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Patrick Stevens

Former mathematics student at the University of Cambridge; now a software engineer.

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(This post is mostly to set up a kind of structure for the website; in particular, to be the first in a series of posts summarising some mathematical results I stumble across.)

EDIT: There is now an Anki deck of this proof, and a collection of poems summarising it.

In Part IB of the Mathematical Tripos (that is, second-year material), there is a course called Groups, Rings and Modules. I took it in the academic year 2012-2013, when it was lectured by Imre Leader. He told us that there were three main proofs of the Sylow theorems, two of which were horrible and one of which was nice; he presented the “nice” one. At the time, I thought this was the most beautiful proof of anything I’d ever seen, although other people have told me it’s a disgusting proof.

Theorem - the Sylow Theorems

Let be a group, of order for some prime , where the HCF . Then:

  1. There is a subgroup of , of order (a Sylow p-subgroup);
  2. All such subgroups are conjugate to each other;
  3. The number of such subgroups, , satisfies and .

Proof

The proof goes as follows: pick a p-subgroup to be of maximal size; then introduce its normaliser , and show that the orbit of under the conjugation action when acts on itself is precisely the set of Sylow p-subgroups.

First Sylow theorem

The proof starts out in a natural way, by naming a subgroup of order for some . Such a subgroup certainly exists, by Cauchy’s Theorem (which has ). If we select to be maximal, then we wish to show that , or equivalently (which seems even easier) that is not a multiple of .

Now, how do we show that  is not a multiple of ? Well, we don’t know anything about such a quotient unless is normal in . But we can’t guarantee this - so let’s introduce a subgroup, , in which is normal. The natural one to pick, because we’re trying to make the subgroup as big as possible, is the normaliser  - that is, , or under the conjugation action. This is the largest subgroup of in which is normal.

Then we want to show that is not a multiple of ; this is true if and only if neither of the multiplicands is divisible by .

The second multiplicand

It looks like it will be easier to start with the second multiplicand, because it’s got a really really obvious interpretation.

We want to show that is not a multiple of . Now, from the First Isomorphism Theorem we have .

Suppose . Then by Cauchy’s Theorem, there is an element such that the order ; let , the group generated by . But we got to this quotient group by applying the projection map , so what happens when we “un-quotient” (that is, apply )? We have has order , because was a -to-one mapping, and so has order . This is a contradiction.

Hence .

The first multiplicand

The first multiplicand, : this is the number of conjugates of , by the Orbit-Stabiliser Theorem (by using the conjugation action: the stabiliser is ; while the orbit of is simply the set of conjugate subgroups). We want to show that this is not divisible by . We can do much more with the conjugates themselves, so let .

We would like to show that . This expression rings a bell - we’ve seen it before, as a key idea in the class equation. In order to use the class equation, we need to act on . There are only three groups we’ve met so far: , and . The group we haven’t yet used is , and it’s a p-group (and we know a bit about actions of p-groups). What’s the only obvious action to use? It has to be conjugation.

Let act on by conjugation. Since the orbits partition the set and have order dividing , the order of each orbit is one of . is clearly in an orbit all of its own (since for every ). What we really want is for to be the only conjugate of which is in its own orbit, because then we have (since the orbits partition the set).

Suppose we have such that is in an orbit of size 1. Then for all , and so (by conjugating with ) we have , and so stabilises and so is in . So is contained within .

Now, we know that is contained within , so we can now use functions defined on . We have that (the quotient map) is a homomorphism with kernel . That is, . Hence considering because is a homomorphism; but so this expression is just .

Hence is contained in the kernel of . But it’s also the same size as which is itself the kernel of . Hence .

So there is only one orbit of size , and hence because orbits partition the set, is not divisible by .

This concludes the proof of the first Sylow theorem.

Second Sylow theorem

Given a Sylow p-subgroup of , we want to show that it is conjugate to .

