## Cauchy’s Theorem

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.