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

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

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

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 $H$ under this action are

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

The class equation theorem states that

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

where the $G_{x_{i}}$ are proper subgroups of $G$, and thus that

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

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

As $G$ is a nontrivial finite $p$-group, it is obvious from Cauchy’s theorem that $|G|=p^{n}$ for $n>0$. Since $H$ and the $G_{x_{i}}$ are subgroups of $G$, each either is trivial or has order a power of $p$, by Lagrange’s theorem. Since $H$ is nontrivial, its order is a nonzero power of $p$. Since each $G_{x_{i}}$ is a proper subgroup of $G$ and has order a power of $p$, it follows that $[G:G_{x_{i}}]$ also has order a nonzero power of $p$.

Title proof that a nontrivial normal subgroup of a finite $p$-group $G$ and the center of $G$ have nontrivial intersection ProofThatANontrivialNormalSubgroupOfAFinitePgroupGAndTheCenterOfGHaveNontrivialIntersection 2013-03-22 14:21:07 2013-03-22 14:21:07 rm50 (10146) rm50 (10146) 9 rm50 (10146) Proof msc 20D20