# Cauchy's Lemma

*This proof is about Cauchy's Lemma on the existence of elements of prime order in a group. For other uses, see Cauchy's Theorem.*

## 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 1

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 $p$.

$\blacksquare$

## Proof 2

By the corollary to the First Sylow Theorem, $G$ has subgroups of order $p^r$ for all $r$ such that $p^r \divides \order G$.

Thus $G$ has at least one subgroup $H$ of order $p$.

As a Prime Group is Cyclic, $H$ is a cyclic group.

Thus by definition $H$ has an element of order $p$.

Hence the result.

$\blacksquare$

## Also see

- Cauchy's Group Theorem, which establishes that, given the same conditions, $G$ also has a subgroup of order $p$.

## Source of Name

This entry was named for Augustin Louis Cauchy.

## Sources

- 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**Cauchy's Lemma**