$G$ is a finite abelian group. $|G| = p^n$ where $p$ is a prime. Show that $G$ has a subgroup of order $p^r, r \leq n$ without Sylow Theorems.

Question

I know that for $G$ to have a subgroup of order $p^r$ it must have an element of order $p^r$.

My approach to this problem was to use the Fundamental Theorem of finite abelian groups. i.e there are, up to isomorphism, only the following abelian groups of order $p^n$:

$\mathbb Z_{p^{a_1}} \times \mathbb Z_{p^{a_2}} \times \mathbb Z_{p^{a_3}} \times ... \times \mathbb Z_{p^{a_k}}$, where $a_i \in \mathbb N$ and $a_1 + a_2 + a_3 + ... + a_k = n$

Then I was going to show that each group of this form has an element of order $p^r$. I started off by using Cauchy's theorem to show that each group of the form $\mathbb Z_{p^{m}}$ where $m \in \mathbb N, m \ne 0$ has an element of order $p$.

However, I am stuck now an do not know how to show that each group of the form $\mathbb Z_{p^{a_1}} \times \mathbb Z_{p^{a_2}} \times \mathbb Z_{p^{a_3}} \times ... \times \mathbb Z_{p^{a_k}}$ has an element of order $p^r$, where r is every integer less than $n$.

Does anyone know how to prove this without using Sylow theorems?

Please see Showing that a finite abelian group has a subgroup of order $m$ for each divisor $m$ of $n$. So we have subgroups of G of order $1, p, p^2, \ldots, p^n$. — amWhy, Dec 01 '16 at 17:52

Stefan4024 · Accepted Answer · 2016-12-01T17:28:12.057

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HINT Use induction on $n$, combined with the Cauchy Theorem and the canonical homomorphism.

Obviously the statement is true for any $n=1$. Now assume that it's true for all $k<n$. We'll prove it holds for $n$ too. By Cauchy Theorem the group $G$ of order $p^n$ has a subgroup $H$ of order $p$. Now $G/H$ is finite abelian subgroup of order $p^{n-1}$. Consider the canonical homomorphism $\gamma: G \to G/H$. As by induction hypothesis $G/H$ has subgroups of order $p^t, 1\le t \le n-1$, we have that for each such subgroup $L$, that $|\gamma^{-1}[L]| = p|L|$. Obviously $\gamma^{-1}[L]$ is a subgroup in $G$ and therefore we have a subgroups of order $p^t, 2\le t \le n$ in $G$. But we already know that there is a subgroup of order $p$ in $G$, hence the proof.

edited Dec 01 '16 at 17:28

answered Dec 01 '16 at 17:20

Stefan4024

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$G/H$ is a subgroup of $G$ if $H$ is normal. How do you know that $H$ is normal? – PiccolMan Dec 01 '16 at 17:42
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@PiccolMan $G$ is an abelian group, so every subgroup is normal. – Stefan4024 Dec 01 '16 at 17:43
Also, I don't understand why $\gamma^{-1}[L]$ is a subgroup of $G$. Could I not just say that $|G| = p^{n}$, hence, $G$ is a subgroup of itself of order $p^{n}$? – PiccolMan Dec 01 '16 at 17:48
@PiccolMan In a homomorphism if $L \le G$, then $\gamma [L] \le \gamma[G]$. Also the converse holds $L \le \phi[G] \implies \gamma^{-1} [L] \le G$. – Stefan4024 Dec 01 '16 at 17:59
@PiccolMan Also you need to prove that $G$ has a subgroup of order $p,p^2,...,p^n$, not just $p^n$ – Stefan4024 Dec 01 '16 at 18:03
I know, but didn't you prove that with induction? – PiccolMan Dec 01 '16 at 19:30
@PiccolMan I did, but I was referring to "Could I not just say that |G|=p^n, hence, G is a subgroup of itself of order p^n?" – Stefan4024 Dec 01 '16 at 19:42
Let us continue this discussion in chat. – PiccolMan Dec 01 '16 at 19:47

$G$ is a finite abelian group. $|G| = p^n$ where $p$ is a prime. Show that $G$ has a subgroup of order $p^r, r \leq n$ without Sylow Theorems.

1 Answers1