Prove Free Abelian Group \mathbb{Z}_{p^r}[p] is Isomorphic to \mathbb{Z}_p

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In summary, the conversation discusses proving that \mathbb{Z}_{p^r}[p] is isomorphic to \mathbb{Z}_p for any r \geq 1 and prime p. The main challenge is finding the elements of the subgroup \{x \in \mathbb{Z}_{p^r} | px = 0 \} and showing that there are no other solutions to px = 0 (mod p^r). The conversation also mentions using Sylow's Theorems and the subgroup generated by p^{r-1} to solve the problem.
  • #1
ehrenfest
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Homework Statement


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]

Homework Equations


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?
 
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  • #2
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
StatusX said:
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)?

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]?
 
  • #4
You don't, but now that you have the right number of elements, it suffices to prove there's nothing else.
 
  • #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?
 

FAQ: Prove Free Abelian Group \mathbb{Z}_{p^r}[p] is Isomorphic to \mathbb{Z}_p

1. What is a free Abelian group?

A free Abelian group is a mathematical structure consisting of a set of elements and a binary operation that satisfies the commutative and associative properties. It is "free" in the sense that its elements can be freely combined without any additional constraints or relations.

2. What is an isomorphism?

An isomorphism is a bijective mapping between two mathematical structures that preserves their algebraic operations and structural properties. In other words, if two structures are isomorphic, they are essentially the same in terms of their underlying mathematical properties.

3. What is the group \mathbb{Z}_{p^r}[p]?

The group \mathbb{Z}_{p^r}[p] is a special type of free Abelian group called a cyclic group. It is generated by the element p^r, where p is a prime number and r is a positive integer. This group is often used in number theory and abstract algebra.

4. How do you prove that \mathbb{Z}_{p^r}[p] is isomorphic to \mathbb{Z}_p?

To prove that two groups are isomorphic, we need to show that there exists a bijective homomorphism (a map that preserves the group operation) between the two groups. In this case, we can define the map f: \mathbb{Z}_{p^r}[p] \rightarrow \mathbb{Z}_p as f(p^k) = k (mod p), where k is an integer. This map is bijective and preserves the group operation, thus proving the isomorphism.

5. Why is it important to prove that \mathbb{Z}_{p^r}[p] is isomorphic to \mathbb{Z}_p?

Proving that \mathbb{Z}_{p^r}[p] is isomorphic to \mathbb{Z}_p allows us to use the properties and theorems of \mathbb{Z}_p in the context of \mathbb{Z}_{p^r}[p]. This can help us simplify calculations and solve problems in number theory and abstract algebra. Additionally, understanding the isomorphism between these two groups provides insight into the underlying structure and properties of both groups.

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