If

is a finite abelian group and

is a prime such that

divides

, then prove that

has a subgroup of order

.

Note: This proof is very similar to the First Sylow Theorem, but in this one, G is abelian.

Attempt at the proof:

If

, then

.

This well may be false: you're only given , not that n is the maximal power of p dividing the order of G. I'd rather go: let by Sylow theorems there exists . Now, it's easy to prove that any p-group of order has a normal sbgp. of order for any (by induction, say) . In the present case normality is for free since G is abelian, and still we're done. Tonio
Consider a group

.

Claim:

is a group such that

.

I know that I have to use Lagrange's Theorem to show that

. I just don't know how to do this. Can anyone help?