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

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

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 $n\in N\cap Z(P)$.

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

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