# Thread: x^3-1=0, soutions really a cyclic group?

1. ## x^3-1=0, soutions really a cyclic group?

"Prove that the cube roots of unity (ie the roots of $x^3-1=0$) form a cyclic group of order three."

I'm assuming that the operation is multiplication, but then that would mean that the set of solutions to the doesn't contain the identity element. Furthermore, it wouldn't be closed under multiplication, since multiplying the two complex roots gives a number outside of the set of solutions. Is there something to this exercise that I'm not getting?

EDIT: Nevermind, I just realized I had a dumb algebra error.