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Finite Abelian Groups

  1. Jul 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Let G be a finite group with N , normal subgroup of G, and a, an element in G.
    Prove that if (a) intersect N = (e), then o(An) = o(a).


    2. Relevant equations



    3. The attempt at a solution
    (aN)^(o(a)) = a^(o(a)) * N = eN = N, but is the least power such that (aN)^m = N. Assume m must divide o(a).

    (aN)^((o(a)) = (aN)^ (mq +r) where 0 <= r < m,
    However, ((aN)^m)-q * a(N)^(o(a)) = (a(N)^r)= N = (a(N)^r).
    r < m so r= 0 and mq= o(a).
    I am not sure how to continue however, am I even going in the right direction?

    Thanks!!
     
  2. jcsd
  3. Jul 28, 2014 #2
    Presumably ##(a)## is meant to indicate the cyclic subgroup of ##G## generated by the element ##a##, ##aN## is the left coset ##aN=\{an:n\in N\}## considered as an element of the quotient group ##G/N##, ##o(a)## is order of ##a## as an element of ##G## and ##o(aN)## the order of the element ##aN## in the quotient group ##G/N##.

    You've correctly identified that ##aN^{o(a)}=a^{o(a)}N=eN=N## and concluded (I think) that ##o(aN)|o(a)##, or at the very least that ##o(aN)\leq o(a)## (which is all that's really needed).

    What remains to be shown is that ##o(a)\leq o(aN)##. This is where you probably want to use the necessary fact that ##(a)\cap N=(e)##.

    Hint:
    For all ##k##, ##a^k\in aN^k##
     
  4. Jul 28, 2014 #3
    That made sense and I think I got it. Thank you very much!
     
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