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.
Sources
- February 1959: James H. McKay: Another Proof of Cauchy's Group Theorem (Amer. Math. Monthly Vol. 66: p. 119) www.jstor.org/stable/2310010
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 55$