Sylow’s Theorem, from the book

The most important theorems of elementary group theory are those of Lagrange and Sylow. I want to describe here what I consider the most beautiful proof of the first part of Sylow’s Theorem, actually based on Sylow’s original proof. But first, some preliminaries.

Lagrange’s Theorem states that the order of a subgroup of the group G divides the order of G. This is closely connected with the theory of group actions, which is crucial to the modern way of looking at these results. If H is a subgroup of G, then G acts transitively on the set of right cosets of H by right multiplication, and every transitive action of G is of this form. So all that is required is to observe that all right cosets of H have the same cardinality, which is true because the map taking h to hx is a bijection from H to Hx. We learn additionally that the quotient |G|/|H| is equal to the size of a set on which G acts transitively.

The converse of this theorem is usually taken to be the assertion that, if n and m are positive integers such that m divides n, then a group of order n has a subgroup of order m. Of course things are not so simple. The smallest counterexample is the alternating group on 4 letters, which has order 12 but has no subgroup of order 6.

Sylow showed that this converse does hold in a special case, namely that when m is a prime power. His theorem is traditionally stated with four parts. Suppose that G is a finite group, and p a prime. We call a subgroup H of G a Sylow p-subgroup if the order of H is the p-part of the order of G, that is, the largest power of p dividing |G|. Now Sylow’s theorem states that Sylow subgroups exist for all choices of G and p; that they are all conjugate in G; that the number of Sylow p-subgroups is congruent to 1 (mod p); and that any subgroup of G of p-power order is contained in a Sylow p-subgroup. A very satisfactory state of affairs!

There are several very different proofs of the first part of the theorem; I will give my favourite shortly. But the other three parts are proved by very similar arguments, and almost no alternative approach to them is known. I outline the proof. It depends on the following claim:

Let P be a Sylow p-subgroup of G, and Q a p-subgroup normalising P. Then Q is contained in P.

For the assumption implies that PQ is a subgroup; its order is |P|.|Q|/|PQ| by a standard formula, so it is a p-group. By Lagrange’s Theorem its order is equal to that of P, so it is equal to P, and the claim is proved.

Now, assuming that Sylow p-subgroups exist, we prove the remaining parts of Sylow’s Theorem by considering the action of G on the set of all Sylow p-subgroups by conjugation.

We restrict the action to P. It clearly fixes itself, but by the above claim it does not fix any other Sylow p-subgroup. So all the orbits except for {P} have size divisible by p (using our observation on the connectiion between Lagrange’s Theorem and group actions). This shows that the number of Sylow p-subgroups is congruent to 1 (mod p). Moreover, the size of the G-orbit containing P is congruent to 1 (mod p), and the sizes of any other orbits are congruent to 0 (mod p). If Q is any other Sylow subgroup, then Q also lies in the (unique) orbit of size congruent to 1 (mod p); so P and Q are in the same orbit, and so are conjugate. (Incidentally, at this point we know that there is just one orbit.) Finally, if H is any p-subgroup, then restrict the action to H; it must fix a point, that is, normalise a Sylow p-subgroup R (say), and by our claim it is contained in R.

So to the first part of the theorem, the existence of Sylow p-subgroups.

Some time ago, I supervised the MSc project of a student who wanted to read and expound Sylow’s original proof of his theorem. This was non-trivial since three kinds of translation were required:

  • from French to English;
  • from the leisurely 19th century style to the terser modern style;
  • from the language of cosets and double cosets to that of group actions.

On the last point, I mentioned in connection with Lagrange’s Theorem that cosets are connected with group actions. It turns out that double cosets are connected with orbits of a group on ordered pairs of elements.

I am afraid that I don’t now even recall the student’s name, since my files from back then are long since lost. Much credit should go to him, altnough of course the main credit goes to Sylow. I claim a small amount; the very last step in the proof was my idea.

Anyway, here we go. The proof of the existence of Sylow p-subgroups proceeds in three steps.

Step 1: Translation. We show the following:

A group G has a Sylow p-subgroup if and only if it has a transitive action on a set X, such that the size of X is coprime to p but the order of any point stabiliser is a power of p.

If P is a Sylow p-subgroup, then the action on the cosets of P has the required property. In the other direction, given such an action, the point stabiliser is a Sylow p-subgroup, since its order is a p-power and its index is coprime to p.

Step 2: The heart of the proof. We show the following:

Suppose that G has a Sylow p-subgroup. Then any subgroup of G has a Sylow p-subgroup.

Take an action of G satisfying the conditions of Step 1. Let H be a subgroup of G, and restrict the action to H. Clearly the number of points is still coprime to p and the orders of the point stabilisers are p-powers. The small difficulty is that the action may not be transitive. But at least some orbit has size coprime to p, since if all sizes were divisible by p then their sum would be as well. Now the action of H on this orbit gives the required conclusion.

Step 3: Conclusion.

Given an arbitrary group, all we have to do is to embed it in a group which has a Sylow p-subgroup.

The first candidate, if you have seen Cayley’s theorem, is a symmetric group. (Cayley’s Theorem asserts that a group of order n is isomorphic to a subgroup of the symmetric group Sn.) It is possible to construct with bare hands a Sylow p-subgroup of Sn. (Write n in base p; the required subgroup is constructed as a direct product of wreath products of copies of the cyclic group of order p.)

But there is an easier way. For any field F, the symmetric group Sn is embeddable in the group GL(n,F) of invertible n×n matrices over F: just take all the permutation matrices, the matrices with entries 0 and 1 having a single 1 in any row or column. Now take F to be the field of integers mod p. A short calculation shows that its order is

(pn−1)(pnp)…(pnpn−1),

and so the exponent of p in its p-part is

0+1+…+(n−1) = n(n−1)/2.

Now consider the group of upper unitriangular matrices (1 on the diagonal, 0 below, and arbitrary entries above). The number in the last display is just the number of arbitrary entries; so the order of this group is the p-part of the order of the general linear group.

We are done.

About Peter Cameron

I count all the things that need to be counted.
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