Let $X$ be a set, $G$ the group of bijection on $X$. Then it is well-known that if $|X|\geq 3$, $Z(G)$ is trivial. However, I cannot see a way of extending this proof to $X$ being an infinite set (other than when a bijection only moves finitely many points). Indeed, I have been somewhat stumped trying to prove that $Z(G)$ is or is not trivial for $X$ infinite.


So, for $X$ an infinite set, $G$ the bijections on $X$, is it true that $Z(G)$ is trivial?



Center of a Symmetric Group on an Infinite Set
I think this is easy. We shall prove that $f=\mathrm{id}$ for every $f\in Z\left(G\right)$. In fact, let $x\in X$ be such that $f\left(x\right)\neq x$ (if no such $x$ exists, we are done anyway). Now let $y\in X$ be such that $y\neq x$ and $y\neq f\left(x\right)$ (such an $y$ exists due to $\left|X\right|\geq 3$). Then, $f$, lying in the center of $G$, must commute with the transposition $\left(x,y\right)$. Hence, $\left(x,y\right)f\left(x\right)=f\left(\left(x,y\right)x\right)$. In other words, $f\left(x\right)=f\left(y\right)$, leading to $x=y$, contradicting $x\neq y$. qed.


