Cauchy's theorem

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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


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 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.