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 product $$G \times G \times \cdots \left({p}\right) \cdots \times G$$.

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 $$r + s p = n^{p-1} \Longrightarrow p \backslash r$$.

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 $$p$$.