Proving o(G) Divisible by p in Abelian Groups

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Discussion Overview

The discussion revolves around proving a property related to the order of elements in abelian groups, specifically addressing the condition that if a prime p divides the order of an abelian group G, then a certain subset G(p) can be defined. Participants explore the implications of this condition and related theorems.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant seeks help to prove that if p divides o(G), then G(p) is the set of elements in G whose order is a power of p.
  • Another participant questions the definition of G(p), suggesting it may refer to a Sylow p-subgroup.
  • A third participant mentions Cauchy's Theorem as potentially relevant to the discussion.
  • A later reply emphasizes the need for a clear definition of G(p) to provide assistance, indicating that the problem may be more complex than initially presented.
  • One participant later claims to have solved the problem, attributing their success to the fundamental theorem of arithmetic and the uniqueness of prime factorization.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the definition of G(p) initially, leading to some confusion. However, one participant eventually claims to have solved the problem, indicating a resolution for themselves but not necessarily for the group.

Contextual Notes

The discussion includes assumptions about the definitions and properties of groups and their elements, which are not fully clarified. The reliance on the fundamental theorem of arithmetic suggests that the proof may hinge on specific mathematical properties that are not universally agreed upon in the thread.

Who May Find This Useful

Readers interested in group theory, particularly those studying properties of abelian groups and the implications of prime divisibility on group structure.

Singularity
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Hey guys. I've been stuck on the following thing for a little while now. Some help would be appreciated.

If p divides o(G) (G an abelian group and p a prime), then show that
G(p) = {g from G | o(g) = p^k for some k }

I keep going round in circles.

P.S. - this is not a homework question, just something I saw in an abstract algebra book that they stated without proof.
 
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Singularity said:
Hey guys. I've been stuck on the following thing for a little while now. Some help would be appreciated.

If p divides o(G) (G an abelian group and p a prime), then show that
G(p) = {g from G | o(g) = p^k for some k }

I keep going round in circles.

P.S. - this is not a homework question, just something I saw in an abstract algebra book that they stated without proof.

What is G(p)? Are you sure this wasn't a definition?
 
What about Cauchy's Theorem?
 
Singularity said:
If p divides o(G) (G an abelian group and p a prime), then show that
G(p) = {g from G | o(g) = p^k for some k }

If you don't define G(p) for us we can't help. (The sylow p subgroup perhaps? but that is an easy exercise - all elements of order p^k lie in some sylow subgroup, and there is only one in an abelian group - so it must be something more difficult than that.)
 
Last edited:
Did it:)

Thanks to all who posted. My apologies for not defining G(p) properly. In any event, I solved the problem. The fundamental theorem of arithmetic did the trick (I underestimated the strength of the uniqueness of prime factorization in my earlier attempts!). Ciao
 

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