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Suppose $G$ is a group where $|G| = p^k$ where $p$ is a prime and $k\gt 0$. Prove that

  1. $|Z(G)| \gt 1$; and
  2. If $N$ is a normal subgroup of $G$ of order $p$, then $N$ is contained in $Z(G)$.
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Dear Connie, could you please read this? Thanks, – Amitesh Datta Nov 20 '11 at 5:08

For 1, use the Class Equation.

For 2, use the Class Equation (if $N$ is normal, then it must be a union of conjugacy classes; conjugacy classes have either order $1$, or order a power of $p$; but $N$ contains at least one conjugacy class of order $1$, hence...)

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Here's the way I'd like to think about it, but it could be just a rewording of what has already been said. Using the orbit stabilizer theorem one can prove that if $G$ is a finite $p$-group acting on a finite set $X$ then $\#(X^G)\equiv \#(X)\text{ mod }p$ where $X^G$ denotes the set $\{x\in X:gx=x\text{ for all }g\in G\}$ of fix points.

So, let $G$ be a finite $p$-group and $N\unlhd G$ be nontrivial. Let $G$ act on $N$ on by conjugation. By the above we see that $\#(N^G)\equiv |N|\text{ mod }p$. But, since $p\mid |N|$ (since it's a nontrivial subgroup of a finite $p$-group) this implies that $p\mid \#(N^G)$. But, write it out and see that $N^G=N\cap Z(G)$ and so this prove that $p\mid |N\cap Z(G)|$ for any nontrivial normal subgroup $N$ of $G$, and so in particular, $N\cap Z(G)$ is nontrivial. Taking $N=Z(G)$ gives 1. immediately. It also gives 2. with a little thought, since we know that $p\mid |N\cap Z(G)|$ and $|N\cap Z(G)|\leqslant p$ we must have that $|N\cap Z(G)|=p$ and so clearly $N\cap Z(G)=N$ and so $N\subseteq Z(G)$.

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This follows directly from the class equation. Since the order of $G$ is a prime power, all of its conjugacy classes are of prime power order. Remember that the center of the group consists entirely of the singleton conjugacy classes (the classes of size $1$). If the center was trivial, the class equation would be $1$ plus positive prime powers. So we would have $1+p(\mathrm{something})=p^k$. This implies that $1=0$ mod $p$ (contradiction). Therefore, the center must contain more than one element.

By a similar line of reasoning, a normal subgroup consists of entire conjugacy classes unioned together (since $N$ is closed under conjugation, if part of a conjugacy class is in $N$, then the whole class must be in $N$). Since conjugacy classes have prime power cardnalities, the identity is in $N$ (whose class has size 1), and $|N|=p$, we have that all of the remaining classes are prime powers smaller than $p$. Thus $N$ is the union of classes of size 1. These are precisely elements of the center.

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