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Free abelian groups

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]\mathbb{Z}_{p^r}[p][/itex] is isomorphic to [itex]\mathbb{Z}_p[/itex] for any [itex]r \geq 1[/itex] and prime p.

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

    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. jcsd
  3. Mar 5, 2008 #2


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    Homework Helper

    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)?
  4. Mar 5, 2008 #3
    Um--but how do I know that the subgroup generated by p^{r-1} is the same as [itex]\mathbb{Z}_{p^r}[p][/itex]?
  5. Mar 5, 2008 #4


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    You don't, but now that you have the right number of elements, it suffices to prove there's nothing else.
  6. Mar 5, 2008 #5
    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?
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