- 11 Jan '08 08:53Since 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. - 11 Jan '08 09:24

I'd serach in the direction of y^3 = 1, those are the special roots from y^9 = 1.*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.

Too much work to do it now though :p