Show G(p) of Order p^t when G is an Abelian Group of Order (p^t)m

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Discussion Overview

The discussion revolves around demonstrating that if G is an abelian group of order (p^t)m, where (p,m)=1, then the subgroup G(p) has order p^t. Participants explore various approaches, theorems, and lemmas related to group theory, particularly focusing on the implications of Cauchy's theorem and properties of abelian groups.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that G(p) consists of elements of G whose order is a power of p, and proposes to show that G(p) has order p^t.
  • Another participant questions what tools are available to tackle the problem and hints at the possibility of a straightforward solution with sufficient background knowledge.
  • A participant references a lemma regarding finite order elements in abelian groups and considers using it to demonstrate properties of G(p), while expressing uncertainty about the implications of (p,m)=1.
  • One participant proposes using Cauchy's theorem for abelian groups to argue that G contains elements of order p, leading to the conclusion that G(p) must contain elements of order p^t.
  • Another participant questions whether Cauchy's theorem implies that if p^m divides |G|, then G contains an element of order p^m, and discusses the implications of G(p) being a subgroup of G.
  • A suggestion is made to prove the existence of an element of order p, particularly when t>0, as a potential step in the argument.

Areas of Agreement / Disagreement

Participants express various viewpoints and approaches, with no clear consensus reached on the method to demonstrate the order of G(p). The discussion remains unresolved, with multiple competing ideas and interpretations of theorems presented.

Contextual Notes

Participants reference the need for specific assumptions regarding the orders of elements and the relationships between the orders of groups and their subgroups, but these assumptions are not fully articulated or resolved.

Thorn
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If G is an abelian group of order (p^t)m, and (p,m)=1, show that G(p) has order p^t

and G(p) = {a e G| |a|=p^m where m is a natural number}

any suggestions?
 
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What do you have to work with? This problem is trivial if you've developed enough tools.

You can also do this with your "bare hands" if you want to. What have you tried so far?
 
Last edited:
Geez...I have no idea where to start...

Ok, I was looking at a lemma...

"Let G be an abelian group and a [tex]\in[/tex] G and is an element of finite order. Then a = a1+a2+...+ak, with ai an element of G(pi), where p1,...,pk are the distinct positive primes that divide the order of a"

... now G(p) is only the set of all elements whose order is some power of p... I was thinking you could use that lemma to show that G(p) has power pt ...seems like that could help somehow.. but it seems I need to use the fact that (p,m)=1 ...this surely means that (pt,m)=1 which means you can write 1=pt*u + m*v...for some numbers u,v... and I have no idea where to go...
 
Well...Actually, I could use Cauchy's theorem for Abelian Groups...

"If G is a finite abelian group and p is a prime that divides |G|, prove that G contains an element of order p...

we know that p divides |G|...and p2 divides the order of G...and so on and so on until we have pt divides the order of G...so all these elements exist in G(p)..then use the (p,m)=1 to show that each one of these elements is uniquely the product of m with some interger...I think that is it...
 
Does Cauchy's theorem imply that if p^m divides |G|, then G contains an element of order p^m?

The fact that G is abelian is vital here. First note that G(p) is a subgroup of G. Next note that G(p) has order a power of p, by Cauchy. If |G(p)| isn't p^t, then we can consider the group G/G(p). G/G(p) is still abelian, and p divides |G/G(p)|. Apply Cauchy's theorem again to get an element g+G(p) of order p. Now what?
 
you might start by trying to prove there is an element of order p, if t>0.
 

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