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

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 $Z / p Z$ by cyclic permutations on the set:

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

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 $p$ .