Cauchy’s Theorem

Posted on May 16, 2017 by tcgrubb

The famous Lagrange’s Theorem states that, in any finite group G, the order of any element g\in G divides the order of G. A natural follow up to this theorem is whether or not the converse holds. That is, if k\in \mathbb{N} divides the order of G, does there exist an element g\in G of order k? Unfortunately the converse is, in general, false. Examples are easy to come by; take, for instance, (\mathbb{Z}/2\mathbb{Z})^j and look for an element of order 2^i, i>1.

While the converse to Lagrange’s Theorem is false in general, Cauchy’s Theorem tells us that there are special circumstances in which the converse is true. Namely, if G is a finite group and p is a prime number dividing |G|, then we can guarantee that some element in G has order p. This means, for example, that in  the symmetric group S_n we can find an element of order p for all p\leq n (it is easy to come up with the elements here, just take a p-cycle. In general Cauchy’s Theorem does not tell you how to find such elements).

There are several proofs of Cauchy’s Theorem available. The standard first proof uses induction on the size of G, and while it is instructive, I find it to be a bit messy. Below I reproduce a (in my opinion) more elegant proof using group actions. The original version of this proof is due to James McKay.

Proof of Cauchy’s Theorem: Let G be a finite group, p a prime with p dividing |G|. Consider the set

S=\{(g_1,g_2,\dots,g_p):g_i\in G, g_1g_2\dots g_p=1\}.

In other words, S consists of tuples from G whose ordered product equals the identity. Note that once we choose g_1,\dots,g_{p-1}, the element g_p is forced to equal (g_1\cdot\dots\cdot g_{p-1})^{-1}, and thus |S|=|G|^{p-1}.

Now we make the following observation: if g_1g_2\dots g_p=1, then any cyclic permutation of the product must also yield the identity (to see this, use the relation that if ab=1 then ba=1 and induct). Thus S admits an action by the cyclic group C_p. If s\in S, then by the Orbit-Stabilizer Theorem the orbit of s must have size 1 or p. What does it mean for s to have an orbit of size 1? If this is the case, then it must be that s=(g_1,\dots,g_p) with g_1=\dots=g_p; in otherwords, s represents an element g with g^p=1. As p is prime, if we can find such a g which is not the identity element, we will have found our element of order p.

To finish, we use the fact that the orbits of a group action partition the set on which the group is acting. Thus we have

|S|=\sum_{\text{orbits}}|O|.

We know that (1,\dots,1) is contained in S since 1^p=1, and thus there is one orbit of size 1. Thus

|S|=1+\sum |O|.

Now |O|\in \{1,p\} as mentioned previously. Further, |S|=|G|^{p-1} is divisible by p, and thus the right hand side above is divisible by p. It follows immediately that the number of orbits of size 1 is a multiple of p, and as we have already found one, there must be at least p. Thus we can guarantee an element of order p as desired.

As always, thanks for reading!

This entry was posted in Group Theory and tagged . Bookmark the permalink.

Leave a comment