Vtyjkyuk

Suppose $G$ is a group where $|G| = p^k$ where $p$ is a prime and $kgt 0$. 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 $Z(G)$.

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...)

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 $xin X:gx=xtext for all gin G$ of fix points.

So, let $G$ be a finite $p$-group and $Nunlhd G$ 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 $pmid |N|$ (since it's a nontrivial subgroup of a finite $p$-group) this implies that $pmid #(N^G)$. But, write it out and see that $N^G=Ncap Z(G)$ and so this prove that $pmid |Ncap Z(G)|$ for any nontrivial normal subgroup $N$ of $G$, and so in particular, $Ncap Z(G)$ is nontrivial. Taking $N=Z(G)$ gives 1. immediately. It also gives 2. with a little thought, since we know that $pmid |Ncap Z(G)|$ and $|Ncap Z(G)|leqslant p$ we must have that $|Ncap Z(G)|=p$ and so clearly $Ncap Z(G)=N$ and so $Nsubseteq Z(G)$.

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 $1+p(mathrmsomething)=p^k$. 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.

Do you know the class equation? Define an action of the group on itself by conjugation:
$$Gtimes Gto G,,,,,gcdot x:=x^g:=g^-1xg$$

Now, under this action, we get

$$forall,,xin G,,,,mathcal O(x):=x^g;:;gin G,,,,Stab(x):=gin G;:;x^g=x=C_G(x)$$

Well, now just use the basic lemma $,|mathcal O(x)|=[G:Stab(x)],$ and , of course, the fact that $,G,$ is a $,p-,$group...