is the number of for
If , then 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 ? What tools do you have at your disposal?
If you assume that is simple then there are simple ways to knock a lot of these out :for example, since otherwise which contradicts simplicity or there is an embedding (by the simplicity of ) with (induced by the action of on the cosets) which would imply that , etc. That said, it is known that the only simple group of order is so you can read the Sylow subgroups from there.
However, order often has a lot to say about the global structure of your group. For example, if has odd order then it is soluble (this is a massive theorem, but has obvious implications - if 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 where 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 -groups" where they show that there are some 10,494,213 groups of order ...
And finally, as an infinite group theorist, I would like to point out that every group of finite order is virtually trivial.
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.