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?
Best Answer
Let $G$ act on itself by conjugation. The orbit stabilizer theorem says $$ |G\cdot x|=\frac{|G|}{|\operatorname{Stab}(x)|}. $$ Here the orbit $G\cdot x$ is precisely the conjugacy class of $x$, denote it $C_G(x)$, and the stabilizer $\operatorname{Stab}(x)$ is precisely the centralizer of $x$, denote it $Z_G(x)$.
Let $x\in G$ have a conjugacy class of order $4$, then necessarily its centralizer $Z_G(x)$ has order $3$ by the above. But clearly $Z(G)\subseteq Z_G(x)$, so the center $Z(G)$ either has order $1$ or $3$. If $|Z(G)|=3$, then $x\in Z(G)$, but then it would have trivial conjugacy class as a central element, contrary to $|C_G(x)|=4$. So $Z(G)$ is trivial.