Sylow Theorems

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  • Topic: Prime number, P-group, Simple group
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The Sylow theoremsThe Sylow Theorems
Here is my version of the proof of the Sylow theorems. It is the result of taking the proof in Gallian and trying to make it as digestible as possible. In particular, I tried to break the long proof into bite-sized pieces. The main goal here is to convey an overview of how the ingredients fit together, so I'll skip lightly over some of the details.

The prerequisites are basically all of the group theory that came before the Sylow theorems in this course, including: Lagrange's theorem, the first and second isomorphism theorems, and the orbit-stabilizer theorem. I'll also use Cauchy's theorem, even though the book lists it as a corollary to the Sylow theorems (more on that later). I'll assume you know the definition of a Sylow-subgroup and all the terms in the statements of the Sylow theorems. From now on, G is a finite group and p is a prime number that divides the order of G. Recall that a p-subgroup of G is a subgroup of G with order equal to a power of p.

Definition. A maximal p-subgroup of G is a p-subgroup of G that is not contained in any larger p-subgroup of G.
This is only a temporary definition, since it will turn out that a "maximal p-subgroup" is just the same thing as a "Sylow p-subgroup". However it would be circular logic to assume this. Hypothetically, you could imagine a group G of order 100, and a subgroup H of order 2 such that H is not contained in any subgroup of G that has order 4, (or indeed where there is no subgroup of G that has order 4). If there were such an H, it would be maximal but not Sylow. Lemma 1. If H is a maximal p-subgroup of G then the number of conjugates of H is equal to 1 modulo p.

Proof. Let C be the collection of all conjugates of H. Suppose, seeking a contradiction, that the number of elements of C is not equal to 1 modulo p. Let H act on C by conjugation. The size of every orbit must divide the order of H, and hence must be a power of p. Clearly {H} is...
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