I was working on a exercise in Michael Artin's Algebra, which stated
A group $G$ of order 12 contains a conjugacy class of order $4$. Prove that the center of $G$ is trivial.
I started by noting that there must always be a conjugacy class of order $1$ (from the identity), and then we much have that the class equation can either be $1 + 1 + 4 + 6$ or $1 + 3 + 4 + 4$, since the order of the conjugacy classes must divide the order of the group. However, how would I eliminate the possibility of $1 + 1 + 4 + 6$ as a class equation, since this has nontrivial center.
EDIT: I just realized that my method was way off. Can anyone tell me a way to approach this problem?