Proving Cyclic Finite Abelian Groups of Order pn

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Homework Help Overview

The problem involves proving that a finite abelian group of order \( p^n \) (where \( p \) is a prime) and containing \( p-1 \) elements of order \( p \) is cyclic. The discussion revolves around the structure of such groups and the implications of the given conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the classification of finite abelian groups and question the implications of the order \( p \) condition on possible group structures. They discuss ruling out certain configurations based on the number of elements of order \( p \) and consider the counting of these elements in various direct product forms.

Discussion Status

The discussion is active, with participants offering insights and questioning each other's reasoning. Some guidance has been provided regarding the counting of elements of order \( p \) in specific group structures, but there is no explicit consensus on the final argument or conclusion yet.

Contextual Notes

Participants are navigating the complexities of group theory, particularly focusing on the implications of the order of elements and the structure of abelian groups. There is an emphasis on ensuring the correctness of counting elements of specific orders, which is crucial for the proof.

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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.


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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.
 
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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.
 

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