# If $G$ is a group of order 12 with conjugacy class of order 4. Show that $G$ has trivial center

Let $$G$$ be a group of order 12. Show that if $$G$$ contains a conjugacy class of order 4, the center of $$G$$ is $$\lbrace 1\rbrace$$.

So I got most of the proof but get stuck at a certain point. By the counting formula, I got that the order of the centralizer $$C_G(x)$$ of $$x$$ is 3. And that because $$Z(G)\leq C_G(x)$$ and $$Z(G)\leq G$$, then the order of $$Z(G)$$ must divide both the order of $$G$$ and the order of $$C_G(x)$$. Hence, we find that the order of $$Z(G)$$ is either 1 or 3.

This is where I get stuck. I know that the order of $$Z(G)\neq 3$$ because this would imply that $$x\in Z(G)$$ which can't happen and is a contradiction.

What I don't understand in the first place is, how do we know that $$x$$ cannot be in $$Z(G)$$? And why does $$Z(G)$$ having order 3 imply that $$x\in Z(G)$$?

To answer your first question, $$x$$ cannot be in $$Z(G)$$ because if it were, it would be in its own conjugacy class of size $$1$$. This is because if $$x \in Z(G)$$, then for any $$g \in G$$ we have $$gxg^{-1} = gg^{-1}x = x$$.
To answer your second question, if $$Z(G)$$ has order $$3$$, then the facts that $$Z(G) \leq C_G(x)$$ and $$|C_G(x)| = 3$$ would imply that $$Z(G) = C_G(x)$$, hence $$x \in C_G(x) = Z(G)$$.