# Burnside’s Theorem

###### Theorem 1 (Burnside’s Theorem).

Let $G$ be a simple group^{}, $\sigma \mathrm{\in}G$. Then the number of conjugates^{} of $\sigma $ is not a prime power (unless $\sigma $ is its own conjugacy class^{}).

Proofs of this theorem are quite difficult and rely on representation theory.

From this we immediately get

###### Corollary 1.

A group $G$ of order ${p}^{a}\mathit{}{q}^{b}$, where $p\mathrm{,}q$ are prime, cannot be a nonabelian^{} simple group.

###### Proof.

Suppose it is. Then the center of $G$ is trivial, $\{e\}$, since the center is a normal subgroup^{} and $G$ is simple nonabelian. So if ${C}_{i}$ are the nontrivial conjugacy classes, we have from the class equation^{} that

$$|G|=1+\sum |{C}_{i}|$$ |

Now, each $|{C}_{i}|$ divides $|G|$, but cannot be $1$ since the center is trivial. It cannot be a power of either $p$ or $q$ by Burnside’s theorem. Thus $pq\mid |{C}_{i}|$ for each $i$ and thus $|G|\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modpq)$, which is a contradiction^{}.
∎

Finally, a corollary of the above is known as the Burnside $p$-$q$ Theorem (http://planetmath.org/BurnsidePQTheorem).

###### Corollary 2.

A group of order ${p}^{a}\mathit{}{q}^{b}$ is solvable.

Title | Burnside’s Theorem |
---|---|

Canonical name | BurnsidesTheorem |

Date of creation | 2013-03-22 16:38:14 |

Last modified on | 2013-03-22 16:38:14 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 4 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 20D05 |