-1
$\begingroup$

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?

$\endgroup$

marked as duplicate by Namaste abstract-algebra Dec 1 '16 at 17:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $G/H$ is a subgroup of $G$ if $H$ is normal. How do you know that $H$ is normal? $\endgroup$ – PiccolMan Dec 1 '16 at 17:42
  • 1
    $\begingroup$ @PiccolMan $G$ is an abelian group, so every subgroup is normal. $\endgroup$ – Stefan4024 Dec 1 '16 at 17:43
  • $\begingroup$ 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 $p^{n}$? $\endgroup$ – PiccolMan Dec 1 '16 at 17:48
  • $\begingroup$ @PiccolMan In a homomorphism if $L \le G$, then $\gamma [L] \le \gamma[G]$. Also the converse holds $L \le \phi[G] \implies \gamma^{-1} [L] \le G$. $\endgroup$ – Stefan4024 Dec 1 '16 at 17:59
  • $\begingroup$ @PiccolMan Also you need to prove that $G$ has a subgroup of order $p,p^2,...,p^n$, not just $p^n$ $\endgroup$ – Stefan4024 Dec 1 '16 at 18:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.