Cauchy's theorem states that if $G$ 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 $p$-[-tuple]s 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).

%%hidden(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 $\bar{x} = (x_1, \dots, x_p) \in X$.

By the Orbit-Stabiliser theorem, the orbit $\mathrm{Orb}_{C_p}(\bar{x})$ of $\bar{x}$ divides $|C_p| = p$, so (since $p$ is prime) it is either $1$ or $p$ for every $\bar{x} \in X$.

Now, what is the size of the set $X$? %%hidden(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 $|\mathrm{Orb}_{C_p}((e, e, \dots, e))| = 1$, 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 $p \mid |X|$. [todo: picture of class equation]

Hence there is indeed another orbit of size $1$; say it is the singleton $\{ \bar{x} \}$ where $\bar{x} = (x_1, \dots, x_p)$.

Now $C_p$ acts by cycling $\bar{x}$ round, and we know that doing so does not change $\bar{x}$, 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 $X$.