# The Center of a p-Group is Not Trivial

## Problem 10

Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$.
(Such a group is called a $p$-group.)

Show that the center $Z(G)$ of the group $G$ is not trivial.

## Hint.

Use the class equation.

## Proof.

If $G=Z(G)$, then the statement is true. So suppose that $G\neq Z(G)$.
Then by the class equation, we have
$|G|=|Z(G)|+ \sum_{i=1}^r |G:C_G(g_i)|,$ where $g_i$ are representatives of the distinct conjugacy class not contained in the center $Z(G)$, and $C_G(g_i)$ is the centralizer of $g_i$.
(Since we are assuming that $G \neq Z(G)$ such $g_i$ exist.)

Since $g_i \not \in Z(G)$, the groups $C_G(g_i)$ are proper subgroups of $G$ and hence $p$ divides $|G: C_G(g_i)|$. Of course $p$ divides $|G|$, thus $p$ should divide $|Z(G)|$ as well.
Therefore the center $Z(G)$ cannot be trivial.

## Comment.

This problems is a simple/nice application of the class equation of  group theory. The only information about the group is that its order is a prime power. From this we can conclude that the center of the group is not trivial.

### 1 Response

1. 07/26/2016

[…] The center of a p-group is not trivial (post 1) […]

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