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Abstract-H is normal to G. Order of G/H is m, g^m exists in H?

  1. Nov 17, 2009 #1
    Let H be a normal subgroup of a finite group G. The order of G/H (quotient/factor group) is m. Show g^m is in H for all g in G.

    Lagrange's Thm says that o(G) = o(H) * o(G/H)
    xH = Hx for all x in G, since H is a normal subgroup of G

    Ok, I've got a lot of statements that i believe to be true, how relevant they are is what i can't decide.

    - let o(G) = k, o(H) = j, o(G/H) = m
    so k = mj by Lagrange

    - the quotient group is broken up into m disjoint partitions of G
    - each partition has j elements
    - I know that the o(g) | o(G)
  2. jcsd
  3. Nov 17, 2009 #2


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    If the order of G/H is m, then what is (g+H)m?
  4. Nov 17, 2009 #3
    If + is the Binary op, then (aH)^m should be equal to (aH)(aH)... m times and since H is normal in G aH = Ha so...

    (aH)^m = a^mH?

    so a^m is in G
  5. Nov 17, 2009 #4
    if a^m was to equal the identity in G, then a^mH = H would imply that a^m is an element of H because hH = H = Hh, but i don't see how a^m could equal the identity in G
    Last edited: Nov 17, 2009
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