Originally Posted by

**crushingyen** If $\displaystyle G$ is a finite abelian group and $\displaystyle p$ is a prime such that $\displaystyle p^n$ divides $\displaystyle |G|$, then prove that $\displaystyle G$ has a subgroup of order $\displaystyle p^n$.

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

Attempt at the proof:

If $\displaystyle p^n|G$, then $\displaystyle G = p^nm, m \in \mathbb{Z}, (p,m) = 1$.

This well may be false: you're only given $\displaystyle p^n\mid |G|$ , not that n is the maximal power of p dividing the order of G.

I'd rather go: let $\displaystyle p^m\mid |G|\,\,\,s.t.\,\,\,p^{m+1}\nmid |G|\Longrightarrow$ by Sylow theorems there exists $\displaystyle H\leq G\,\,\,s.t.\,\,|H|=p^m$ . Now, it's easy to prove that any p-group of

order $\displaystyle p^m$ has a normal sbgp. of order $\displaystyle p^k$ for any $\displaystyle 0\leq k\leq m-1$ (by induction, say) . In the present case normality is for free since G is abelian, and still

we're done.

Tonio

Consider a group $\displaystyle H$.

Claim: $\displaystyle H \subset G$ is a group such that $\displaystyle |H| = p^n$.

I know that I have to use Lagrange's Theorem to show that $\displaystyle [G:H] = m$. I just don't know how to do this. Can anyone help?