Group order

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 group

Proof via permutation groups (James McKay)

Let

Ω={(𝑔1,…,𝑔𝑝)∈𝐺𝑝:𝑔1⋯𝑔𝑝=1}

Note Ω is closed under the natural action of C𝑝 ≤S𝑝, since if 𝑔1⋯𝑔𝑝 =1, then 𝑔−11𝑔1⋯𝑔𝑝𝑔1 =1.

By the Orbit-stabilizer theorem, a C𝑝-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.

---

tidy | en | SemBr

Footnotes

1. MATH4031 ↩