Groups whose order is a power of a prime

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Every group whose order is a power of a prime p contains an element of order p, as established by Cauchy's theorem. Participants in the discussion explored the implications of Lagrange's theorem and the construction of elements of specific orders from arbitrary group elements. Initial confusion about the order of the group was clarified when it was recognized that the order is p^n, allowing for the application of relevant group theory principles. The conversation highlighted the importance of understanding Cauchy's theorem and its role in proving the existence of elements of order p. Overall, the conclusion affirms that such groups indeed contain elements of the specified 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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