Originally Posted by

**Boysilver** I see, thanks. I've had a look at the Sylow theorems and they will solve my problem. From what I gather, my original question seems to be a particular case of a standard result. Just to check though, is there a quick and simple proof to show that when $\displaystyle G$ is a group of order $\displaystyle 2^m$, then there exist subgroups

$\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n $

such that $\displaystyle |G_j| = 2^j $? Or really is the use of the Sylow theorems about the simplest argument we can give? Just because if there is indeed something simple that I could use instead, I'd much rather do it that way. I should have said: we can also use the fact that the centre of any group of order$\displaystyle p^n$ (p prime) is not trivial, and we can use Cauchy's theorem (if this helps, but I get the feeling they only help to prove the Sylow theorems anyway...)