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Let $N\neq 1$ be a normal subgroup of G, where G has order $p^n$, $p$ prime, $n\geq 1$. I need to show that $Z(G)\cap N\neq \{1\}$.

I have tried to do it this way: Suppose $|N|=p^m$, $m<n$. Since $N\cap Z(G)$ is normal in $Z(G)$, in both cases ($N\cap Z(G)=\{1\}$ and $N\cap Z(G)\neq\{ 1\}$), I used the Second Group Isomorphisms Theorem to get:

$G/N\cong Z(G)/N\cap Z(G)$

so $|G/N| = |Z(G)/N\cap Z(G)|$

Since N is normal in G, $|G/N|=p^{n-m}$, with $n-m>0$. So $|Z(G)/N|\cap Z(G)|=p^{n-m}\Rightarrow [Z(G):N\cap Z(G) ]=p^{n-m}$

and this means that $N\cap Z(G)\neq \{1\}$.

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This is the same as this question. – Mikko Korhonen May 4 '12 at 7:07
up vote 9 down vote accepted

Since $N$ is normal in $G$, then $G$ acts on $N$ by conjugation. So,

$$N = \sum \operatorname{Orbit}_{G}(x),$$ ($x\in N $) $ =$ 1 + $\sum \textrm{Orbit}_{G}(x)$, $x\in N-\{1\}$. Now, since the order of $\textrm{Orbit}_{G}(x)$ is a non negative power of $p$ and $p$||N|, then we see one orbit has order of 1. So, there is an non trivial element $s\in N$ that $\operatorname{Orbit}_{G}(s)$=${s}$. Therefore for all $g\in G$, $s^g=s$ and this means that $N \cap Z(G)$ is not 1.

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Nicely presented! :+) – amWhy Mar 5 '13 at 0:18

Since $N$ is normal, it is an union of disjoint conjugacy classes. It contains the conjugacy class $\{1\}$, and thus not all of the other conjugacy classes can have order divisible by $p$. This means that besides the conjugacy class $\{1\}$, the subgroup $N$ contains also some other conjugacy class $\{z\}$ of size $1$. Then $z$ is a nontrivial central element in $N$.

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