $p$-nilpotent, so it has a $p$-normal complement by definition. Let $N$ be a $p$-normal complement for $G$, i.e., a normal subgroup $N$ of $G$ such that $G=NP$ and $N\cap P=1$.

So again, we need to show $Z(P)^g=Z(P)$. We could write $g=np$ and this becomes $Z(P)^n=Z(P)$, for the center $Z(P)$ is a normal subgroup of $P$. But I'm not sure what to do now.

Any help? Thank you.

Suppose $Q = P^{g}$ is another Sylow $p$-subgroup such that $Z(P) \le Q$. We show that $Z(P) = Z(Q)$.
Note first that we may take $g \in N$.
Let $z \in Z(P)$, and $y \in Q$, so that $y = x^{g}$ for some $x \in P$.
Then \begin{equation} [z, y] = [z, x^{g}] = [z, x [x, g]]= [z, [x, g]] [z, x]^{[x, g]} = [z, [x, g]]. \end{equation} Now $[z, y] \in Q$, but also $[z, [x, g]] \in N$, as $g \in N\trianglelefteq G$, so $[z, y] = 1$ as $Q \cap N = 1$
This shows that $Z(P) \le Z(Q)$, equality follow from the fact that they have the same order.