# Finite Abelian Groups

In summary, an abelian group of order pn (where p is a prime) with p-1 elements of order p can be proven to be cyclic by ruling out certain possibilities for its structure and using the theorem for classifying finite abelian groups. By considering the number of factors of p in an element of order p in Zp^x, it can be shown that there are exactly p-1 elements of order p in the group, leading to the conclusion that it is cyclic.

## Homework Statement

An abelian group has order pn (where p is a prime) and contains p-1 elements of order p. Prove that this group is cyclic.

## The Attempt at a Solution

I know I should use the theorem for classifying finite abelian groups, which I understand, and I feel like I have all the pieces but I don't know how to put them together.

Ok, so what are some possibilities for the structure of an abelian group of order p^n? Can you rule some of them out based on the order p condition?

The only case I've managed to definitively rule out is the case where in
Zm1 x Zm2 x ... x Zmk we set k=n and m1=m2=...=p. In each of Zm1 through Zmk there are p-1 elements of order p so taking the direct product of all these you certainly end up with more than p-1 elements of order p.
In the case where any of m1,..., mk are equal to some p^x where x>1 and k>1 then in some Zmi there will be p^(x-1) - 1 elements with order not equal to p. I'm not sure that is entirely correct but if so then taking direct product we will end up with more than p-1 elements of order p and we'll conclude that the group has to be cyclic.

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You've got the right idea but I'm not sure you are counting correctly. How many elements of order p in Zp^x??

Well, aren't the elements of order p the ones that are not divisible by p? That was what I was trying to count before. Although looking it over again I would say that there are px-px-1 elements of order p.

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An element a of order p in Zp^x satisfies the conguence a*p=p^x. How many factors of p does a have to have?

Oh okay. So there are x-1 factors of p in a. So I guess this gives you that there are p-1 elements of order p since you can only multiply p^(x-1) by numbers less than p until you get p^x and the rest of the argument would be similar to the first one for setting all of the m's equal to p.

Thanks.