1. Joined
    26 Nov '07
    Moves
    1085
    11 Jan '08 08:53
    Since people are asking weird maths functional questions, I though I could add some algebra to the mix.

    G=(x,y| x^13=y^9=1, y.x.(y^-1)=x^3> is a group of order 117. Prove that it has a cyclic group of index 3 (order 39).

    Optional Hint: try classifying all non-abelian groups of order 117. But that's quite advanced, so really it's a pointless hint. I thought I would add it anyway though.
  2. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
    Moves
    20443
    11 Jan '08 09:24
    Originally posted by Swlabr
    Since people are asking weird maths functional questions, I though I could add some algebra to the mix.

    G=(x,y| x^13=y^9=1, y.x.(y^-1)=x^3> is a group of order 117. Prove that it has a cyclic group of index 3 (order 39).

    Optional Hint: try classifying all non-abelian groups of order 117. But that's quite advanced, so really it's a pointless hint. I thought I would add it anyway though.
    I'd serach in the direction of y^3 = 1, those are the special roots from y^9 = 1.

    Too much work to do it now though :p

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