# 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

$$g\cdot h=gh{g}^{-1}$$ |

Note that $g\cdot h\in H$ since $H\u25c1G$. This is easily seen to be a well-defined group action^{}.

Now, the set of invariants of $H$ under this action are

$${G}_{H}=\{h\in H|g\cdot h=h\forall g\in G\}=\{h\in H|gh{g}^{-1}=h\forall g\in G\}=H\cap Z(G)$$ |

The class equation theorem states that

$$|H|=|{G}_{H}|+\sum _{i=1}^{r}[G:{G}_{{x}_{i}}]$$ |

where the ${G}_{{x}_{i}}$ are proper subgroups^{} of $G$, and thus that

$$|{G}_{H}|=|H|-\sum _{i=1}^{r}[G:{G}_{{x}_{i}}]$$ |

We now use elementary group theory to show that $p$ divides each term on the right, and conclude as a result that $p$ divides $|{G}_{H}|$, 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^{} |
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Canonical name | ProofThatANontrivialNormalSubgroupOfAFinitePgroupGAndTheCenterOfGHaveNontrivialIntersection |

Date of creation | 2013-03-22 14:21:07 |

Last modified on | 2013-03-22 14:21:07 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 9 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 20D20 |