Group Actions, Third Sylow Theorem

Third Sylow Theorem

Let G have order mpⁿ, where p does not divide m. The number of sylow subgroups divides m, and is equal to 1 mod p.

By the second Sylow theorem, the action of conjugation on sylow subgroups is transitive. Let H be any of these sylow subgroups. The size of the orbit is the index of the stabilizer of H in G, and since the action is conjugation, the stabilizer is the normalizer, which is a subgroup somewhere between H and G. The size of the orbit, which is the number of sylow subgroups, equals the index of the normalizer of H in G. Therefore the number of sylow subgroups divides m.

Instead of G, let H act on sylow subgroups by conjugation. The sylow subgroup J is fixed by H iff yJ/y = J for all y in H, iff H is contained in the normalizer of J. Call this normalizer K. Now K contains both sylow subgroups, H and J. By the second sylow theorem, H and J are conjugate in K. Since J is normal in K, it is the unique sylow subgroup in K, and H = J.

If H fixes J then H = J, so H is the only sylow subgroup fixed by H. By the fixed point principle, the number of sylow subgroups is congruent to 1 mod p.

Disjoint Variation

Assume there is an element z in H that can't be in any other sylow subgroup. Perhaps the sylow subgroups are disjoint, or perhaps they share some p cycles, but z has order p². In any case, z seeds a group Z whose order is some power of p, and Z acts on sylow subgroups by conjugation. Since z belongs to H Z fixes H, but suppose Z also fixes J. Now Z belongs to the normalizer K of J. J is a sylow subgroup of K, and by the first sylow theorem, Z belongs to a sylow subgroup of K, which cannot be J. These two subgroups are conjugate, yet J is normal in K. This is a contradiction, hence Z does not fix J after all. Z moves every sylow subgroup (other than H) through an orbit that is a power of p.

Sylow Applications

Here are a couple of ways you can use the sylow theorems. We'll use them a lot when we start classifying finite groups.

A simple group G with |G| = 10000 has a subgroup of order 625, and using the strong caley theorem, a homomorphism embeds G/1 into S₁₆. Yet 10000 does not divide 16!, so there is no simple group of order 10000.

Suppose G is a simple group of order 350, with j subgroups of order 25. By the third sylow theorem, j = 1 mod 5, and j divides 14, hence j = 1. The sylow subgroup is normal, and G is not simple.

The Normalizer of the Normalizer of a Sylow Subgroup

Let J be a sylow subgroup with H as its normalizer. Since all sylow subgroups are conjugate, and J is normal in H, H contains no other sylow subgroups.

Now consider the normalizer of H, the set of elements x such that xH/x = H. If this moved J somewhere else there would be two sylow subgroups in H. Thus xJ/x = J, and we already have x in H. The normalizer of H is H.