# Cauchy's theorem on subgroup existence

Cauchy’s the­o­rem states that if $$G$$ is a finite group and $$p$$ is a prime di­vid­ing $$|G|$$ the or­der of $$G$$, then $$G$$ has a sub­group of or­der $$p$$. Such a sub­group is nec­es­sar­ily cyclic (proof).

# Proof

The proof in­volves ba­si­cally a sin­gle magic idea: from thin air, we pluck the defi­ni­tion of the fol­low­ing set.

Let $$X = \{ (x_1, x_2, \dots, x_p) : x_1 x_2 \dots x_p = e \}$$$the col­lec­tion of $$p$$-tu­ples of el­e­ments of the group such that the group op­er­a­tion ap­plied to the tu­ple yields the iden­tity. Ob­serve that $$X$$ is not empty, be­cause it con­tains the tu­ple $$(e, e, \dots, e)$$. Now, the cyclic group $$C_p$$ of or­der $$p$$ acts on $$X$$ as fol­lows: $$(h, (x_1, \dots, x_p)) \mapsto (x_2, x_3, \dots, x_p, x_1)$$$ where $$h$$ is the gen­er­a­tor of $$C_p$$. So a gen­eral el­e­ment $$h^i$$ acts on $$X$$ by send­ing $$(x_1, \dots, x_p)$$ to $$(x_{i+1}, x_{i+2} , \dots, x_p, x_1, \dots, x_i)$$.

This is in­deed a group ac­tion (ex­er­cise).

• It cer­tainly out­puts el­e­ments of $$X$$, be­cause if $$x_1 x_2 \dots x_p = e$$, then $$x_{i+1} x_{i+2} \dots x_p x_1 \dots x_i = (x_1 \dots x_i)^{-1} (x_1 \dots x_p) (x_1 \dots x_i) = (x_1 \dots x_i)^{-1} e (x_1 \dots x_i) = e$$\$

• The iden­tity acts triv­ially on the set: since ro­tat­ing a tu­ple round by $$0$$ places is the same as not per­mut­ing it at all.

• $$(h^i h^j)(x_1, x_2, \dots, x_p) = h^i(h^j(x_1, x_2, \dots, x_p))$$ be­cause the left-hand side has performed $$h^{i+j}$$ which ro­tates by $$i+j$$ places, while the right-hand side has ro­tated by first $$j$$ and then $$i$$ places and hence $$i+j$$ in to­tal. <div><div>

Now, fix $$\bar{x} = (x_1, \dots, x_p) \in X$$.

By the Or­bit-Sta­bil­iser the­o­rem, the or­bit $$\mathrm{Orb}_{C_p}(\bar{x})$$ of $$\bar{x}$$ di­vides $$|C_p| = p$$, so (since $$p$$ is prime) it is ei­ther $$1$$ or $$p$$ for ev­ery $$\bar{x} \in X$$.

Now, what is the size of the set $$X$$?

It is $$|G|^{p-1}$$.

In­deed, a sin­gle $$p$$-tu­ple in $$X$$ is speci­fied pre­cisely by its first $$p$$ el­e­ments; then the fi­nal el­e­ment is con­strained to be $$x_p = (x_1 \dots x_{p-1})^{-1}$$. <div><div>

Also, the or­bits of $$C_p$$ act­ing on $$X$$ par­ti­tion $$X$$ (proof). Since $$p$$ di­vides $$|G|$$, we must have $$p$$ di­vid­ing $$|G|^{p-1} = |X|$$. There­fore since $$|\mathrm{Orb}_{C_p}((e, e, \dots, e))| = 1$$, there must be at least $$p-1$$ other or­bits of size $$1$$, be­cause each or­bit has size $$p$$ or $$1$$: if we had fewer than $$p-1$$ other or­bits of size $$1$$, then there would be at least $$1$$ but strictly fewer than $$p$$ or­bits of size $$1$$, and all the re­main­ing or­bits would have to be of size $$p$$, con­tra­dict­ing that $$p \mid |X|$$. pic­ture of class equation

Hence there is in­deed an­other or­bit of size $$1$$; say it is the sin­gle­ton $$\{ \bar{x} \}$$ where $$\bar{x} = (x_1, \dots, x_p)$$.

Now $$C_p$$ acts by cy­cling $$\bar{x}$$ round, and we know that do­ing so does not change $$\bar{x}$$, so it must be the case that all the $$x_i$$ are equal; hence $$(x, x, \dots, x) \in X$$ and so $$x^p = e$$ by defi­ni­tion of $$X$$.

Children:

Parents:

• Group

The alge­braic struc­ture that cap­tures sym­me­try, re­la­tion­ships be­tween trans­for­ma­tions, and part of what mul­ti­pli­ca­tion and ad­di­tion have in com­mon.