Cauchy's Lemma (Group Theory)

Theorem
Let $G$ be a group of finite order whose identity is $e$.

Let $p$ be a prime number which divides the order of $G$.

Then $G$ has an element of order $p$.

Proof
Let $\left|{G}\right| = n$.

Let $X = \left\{{\left({a_1, a_2, \ldots, a_p}\right) \in G^p: a_1 a_2 \cdots a_p = e}\right\}$

where $G^p$ is the cartesian space $G \times G \times \cdots \left({p}\right) \cdots \times G$.

The first $p-1$ co-ordinates of an element of $X$ can be chosen arbitrarily, the last one being determined by the fact that $a_1 a_2 \cdots a_{p-1} = a_p^{-1}$.

So from the Product Rule For Counting, it follows that $\left|{X}\right| = n^{p-1}$.

Let $C_p$ be a cyclic group order $p$ generated by the element $c$.

Let $C_p$ act on the set $X$ by the rule $c \wedge \left({a_1, a_2, \ldots, a_p}\right) = \left({a_2, a_3, \ldots, a_p, a_1}\right)$

By the Orbit-Stabilizer Theorem, the number of elements in any orbit is a divisor of the order of $C_p$, which is $p$.

Thus an orbit has either $p$ elements or $1$ element.

Let $r$ be the number of orbits with one element, and $s$ be the number of orbits with $p$ elements.

Then by the Partition Equation, $r + s p = n^{p-1} = \left|{X}\right|$.

By hypothesis, $p \backslash n$, so $r + s p = n^{p-1} \implies p \backslash r$.

We know that $r \ne 0$ because, for example, the orbit of $\left({e, e, \ldots, e}\right) \in X$ has only one element.

So there must be at least $p$ orbits with only one element.

Each such element has the form $\left({a, a, \ldots, a}\right) \in X$ so $a^p = e$.

So $G$ contains at least $p$ elements $x$ satisfying $x^p = e$.

So $G$ contains an element $a \ne e$ such that $a^p = e$, and $a$ must have order $p$.