Cauchy's theorem

Statement

If $G$ is a finite group, and $p$ is a prime number dividing the order of $G$, then $G$ has a subgroup of order exactly $p$. In particular, $G$ has an element of order exactly $p$.

Related facts

Stronger facts

• Sylow's theorem

Applications

• Exponent of a finite group has precisely the same prime factors as order

Facts used

1. Sylow subgroups exist: For any finite group $G$ and any prime $p$, $G$ has a subgroup whose order is the largest power of $p$ dividing $G$. Such a subgroup is termed a $p$-Sylow subgroup.

Proof

Proof using existence of Sylow subgroups

Given: A finite group $G$, a prime divisor $p$ of the order of $G$.

To prove: $G$ has a subgroup of order exactly $p$.

Proof: By fact (1), $G$ has a $p$-Sylow subgroup, say $P$. Since $p$ divides the order of $G$, $P$ is nontrivial. Pick a non-identity element $x\in P$. The order of $x$ divides the order of $P$, and is hence a power of $p$, say $p^{r}$, where $r\geq 1$. Thus, the element $x^{p^{r-1}}$ is an element of order $p$, and the cyclic subgroup it generates thus has order exactly $p$.

Proof without using existence of Sylow subgroups

The idea is to consider the action of $\mathbb {Z} /p\mathbb {Z}$ by cyclic permutations on the set:

$\{(g_{1},g_{2},\dots ,g_{p})\in G\times G\times \dots \times G\mid g_{1}g_{2}\dots g_{p}={\mbox{identity element of }}G\}$

First, note that the action does indeed send the set to itself: cyclic permutation of a word corresponds to a conjugation operation on the product, which preserves the identity. Explicitly:

$g_{p}g_{1}g_{2}\dots g_{p-1}=g_{p}(g_{1}g_{2}\dots g_{p})g_{p}^{-1}$

Hence, if the original word gives the identity element, so does the new word.

The set has size $|G|^{p-1}$. Using the class equation of a group action (or otherwise), note that the orbits all have size 1 or $p$, so the number of orbits of size 1 (i.e., the number of fixed points) is congruent to $|G|^{p-1}$ mod $p$. The latter number is a multiple of $p$, hence the number of fixed points is a multiple of $p$. Note that the point with all coordinates the identity element is a fixed point, so there is at least one fixed point. Thus, there must be at least $p$ fixed points. Any fixed point other than the "all-identity fixed point" is of the form $(g,g,\dots ,g)$ where $g^{p}$ is the identity element, and we thus get an element of order $p$.