Cauchy's Lemma (Group Theory)/Proof 1

Theorem

Let $\struct {G, \circ}$ be a group of finite order whose identity is $e$.

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

Then $\struct {G, \circ}$ has an element of order $p$.

Proof

Let $\order G = n$ such that $p \divides n$.

Let:

$X = \set {\tuple {a_1, a_2, \ldots, a_p} \in G^p: a_1 a_2 \cdots a_p = e}$

where $G^p$ is the cartesian product $\underbrace {G \times G \times \cdots \times G}_p$.

The first $p - 1$ coordinates of an element of $X$ can be chosen arbitrarily.

The last coordinate is 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:

$\card X = n^{p - 1}$

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

Let $C_p$ act on the set $X$ by the rule:

$c * \tuple {a_1, a_2, \ldots, a_p} = \tuple {a_2, a_3, \ldots, a_p, a_1}$

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

As $p$ is prime, an orbit has either $p$ elements or $1$ element by definition.

Let $r$ be the number of orbits with one element.

Let $s$ be the number of orbits with $p$ elements.

Then by the Partition Equation:

$r + s p = n^{p - 1} = \card X$

By hypothesis, $p \divides n$, so:

$r + s p = n^{p - 1} \implies p \divides r$

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

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

Each such element has the form $\tuple {a, a, \ldots, a} \in X$ 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 $a^p = e$.

That is, $a$ must have order $p$.

$\blacksquare$

Source of Name

This entry was named for Augustin Louis Cauchy.