Question about a proof of: If $P$ is a finite $p$-group and $\langle 1\rangle\neq N\trianglelefteq P$, then $N\cap Z(P)\neq\langle 1\rangle$.

Question

Question: If $P$ is a finite $p$-group and $\langle 1\rangle\neq N\trianglelefteq P$, then $N\cap Z(P)\neq\langle 1\rangle$.

I have a question about a proof I saw in (I believe, Isaacs book). It goes: Since $N$ is normal in $P$, $P$ acts by conjugation on $N$ with a set of fixed points in this action precisely $N\cap Z(P)$. Thus $|N\cap Z(P)|=|N|$ mod $p$. Since $N$ is a non-trivial $p$ group, $|N|=0$ mod $p$, so $|N\cap Z(P)|=0$ mod $p$, hence the result.

My question is how do we know that the set of fixed points in the action is PRECISELY $N\cap Z(P)$?

Answer 1

By definition $x\in Z(P)$ if and only if $x\in P$ is fixed by the conjugation action of $P$.

Answer 2

We know this because $n\in N$ is a fixed point of conjugation by $P$ if and only of $n\in N$ and $pnp^{-1}=n$ for all $p\in P$, if and only if $n\in N$ and $pn=np$ for all $p\in P$, if and only if $n\in N\cap Z(P)$.