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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?

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up vote 5 down vote accepted

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.

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Hmm. Yes. I think I was making it slightly harder than it was... – ADL Dec 9 '10 at 11:09

For the record, here is a somewhat different proof: For $x \in X$ there is a permutation $g_x \in G$ which has only $x$ as a fixed point (as $|X|\geq 3)$. If two permutations $a,b$ commute, then $a$ sends fixed points of $b$ to fixed points of $b$ again.

Now if $g$ is central, it commutes in particular with every $g_x$ for $x \in X$, hence it must fix $x$ as $x$ is the only fixed point of $g_x$. Hence $g$ is the identiy.

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Let $\sigma$ be any permutation on $X$ that moves a point $a\in X$. Let $b=\sigma(a)$ and let $c\in X$ be different from both $a$ and $b$. Choose a permutation $\tau$ of $X$ with $\tau(a)=c$ and $\tau(b)=b$. Now $(\sigma\circ\tau)(a)=\sigma(c)\not=\sigma(a)=b$ but $(\tau\circ\sigma)(a)=\tau(b)=b$. Hence $\sigma$ is not in the center of the group of all permutations of $X$.

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