Groups whose order is a power of a prime

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Homework Help Overview

The discussion revolves around whether every group whose order is a power of a prime \( p \) contains an element of order \( p \). Participants explore concepts related to group theory, particularly focusing on the implications of group order and relevant theorems.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the potential applicability of Cauchy's theorem and Lagrange's theorem in addressing the problem. Some express uncertainty about their reasoning and seek clarification on the implications of group order.

Discussion Status

Several participants have shared their thoughts and attempted to reason through the problem. There is acknowledgment of Cauchy's theorem as a relevant concept, with some participants indicating they have made progress in their understanding. However, no consensus has been reached regarding the question posed.

Contextual Notes

Some participants mention initial confusion regarding the order of the group and its implications, indicating a need for clarity on definitions and theorems related to group order.

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Homework Statement



Does every group whose order is a power of a prime p contain an element of order p?

Homework Equations





The Attempt at a Solution


I know it certainly can contain an element of order p. I also feel that
|G|=|H|[G:H] might be useful. Any help is appreciated!
 
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Hmm, cool problem. I think the answer is yes, but I haven't thought it through completely, and more importantly I want to know your ideas. Have you tried using any of the corollaries of Lagrange's theorem that deal with the order (of a group or of an element in a group).
 
It's called Cauchy's theorem.
 
Cauchy's theorem is overkill for this.

You can pick an arbitrary element of the group and construct an element of order p from it
 
Dick said:
It's called Cauchy's theorem.

True. But we haven't got to the part about the center of a group yet.
 
Office_Shredder said:
Cauchy's theorem is overkill for this.

You can pick an arbitrary element of the group and construct an element of order p from it

Yes I figured it out! That's what I did! Thanks!
 
snipez90 said:
Hmm, cool problem. I think the answer is yes, but I haven't thought it through completely, and more importantly I want to know your ideas. Have you tried using any of the corollaries of Lagrange's theorem that deal with the order (of a group or of an element in a group).

I figured it out. I just picked an element g of the group and proved that there is an element in <g> that has order p.
 
Okay yeah I was slightly confused. At first I kept thinking that G had an order that was a multiple of p, and the tools I had did not solve it completely. When I realized that you actually stated ord(G) = p^n, I could do it (namely by using a^ord(G) = e and the fact that if a^m = e then ord(a)|m).

Then Dick mentioned Cauchy's Theorem, so I briefly glanced at a proof on wikipedia and noticed that the tools used were more advanced. Finally I realized that Cauchy's Theorem is what I originally had in mind. So I guess the only thing left to do now is learn Cauchy's Theorem.
 

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