Cauchy's theorem states that if is a finite group and $p$ is a prime dividing $| G |$ the order of $G$ , then $G$ has a subgroup of order $p$ . Such a subgroup is necessarily cyclic (proof).

Proof

The proof involves basically a single magic idea: from thin air, we pluck the definition of the following set.

Let $X = {(x_{1}, x_{2}, \dots, x_{p}) : x_{1} x_{2} \dots x_{p} = e}$ the collection of $p$ -tuples of elements of the group such that the group operation applied to the tuple yields the identity. Observe that $X$ is not empty, because it contains the tuple $(e, e, \dots, e)$ .

Now, the cyclic group $C_{p}$ of order $p$ acts on $X$ as follows: $(h, (x_{1}, \dots, x_{p})) \mapsto (x_{2}, x_{3}, \dots, x_{p}, x_{1})$ where $h$ is the generator of $C_{p}$ . So a general element $h^{i}$ acts on $X$ by sending $(x_{1}, \dots, x_{p})$ to $(x_{i + 1}, x_{i + 2}, \dots, x_{p}, x_{1}, \dots, x_{i})$ .

This is indeed a group action (exercise).

Show solution

It certainly outputs elements of $X$ , because if $x_{1} x_{2} \dots x_{p} = e$ , then $x_{i + 1} x_{i + 2} \dots x_{p} x_{1} \dots x_{i} = (x_{1} \dots x_{i})^{- 1} (x_{1} \dots x_{p}) (x_{1} \dots x_{i}) = (x_{1} \dots x_{i})^{- 1} e (x_{1} \dots x_{i}) = e$
The identity acts trivially on the set: since rotating a tuple round by $0$ places is the same as not permuting it at all.
$(h^{i} h^{j}) (x_{1}, x_{2}, \dots, x_{p}) = h^{i} (h^{j} (x_{1}, x_{2}, \dots, x_{p}))$ because the left-hand side has performed $h^{i + j}$ which rotates by $i + j$ places, while the right-hand side has rotated by first $j$ and then $i$ places and hence $i + j$ in total.

Now, fix $¯ x = (x_{1}, \dots, x_{p}) \in X$ .

By the Orbit-Stabiliser theorem, the orbit ${O r b}_{C_{p}} (¯ x)$ of $¯ x$ divides $| C_{p} | = p$ , so (since $p$ is prime) it is either $1$ or $p$ for every $¯ x \in X$ .

Now, what is the size of the set $X$ ?

Show solution

It is $| G |^{p - 1}$ .

Indeed, a single $p$ -tuple in $X$ is specified precisely by its first $p$ elements; then the final element is constrained to be $x_{p} = (x_{1} \dots x_{p - 1})^{- 1}$ .

Also, the orbits of $C_{p}$ acting on $X$ partition $X$ (proof). Since $p$ divides $| G |$ , we must have $p$ dividing $| G |^{p - 1} = | X |$ . Therefore since $| {O r b}_{C_{p}} ((e, e, \dots, e)) | = 1$ , there must be at least $p - 1$ other orbits of size $1$ , because each orbit has size $p$ or $1$ : if we had fewer than $p - 1$ other orbits of size $1$ , then there would be at least $1$ but strictly fewer than $p$ orbits of size $1$ , and all the remaining orbits would have to be of size $p$ , contradicting that $p ∣ | X |$ .

picture of class equation

Hence there is indeed another orbit of size $1$ ; say it is the singleton ${¯ x}$ where $¯ x = (x_{1}, \dots, x_{p})$ .

Now $C_{p}$ acts by cycling $¯ x$ round, and we know that doing so does not change $¯ x$ , so it must be the case that all the $x_{i}$ are equal; hence $(x, x, \dots, x) \in X$ and so $x^{p} = e$ by definition of $X$ .

Cauchy's theorem on subgroup existence