Free abelian groups

1. Mar 4, 2008

ehrenfest

1. The problem statement, all variables and given/known data
Show that $\mathbb{Z}_{p^r}[p]$ is isomorphic to $\mathbb{Z}_p$ for any $r \geq 1$ and prime p.

$\mathbb{Z}_{p^r}[p]$ is defined as the subgroup $\{x \in \mathbb{Z}_{p^r} | px = 0 \}$

2. Relevant equations

3. The attempt at a solution

I don't think I should need to use Sylow's Theorems for this since it is in a different section. I can only think of two elements in that subgroup p^{r-1} and 0 and am not really sure how to find the rest or figure out their subalgebra. Actually I guess I just need to prove that the group has p elements and then the only possibility will be Z_p. But how to do that?

Last edited: Mar 5, 2008
2. Mar 5, 2008

StatusX

First of all, they call the set a subgroup, and you should try to see why it is one. Assuming it is, you can't have found all the elements since what you have isn't a subgroup. What is the smallest subgroup containing those elements (also known as the subgroup generated by them)?

3. Mar 5, 2008

ehrenfest

Um--but how do I know that the subgroup generated by p^{r-1} is the same as $\mathbb{Z}_{p^r}[p]$?

4. Mar 5, 2008

StatusX

You don't, but now that you have the right number of elements, it suffices to prove there's nothing else.

5. Mar 5, 2008

ehrenfest

It is really easy to calculate the subgroup generated by p^{r-1}:

It is {0, p^{r-1},2p^{r-1},...,(p-1)p^{r-1}}.

Now I need to show that there are no other solutions to px = 0 (mod p^r). Seems like a number theory problem. That congruence implies x = 0 (mod p^{r-1}) and I just listed all the multiples of p^(r-1) that are less than p^r.
Is that right?