|G|=p^k. Prove G has an element of order p.

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Discussion Overview

The discussion revolves around proving that a group G, with order |G|=p^k where p is a prime and k is a positive integer, contains an element of order p. The scope includes mathematical reasoning and exploration of group theory concepts.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes that for an element x in G, the expression x^{\frac{m}{p}} can be considered, where m is the order of x.
  • Another participant expresses confusion regarding the implications of m/p being smaller than m, suggesting it leads to a contradiction since x^m equals the identity element.
  • A later reply questions the meaning of (x^m)^(1/p), seeking clarification on the notation.
  • Another participant suggests that (x^m)^(1/p) refers to the pth root of x^m, but expresses uncertainty about the direction of the argument.
  • One participant asserts that the expression x^{\frac{m}{p}} is valid, noting that m divides p^k, yet does not clarify the implications of this statement.

Areas of Agreement / Disagreement

Participants express confusion and disagreement regarding the interpretation of the mathematical expressions and their implications, indicating that the discussion remains unresolved.

Contextual Notes

There are limitations in the clarity of the mathematical expressions used, particularly regarding the interpretation of fractional powers and their implications in the context of group theory.

mathmajor2013
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Let p be a prime, k a pos. int., and G a group with |G|=p^k. Prove that G has an element of order p.
 
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Consider [tex]x^{\frac{m}{p}}[/tex] for some element x not equal to the identity, where m is the order of x.
 
This kind of seems like a contradiction because m/p is smaller than m, yet x^m/p=(x^m)^1/p=e^1/p=e.
 
mathmajor2013 said:
This kind of seems like a contradiction because m/p is smaller than m, yet x^m/p=(x^m)^1/p=e^1/p=e.

What does (x^m)^(1/p) mean?
 
The pth root of x^m? I'm sorry I cannot see where this one is going
 
mathmajor2013 said:
The pth root of x^m? I'm sorry I cannot see where this one is going

You have written it yourself, but it doesn't make any sense. [tex]x^{\frac{m}{p}}[/tex] makes perfect sense, since m divides p^k.
 

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