Cauchy's order theorem

Let $𝐺$ be a finite group and $𝑝$ be a prime dividing $| 𝐺 |$ . Then $𝐺$ has an element of order $𝑝$ .¹ #m/thm/group

Proof via permutation groups (James McKay)

Let

Ω = {(𝑔 1, \dots, 𝑔 𝑝) \in 𝐺 𝑝 : 𝑔 1 \dots 𝑔 𝑝 = 1}

Note $Ω$ is closed under the natural action of $C 𝑝 \leq S 𝑝$ , since if $𝑔 1 \dots 𝑔 𝑝 = 1$ , then $𝑔 - 1 1 𝑔 1 \dots 𝑔 𝑝 𝑔 1 = 1$ .

By the Orbit-stabilizer theorem, a $C 𝑝$ -orbit in $Ω$ has size 1 or $𝑝$ . For an element to have an orbit of size 1, it must have order 1 or $𝑝$ .

Furthermore, $| Ω | = | 𝐺 | 𝑝 - 1$ , by basic combinatorics (the first $𝑝 - 1$ choices are free).

It follows that the number of orbits of size 1 is divisible by $𝑝$ , and hence there exists more than 1 orbit of size 1. Since only one of these may be the repeated identity, it follows there exists at least one element of order $𝑝$ .

#state/tidy | #lang/en | #SemBr

MATH4031 ↩

Cauchy's order theorem

Footnotes