# 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?