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Cyclic groups

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A. Let |g| = 20 in a group G. Compute
    |g^2|, |g^8|,|g^5|, |g^3|

    B. In each case find the subgroup H = <x,y> of G.
    a) G = <a> is cyclic, x = a^m, y = a^k, gcd(m,k)=d
    b) G=S_3, x=(1 2), y=(2 3)
    c) G = <a> * <b>, |a| = 4, |b| = 6, x = (a^2, b), y = (a,b^3)

    3. The attempt at a solution
    A. I know |g^2| = 20/2 = 10 and |g^5| = 20/5 = 4
    But |g^8|, |g^3| don't know..

    B. a)H=<a^d> , right?
    but
    I don't know how to solve b) and c)
    Thanks!
     
  2. jcsd
  3. Mar 1, 2009 #2

    Tom Mattson

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    Don't forget that if [itex]g^{20}=e[/itex] then [itex]g^{40}=e[/itex] also.

    Yes.

    b should be easy, because you've got a concrete group to play with. Just get in there and start computing. As for c, what does <a>*<b> mean?
     
  4. Mar 1, 2009 #3

    HallsofIvy

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    The least common multiple of 20 and 8 is 2*4*5= 40. [itex](g^8)^5= (g^20)^2= e[/itex].

    The least common multiple of 3 and 20 is 60. [itex](g^3)^20= (g^20)^3= e[/itex].
     
  5. Mar 1, 2009 #4
    so.. for b) is H=(1 2) * (2 3) = (1 2 3)..?
     
  6. Mar 1, 2009 #5

    Tom Mattson

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    Yes, that's right.
     
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