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.