PlanetMath: proof that a nontrivial normal subgroup of a finite $p$-group $G$ and the center of $G$ have nontrivial intersection

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proof that a nontrivial normal subgroup of a finite $p$

-group

and the center of $G$

have nontrivial intersection

(Proof)

Define to act on by conjugation; that is, for $g\in G$ , $h\in H$ , define

$\displaystyle g \cdot h = ghg^{-1}$

Note that $g\cdot h\in H$ since $H\triangleleft G$ . This is easily seen to be a well-defined group action.

Now, the set of invariants of under this action are

$\displaystyle G_H=\{h \in H \ \lvert \ g \cdot h = h \ \forall g\in G\} = \{h \in H \ \lvert \ ghg^{-1} = h \ \forall g\in G\}=H \cap Z(G)$

The class equation theorem states that

$\displaystyle \lvert H\rvert = \lvert G_H\rvert + \sum_{i=1}^{r}[G:G_{x_{i}}]$

where the $G_{x_{i}}$ are proper subgroups of

, and thus that

$\displaystyle \lvert G_H\rvert = \lvert H\rvert - \sum_{i=1}^{r}[G:G_{x_{i}}]$

We now use elementary group theory to show that divides each term on the right, and conclude as a result that divides $\lvert G_H\rvert$ , so that $G_H=H\cap Z(G)$ cannot be trivial.

As is a nontrivial finite -group, it is obvious from Cauchy's theorem that $\vert G\vert=p^{n}$ for . Since and the $G_{x_i}$ are subgroups of , each either is trivial or has order a power of , by Lagrange's theorem. Since is nontrivial, its order is a nonzero power of . Since each $G_{x_i}$ is a proper subgroup of and has order a power of , it follows that $[G:G_{x_i}]$ also has order a nonzero power of .