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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 $x\in Z(G)$?

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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 $x \in C_G(x) = Z(G)$.

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