Cauchy's Lemma (Group Theory)
This proof is about Cauchy's Lemma in the context of Group Theory. 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): Cauchy's Lemma