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Show H is a subgroup of N

  1. Sep 17, 2011 #1
    Let G be a finite group, let H be a subgroup of G and let N be a normal subgroup of G. Show that if |H| and |G:N| are relatively prime then H is a subgroup of N.

    I have tried using the fact that since N is normal, HN is a subgroup of G.
    Suposing that H is not contained in N, I tried finding a common factor for |H| and |G:N|.
    Numbers that divide |H| are |HnN|, |H:HnN| and |H|.
    Numbers that divide |G:N| are |G:HN| and |HN:N|.
    I'm stuck.

    I would appreciate any suggestions.
    Thanks
     
  2. jcsd
  3. Sep 17, 2011 #2

    micromass

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    Take a [itex]h\in H[/itex]. You want to show that [itex]h\in N[/itex]. Now, what can you say about [itex]h+N\in G/N[/itex]?? What is its order??
     
  4. Sep 17, 2011 #3
    The order of hN in G/N is the smallest integer k such that h^k = n for some n in N.
    We know that k divides |G/N|. I don't know if k divides |H|. We know that h^k is in the intersection of H and N.
     
  5. Sep 17, 2011 #4

    micromass

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    What I meant was: let k be the order of h. Then we know that [itex]h^k=e[/itex]. And also [itex](hN)^k=N[/itex]. So the order of hN divides k. What can you conclude?
     
  6. Sep 17, 2011 #5
    Thank you micromass,
    I got it.
     
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