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.
Originally posted by SwlabrI'd serach in the direction of y^3 = 1, those are the special roots from y^9 = 1.
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