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Prove that if $G$ is a finite nilpotent group and $N\trianglelefteq G$, then $N\cap Z(G)\ne\{e\}$.

This is true for all $p$-groups.

Suppose $|G|=p_1^{\alpha_1}\cdots p_s^{\alpha_s}$ where $p_1,\cdots,p_s$ are distinct primes, then $G\cong P_1\times\cdots\times P_s$ where $P_i$ is a Sylow $p_i$-subgroup.

Let $\varphi$ be an isomorphism from $G$ to $P_1\times\cdots\times P_s$. Since $Z(G)\cong Z(P_1)\times\cdots\times Z(P_s)$, I want to prove that $\varphi(N)\cap[Z(P_1)\times\cdots\times Z(P_s)]\ne\{e\}$, then the inequality is hold. But I don't know how to do this.

Knt
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    It suffices to handle the $p$-group case, but that is fairly standard. Try Google or searching MSE. – the_fox Jul 14 '19 at 02:07
  • I know the proof when $G$ is a $p$-group. But I don't know how to apply it to this exercise. I have already done some internet searching. – Knt Jul 14 '19 at 04:57
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    Consider a (non-trivial) Sylow $p$-subgroup of $N$. Since $G$ has a unique Sylow $p$-subgroup, that subgroup contains the $p$-Sylow of $N$ as a normal subgroup. You can apply the $p$-group case now. – the_fox Jul 14 '19 at 05:08

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