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 $$p$$.

Stronger facts

 * Sylow's theorem

Applications

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

Facts used

 * 1) uses::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 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 $$r \ge 1$$. 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:

$$\{ (g_1,g_2,\dots,g_p) \in G \times G \times \dots \times G \mid g_1g_2 \dots g_p = \mbox{identity element of } G \}$$

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:

$$g_pg_1g_2\dots g_{p-1} = g_p(g_1g_2\dots g_p)g_p^{-1}$$

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 $$p$$.