Order Problem

  • #1

Main Question or Discussion Point

Hey guys. Ive 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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  • #2
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Hey guys. Ive 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?
 
  • #3
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What about Cauchy's Theorem?
 
  • #4
matt grime
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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:
  • #5
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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