Sylow’s Theorem, from group theory, states that:

- If then contains a subgroup of order .
- Any two such subgroups are conjugate.
- The number of such subgroups satisfies .

The standard proof is to attack 1 using a clever binomial coefficient trick, and then use different arguments to tackle 2 and 3. The following proof, shown to the author by Prof. Leader of Trinity College, Cambridge, more or less proves all three parts simultaneously, and is very natural. The proof is probably slightly different from that I was originally shown, but is hopefully identical in spirit.

**Proof:** For a finite group , let be a maximal -subgroup (a subgroup of largest cardinality where every element has order a power of ).

**Proof of part 1: **It will suffice if we can prove that . But , where is the normaliser of in . Splitting the fraction in this way allows us to assign meaning to the quantities.

Since is normal in , we can measure the second quantity as simply the order of the group . Suppose divides this order. Then, by Cauchy’s Theorem, we can find of order . But then would be a -subgroup and have order greater than , contradicting its maximality.

Now, by the Orbit-Stabiliser theorem considering the action of conjugation on by , we can see that the first quantity must be the size of the orbit of under this action. Let us call the set of these orbits , so we need to show that . In fact, we can do better and show that , which will immediately give part 3 of the theorem as a consequence of part 2.

To do this we consider acting on by conjugation, and aim to show that is the only singleton orbit (and since is a -subgroup, all the other orbits must have sizes divisible by , so this would give the required congruence). Suppose is another singleton orbit (with ). Then normalises , which implies normalises , so . But in this case, it is clear that is a -subgroup of order greater than that of (having nontrivial image of order under the quotient map ). Thus the only singleton orbit is , whence .

So it remains to prove statement 2 (essentially that ). Suppose is a Sylow -subgroup. If we let act by conjugation on , then there must be some (unique) singleton orbit which is normalised by . But considering that the image of under the quotient map must be a single element (by maximality of and the fact all elements of have order divisible by ), we must have . So , as required.

As remarked above, the third part is now immediate.

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February 10, 2009 at 3:47 pm

Yimin GeThis is indeed a very nice proof of Sylow’s theorems. I myself however wouldn’t have said that the ‘clever binomail coefficient’-proof of the first Sylow’s theorem is the standard proof (but only because of the simple reason that I have actually never heard of that one *blush*^^) – but ok, it probably still is^^. The proof I found some time ago is purely inductive, using the theorem that if divides the order of a finite group ( prime, natural), then either divides the order of some proper subgroup or divides the order of the center.

Part 2 and 3 can more or less simultanously be killed by the fact that if a -group acts on a finite set , then the number of singleton orbits is – just let a -subgroup of act on the left cosets of a Sylow -subgroup (and choose itself for part 3) and everything else follows automatically. This way, you immediately get another result ‘for free’, namely that every -subgroup is contained in some Sylow -subgroup (but ok…you get that result as well if in your last part, you let be an arbitrary -subgroup).