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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.