Use as before, the set of . In the first theorem, we had acting on ; now let’s use in the same way. We want to show that there is some such that , or equivalently that .

Let act on by conjugation. We have that is not a multiple of by the earlier part, but is a union of orbits which are of size for some . Hence there is a such that is the entire orbit of when acts on that conjugate. (That is, there is such that for all .) Hence, as before, all elements of fix under conjugation, and hence .

Now, so we can apply the projection map to it. We show that . Indeed, suppose it isn’t. Then is a non-trivial subgroup of , because was a subgroup of . It has order dividing that of , because applying a homomorphism to a subgroup yields a subgroup of order dividing that of the original - and so its order is a multiple of . Also, its order divides that of , by Lagrange, because it’s a subgroup of - and this is not a multiple of . But now we have a multiple of which divides a non-multiple of - contradiction.

Then ; and hence we must have . So and hence .

This concludes the proof of the second Sylow theorem.

Third Sylow theorem

We now want to show that the number of Sylow p-subgroups is and divides .

We certainly have that , because every Sylow p-subgroup is a conjugate of , but also every conjugate of (that is, every member of ) is itself a subgroup of , and has the same size as , so is also a Sylow p-subgroup. Hence, just as before, .

Also, is the size of an orbit under conjugation, and hence by the Orbit/Stabiliser Theorem, it divides ; but does not have a factor of , so it must divide .

This concludes the proof of the third Sylow theorem.

Summary

So the proof went as follows:

  1. We’re looking for information about Sylow p-subgroups, so we pick the maximum possible p-subgroup and hope that it’s a Sylow one.
  2. How do we know whether this p-group is Sylow? If is not divisible by .
  3. What can we do with a quotient? Not much, but we can use a quotient of a normal subgroup. We can’t guarantee that is normal in , so we split up the fraction into and .
  4. What’s a good normal subgroup to use? We have a choice. We’ll go for the normaliser , because that gives a nice interpretation to . (But otherwise, this step seems a bit arbitrary to me.)
  5. Now we’ll go for ; this is definitely something to do with the quotient group . Let’s imagine its size were divisible by ; then we can use Cauchy on and get a contradiction on moving back to .
  6. Let’s now consider ; the normaliser is something to do with conjugates, so we’ll consider the conjugation action. Happily, this expression then becomes the size of the orbit of under the conjugation action; call that orbit .
  7. We need . Remember the class equation; we want to act on using a p-group. is such a p-group, so we’ll let act on . The only natural action to use is conjugation. We know straight away that is in an orbit all to itself; we need it to be the only one.
  8. Name a different conjugate of ; call it . We need this to be exactly . It’s got the right size already, so we just need it to be contained in . Here a leap of faith: what’s special about ? It’s the kernel of a homomorphism (because it’s a normal subgroup of ). So, after proving that is defined on what we want to give as its arguments (that is, after showing that is contained in , or equivalently that all elements of stabilise under conjugation), consider . This is clearly , and hence is in the kernel of , and hence is a subset of , as required.
  9. Now the second theorem: all the Sylow p-subgroups need to be conjugate. Name a Sylow p-subgroup , and have it act on as above. Then in exactly the same way as in step 7, since is not a multiple of , we have that there is some such that is an entire orbit under conjugation by .
  10. Exactly as in step 8, a conjugate is on its own in an orbit, so it is fixed under conjugation by every element in . Hence is contained within and we can use . Suppose that is not fully contained in the kernel of ; then applying to it gives us a subgroup, which must have prime power order (from the fact that had prime power order); it also has order dividing that of , which is not a multiple of : contradiction.
  11. , a conjugate of , is hence contained in the kernel of . Then since it is of the same size as the kernel, it must be the kernel, but that is .
  12. Now the third theorem: we’ve just shown that is precisely the set of Sylow p-subgroups, so is just what we want (but we’ve already shown it back in step 8); and since it is also precisely an orbit when acts on by conjugation, it must have order dividing that of .