1. ## Simple Groups

$|G|=60=2^2\times 3\times 5$

$n_i$ is the number of $Syl_p(G)$ for $i=2,3,5$

$n_2\equiv 1 \ (\text{mod} \ 2) \ \ \text{and} \ \ n_2|15$

$n_2=1,3,4,15$

$n_3\equiv 1 \ (\text{mod} \ 3) \ \ \text{and} \ \ n_3|20$

$n_3=1,4,10$

$n_5\equiv 1 \ (\text{mod} \ 5) \ \ \text{and} \ \ n_5|12$

$n_5=1,6$

If $n_2=15$, then $n_3 \ \text{or} \ n_5$ is normal in G so G is not simple.

But if I keep going, I will get cases of G being simple. Is it correct to have times when G can be simple and not?

If not, how should I do this?

2. ## Re: Simple Groups

Originally Posted by dwsmith
$|G|=60=2^2\times 3\times 5$

$n_i$ is the number of $Syl_p(G)$ for $i=2,3,5$

$n_2\equiv 1 \ (\text{mod} \ 2) \ \ \text{and} \ \ n_2|15$

$n_2=1,3,4,15$

$n_3\equiv 1 \ (\text{mod} \ 3) \ \ \text{and} \ \ n_3|20$

$n_3=1,4,10$

$n_5\equiv 1 \ (\text{mod} \ 5) \ \ \text{and} \ \ n_5|12$

$n_5=1,6$

If $n_2=15$, then $n_3 \ \text{or} \ n_5$ is normal in G so G is not simple.

But if I keep going, I will get cases of G being simple. Is it correct to have times when G can be simple and not?

If not, how should I do this?

What precisely are you trying to do? Figure out the Sylow subgroup structure of $G$? What tools do you have at your disposal?

If you assume that $G$ is simple then there are simple ways to knock a lot of these out :for example, $n_i>4$ since otherwise $n_i=1$ which contradicts simplicity or there is an embedding (by the simplicity of $G$) $G\to S_k$ with $k<4$ (induced by the action of $G$ on the cosets) which would imply that $60\mid k!$, etc. That said, it is known that the only simple group of order $60$ is $A_5$ so you can read the Sylow subgroups from there.

3. ## Re: Simple Groups

for some orders of finite groups, you can make a lot of arguments from the order of the group alone. but, the order of a group does not usually uniquely determine its structure (an exception is groups of prime order).

4. ## Re: Simple Groups

Originally Posted by Deveno
for some orders of finite groups, you can make a lot of arguments from the order of the group alone. but, the order of a group does not usually uniquely determine its structure (an exception is groups of prime order).
In fact, it's not hard to prove that the only groups whose structure is determined by their order are square free integers where none of the prime factors are equivalent to 1 modulo the other.

5. ## Re: Simple Groups

However, order often has a lot to say about the global structure of your group. For example, if $G$ has odd order then it is soluble (this is a massive theorem, but has obvious implications - if $G$ is simple and non-cyclic then it is of even order). This is why a lot of research is done into involutions - elements of order 2 - because simple groups must contain them.

Also, any group of order $p^aq^b$ where $p, q$ are primes is soluble.

Another very well-known result is that groups of prime-power order are nilpotent, etc. So a lot is known about their global structure. However, there is a paper of O'Brien and Eick called "Enumerating $p$-groups" where they show that there are some 10,494,213 groups of order $2^9$...

And finally, as an infinite group theorist, I would like to point out that every group of finite order is virtually trivial.

6. ## Re: Simple Groups

Originally Posted by Drexel28
In fact, it's not hard to prove that the only groups whose structure is determined by their order are square free integers where none of the prime factors are equivalent to 1 modulo the other.
i like the way you say "not hard". i got most of the way through proving this for myself, and then i ran into a snag: i needed burnside's theorem on normal complements (basically to show that a group of square-free order is solvable). and every proof i read of that, used the transfer homomorphism, and that's...ugly (it's a bit nicer because the sylow subgroups of a group of square-free order are all cyclic, but still...).

basically, i got this far:

use induction, minimal counter-example, number of primes > 2 (easy part).

by induction hypothesis, if we can decompose G into a semi-direct product, boom! we're done, because the only semi-direct product possible is trivial (i.e., direct).

but...to get a normal subgroup of G, i need to construct factor groups using the commutator subgroup, and the case G = G' needs to be ruled out (which i can if G is solvable, hence burnside's complement theorem).

i know a simpler proof if i can prove this lemma: if none of the sylow subgroups of G are normal, then the number of elements of prime order > |G|.

i spent a long-time reading NonCommAlg's blog (and some of yours, too). his proof of what you say here is 1/2 a page....with 5 or 6 pages of referenced material (maybe more, if you back-track all the links).

the thing is, in this case, it will turn out (in this case of square-free orders where gcd(n,φ(n) = 1) that EVERY sylow subgroup is normal, because G is actually cyclic. and using burnside (or character theory, which amounts to the same thing, i think), seems like mashing potatoes with a steamroller. i'd like to see a more "elementary proof" that is a bit more direct. i feel sure there is one, just haven't come across it.

7. ## Re: Simple Groups

Originally Posted by Deveno
i like the way you say "not hard". i got most of the way through proving this for myself, and then i ran into a snag: i needed burnside's theorem on normal complements (basically to show that a group of square-free order is solvable). and every proof i read of that, used the transfer homomorphism, and that's...ugly (it's a bit nicer because the sylow subgroups of a group of square-free order are all cyclic, but still...).

basically, i got this far:

use induction, minimal counter-example, number of primes > 2 (easy part).

by induction hypothesis, if we can decompose G into a semi-direct product, boom! we're done, because the only semi-direct product possible is trivial (i.e., direct).

but...to get a normal subgroup of G, i need to construct factor groups using the commutator subgroup, and the case G = G' needs to be ruled out (which i can if G is solvable, hence burnside's complement theorem).

i know a simpler proof if i can prove this lemma: if none of the sylow subgroups of G are normal, then the number of elements of prime order > |G|.

i spent a long-time reading NonCommAlg's blog (and some of yours, too). his proof of what you say here is 1/2 a page....with 5 or 6 pages of referenced material (maybe more, if you back-track all the links).

the thing is, in this case, it will turn out (in this case of square-free orders where gcd(n,φ(n) = 1) that EVERY sylow subgroup is normal, because G is actually cyclic. and using burnside (or character theory, which amounts to the same thing, i think), seems like mashing potatoes with a steamroller. i'd like to see a more "elementary proof" that is a bit more direct. i feel sure there is one, just haven't come across it.
I don't have quite the time to fully reply, but I can at least give some advice how to prove this in a more elementary way (i.e. no hom. alg. or rep theory). I first came across this result in Dummit and Foote where one is prompted to prove it in a series of exercises. I did this here.

As a remark, while I'm absolutely positive your route of attack is valid, Burnside's complement theorem is an elephant gun--I don't quite think it's needed here.