Post #19 can be simplified quite a bit.  You know that the number of Sylow p-subgroups, call this n_p, must be congruent to 1 modulo p.  Further, n_p must divide k, where our group is of order p*k with p > k.  But if n_p is not 1, then it must be greater than p, hence greater than k, and therefore n_p can not divide k.

We conclude n_p must be 1.

Post #22 has the typo

But is fine otherwise.

If you want some help, there is a theorem which is quite useful for both Sylow theory and in general.  I've scanned (briefly) through what I could get my hands on in Humphrey, but could not find this theorem.

Such a group

has 1 or 4 Sylow 3-subgroups, each of order 9. Suppose 4 and let

be one of them. Then

and so by Humphreys’s corollary, there is a normal subgroup

of

contained in

such that 4 divides

and

divides 4! = 24.

showing that

is not simple.

Wow, it really is simple when you know how.

Last edited by JaneFairfax (2009-05-16 23:31:42)

And now, how about a bonus extra?

I’ve just seen two proofs of this on the Web. Here is my own adaptation of them. Note that it assumes the knowledge that

is simple.

Suppose there is a simple group

of order 120. The number of Sylow 5-subgroups must be 6. If

is one of them, then

and so there is a homomorphism

with

. (Indeed

is none other than the

in my oft-quoted Humphreys corollary. ) As

is simple,

. Hence

is injective and so there is a simple subgroup

of

of order 120.

As

, we have

and so by simplicity of

either

or

. The former is impossible as it would imply that the number of elements of the form

for

would be

– way above the order of

. Thus

. Then

and so there is a homomorphism

with

. By simplicity of

,

meaning that

is injective and leading to the utter balderdash that

is isomorphic to a subgroup of

.

This proves by contradiction that

cannot be simple.

In fact, this method can also be used to show that groups of order 90 are not simple as well. However, the method I used has the advantage of not assuming that the alternating group of degree 6 is simple.

The fact that

is simple for

is established in several stages. First it is shown that that

is simple. After a couple of other results are proved, the job is completed using mathematical induction. The proof of the simplicity of

is readably done in not more that two pages by Humphreys. Humphreys first demonstrates that

has precisely five conjugacy classes, one of which is the singleton class containing just the identity. Any normal subgroup of

would have to be a union of this trivial conjugacy class with none, some or all of the other conjugacy classes. By determining the sizes of all the conjugacy classes, Humphreys is able to show that the number of elements in any union of the trivial class with some but not all of the other classes cannot be a divisor of 60, thus demonstrating by Lagrange that

cannot have any nontrivial proper normal subgroup.

Last edited by JaneFairfax (2009-05-17 00:36:04)

Yeah, I wasn’t sure if I should really use the simplicity of the alternating groups. I didn’t know what counted as “heavy machinery”. You are obviously accustomed to a lot of heavy-duty stuff in group theory whereas I am only picking up from where I left off years ago, so what may be tough going for me may be feather light for you.

So, now that we have completed this assignment, what’s next?

I have a question for JaneFairfax or anyone who want it to answer, in the #22 in the case that the order of G is 56 why the union of the the Sylow 7-subgroups has 48 non-identity elements? i mean why the intersection is only the indentity? sorry maybe that is very easy to see but im not quite familiar with this definitions..

Thank you! the intersection can't be all of P because in that case the intersection of P and Q must be P, this is, P C Q and that it's a contradiction since P is a Sylow P-subgroup a maximal p-soubgroup.......?????????????