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Homework Help: Lagrange theorem and subgroup help

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Let G be group, H<G , K<G, if gcd(lHl,lKl)=1, prove that H[tex]\bigcap[/tex]K={1}

    2. Relevant equations



    3. The attempt at a solution

    so Lagrange theorem says that lHl l lGl, lKl l lGl,
    and of course 1 is inside both H and K, but how when they are coprime, the element are all different except their identity? i cannot see T_T
     
  2. jcsd
  3. Aug 18, 2010 #2

    Office_Shredder

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    Re: SubGroup

    What must be true about the order of an element of a group
     
  4. Aug 19, 2010 #3
    Re: SubGroup

    i don't know much about, order of an element, but i noticed the corollary from lagrange theorm, state that order on an element divides order of the group, provided that it is a finite group,

    i use oG for order of G for simplicity.

    so, gcd(oH,oK)=1 , let h in H, k in K,

    so oh l oH and ok l oK, would imply gcd(oh,ok)=1

    so their orders are coprime, still cannot see, why their element are different, T_T,

    is, cyclic group involved? i still didn't go through cyclic yet
     
  5. Aug 19, 2010 #4

    Office_Shredder

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    Re: SubGroup

    Well, suppose we have an element g which is in both H and K.

    What is the order of g? You almost calculated it in your post
     
  6. Aug 19, 2010 #5
    Re: SubGroup

    yeaaa, i can see now, the order is 1, therefore 1 is the element in both H and K, correct?
     
  7. Aug 19, 2010 #6
    Re: SubGroup

    No, 1 is in H and K because H and K are subgroups of G.

    Are you sure you can assume finiteness here?

    Here is a different way to approach. Pick an arbitrary element,k, in K. Now state something about the order of k. Now what happens if you assume k is in H? What is true about all the elements of H?
     
  8. Aug 19, 2010 #7
    Re: SubGroup

    i mean, 1 is the "only" element in H and K, if it is finite...

    hmmmm, so maybe can i divide it into 2 cases? but i do no know what to do if they are infinite, cause i can't use lagrange.

    anyway,

    let order of k is n, so kn=1, Suppose k also in H, then kn also in H, is that what you mean? T_T and i don't know how to continued
     
  9. Aug 19, 2010 #8
    Re: SubGroup

    You can get there!!!

    Yup, okay, we have k^n = 1. Now what can you tell me, using Lagrange about n and |K|?

    Write that down. Now if you assume k is also in H, what does that imply?
     
  10. Aug 19, 2010 #9
    Re: SubGroup

    yes, but i thought i can only use lagrange for finite group,

    and if it is finite, and if k also in H k^n also in H, so n divides both l K l and l H l, will imply gcd(n,n)=1, so n must equal to 1, so order of k is 1, will imply k=1, so 1 is the only element in K and H,
    but i did it in the last post, T_T what if it is infinite? i cannot use Lagrange right?
     
  11. Aug 19, 2010 #10
    Re: SubGroup

    Yeah, I was thinking about that...
    Does taking the gcd of a group with infinite order make sense?
     
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