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 \(p\)-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).

• 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. <div><div>

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 \(X\)?

Indeed, a single \(p\)-tuple in \(X\) is specified precisely by its first \(p\) elements; then the final element is constrained to be \(x_p = (x_1 \dots x_{p-1})^{-1}\). <div><div>

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 \(p \mid |X|\). 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 \(X\).