# 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 $$X$$.

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• Group

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