Prove that G has a subgroup of order p^n

**crushingyen** · February 23rd 2010, 09:53 PM

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

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

Attempt at the proof:

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

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

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

**tonio** · February 24th 2010, 01:33 AM

Originally Posted by crushingyen

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

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

Attempt at the proof:

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

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

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

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

we're done.

Tonio

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

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

.

**crushingyen** · February 24th 2010, 01:36 AM

Aw man, that would've been perfect. I forgot to say that you're not allowed to use the Sylow Theorem to answer this.

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