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Help please. proof in algebra

  1. Jul 14, 2011 #1
    Can someone help to understand the under lined part of the proof:
    Let G be a group and let N be a normal subgroup of G of finite index. Suppose that H is a finite subgroup of G and that the order of H is relatively prime to the index of N in G. Prove that H is contained in N

    Proof:
    Let f : G -> G/N be the natural projection. Then f(H) is a subgroup of G/N, so its order must be a divisor of |G/N|. On the other hand, |f(H)| must be a divisor of |H|. Since gcd (|H|,[G:N]) = 1, we must have |f(H)| = 1, which implies that H c ker (f ) = N.
    Thanks in advance
     
  2. jcsd
  3. Jul 14, 2011 #2

    micromass

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    This follows from the isomorphism theorem: for every group G and every homomorphism f, we have an isomorphism

    [tex]f(G)\cong G/Ker(f)[/tex]

    thus

    [tex]|G|=|f(G)||Ker(f)|[/tex]
     
  4. Jul 15, 2011 #3
    Thank you micromass, you are helpful.
     
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