Burnside's Theorem for p-Groups

# Burnside's Theorem for p-Groups

Recall from the p-Groups page that if $p$ is a prime, then a group $G$ is said to be a $p$-group if $G$ has order $p^k$ for some $k \geq 1$.

The following theorem is very important. It tells us that the center of any $p$-group is nontrivial.

Theorem 1 (Burnside's Theorem for $p$-Groups): Let $p$ be a prime. If $G$ is a $p-group$ then $Z(G)$ is not the trivial group. |

**Proof:**Let $G$ be a $p$-group. Then $|G| = p^k$ for some $k \geq 1$. So the class equation for $G$ is:

\begin{align} \quad p^k = |Z(G)| + \sum [G:C_G(g)] \quad (*) \end{align}

- Where the sum on the righthand side runs over one element $g$ for each nontrivial conjugacy class.

- Let $[g]$ be a nontrivial conjugacy class, i.e., $g \not \in Z(G)$. Then $|[g]| = |[G:C_G(g)]| > 1$.

- Also, by Lagrange's Theorem that:

\begin{align} \quad |G| &= [G:C_G(g)] |C_G(g)| \\ p^k &= [G:C_G(g)] |C_G(g)| \end{align}

- So $1 < [G:C_G(g)] \mid p^k$. In particular, $p | [G:C_G(g)]$.

- So if $[g]$ is a nontrivlal conjugacy class we have that $p \mid [G:C_G(g)]$. So from the class equation at $(*)$, since $p \mid p^k$ and $p \mid \sum [G:C_G(g)]$ (where the sum runs over one element $g$ for each nontrivial conjugacy class), we have that [[$ p \mid |Z(G)|

- Thus $|Z(G)| \geq p$, i.e., the center of $G$, $Z(G)$, is nontrivial. $\blacksquare$