Cauchy's theorem states that if is a finite Group and is a prime dividing the order of , then has a subgroup of order . 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 the collection of -[-tuple]s of elements of the group such that the group operation applied to the tuple yields the identity. Observe that is not empty, because it contains the tuple .
Now, the cyclic group of order acts on as follows: where is the generator of . So a general element acts on by sending to .
This is indeed a group action (exercise).
%%hidden(Show solution):
- It certainly outputs elements of , because if , then
- The identity acts trivially on the set: since rotating a tuple round by places is the same as not permuting it at all.
- because the left-hand side has performed which rotates by places, while the right-hand side has rotated by first and then places and hence in total. %%
Now, fix .
By the Orbit-Stabiliser theorem, the orbit of divides , so (since is prime) it is either or for every .
Now, what is the size of the set ? %%hidden(Show solution): It is .
Indeed, a single -tuple in is specified precisely by its first elements; then the final element is constrained to be . %%
Also, the orbits of acting on partition (proof). Since divides , we must have dividing . Therefore since , there must be at least other orbits of size , because each orbit has size or : if we had fewer than other orbits of size , then there would be at least but strictly fewer than orbits of size , and all the remaining orbits would have to be of size , contradicting that . [todo: picture of class equation]
Hence there is indeed another orbit of size ; say it is the singleton where .
Now acts by cycling round, and we know that doing so does not change , so it must be the case that all the are equal; hence and so by definition of .