An elementary problem related to $Z(G)$

Question

Let $G$ be a group,

If $xy\in Z(G)$ then $C_G(x)=C_G(y)$.

Note: Even it is very elementary, I liked it.

Edit: Thanks for different solutions, you may want to examine the case if $xyz\in Z(G)$ then what happens ? By using the first result, we can say $C_G(x)=C_G(yz)$ or $C_G(xy)=C_G(z)$. And with a little work, we can also show that $C_G(y)=C_G(xz)$ but what does these relation mean ? If you can explore this case, please add as an answer.

Answer 1

Let $r\in C_G(x)$ then

$(xy)^r=xy$ as $xy \in Z(G)$ and $$xy=(xy)^r=x^r y^r$$

since $r\in C_G(x)$, $x^r=x$ we have $$xy^r=xy\implies y^r=y\implies r\in C_G(y) $$

And converse inclusion can be done in a similiar way.

Corallary $1$: The sizes of conjugacy classes of $x$ and $y$ are equal.

Corallary $2$: if $x\notin Z(G)$ then $|C_G(x)|\geq 2|Z(G)|$.

Answer 2

A shorter one (but essentially the same as yours):

$rxr^{-1}=rx(yr^{-1}y^{-1})=(rxyr^{-1})y^{-1}=xyy^{-1}=x$. (Parentheses for clarity.)

Answer 3

Suppose $r\in C_G(x)=C_G(x^{-1})$; then \begin{align} ry&=rx^{-1}xy \\ &=x^{-1}rxy && \text{because $r\in C_G(x^{-1})$}\\ &=x^{-1}xyr && \text{because $xy\in Z(G)$}\\ &=yr \end{align} The other inclusion is similar.

Answer 4

The claim is equivalent to $C_G(x)=C_G(xz)$ for $z \in Z(G)$. It suffices to prove $\subseteq$ (consider $z^{-1}$ for $\supseteq$). If $g$ commutes with $x$, then $g$ commutes with $xz$ because $g \cdot xz=x \cdot gz=xz \cdot g$.