ProofOfANontrivialNormalSubgroupOfAFinitePGroupGAndTheCenterOfGHaveNontrivialIntersection 2004-05-02 06:49:32 2007-01-23 22:15:51 Proof a nontrivial normal subgroup of a finite $p$-group $G$ and the center of $G$ have nontrivial intersection 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 $p$.