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Is Thompson's group F residually finite?

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The answer is "no" -- the quickest way to see this appeals to the following nontrivial fact: the commutator subgroup of $F$, denoted by $F'$ as usual, is infinite and simple.

Armed with this, we argue as follows. Let $N$ be a normal subgroup in $F$ of finite-index; then $N\cap F'$ is going to be normal in $F'$ and of finite index in $F'$. Hence $N\cap F'=F'$, that is, $N$ contains $F'$. So the intersection of all finite-index normal subgroups of $F$ must contain $F'$. But if $F$ were residually finite then this intersection would only contain the identity element, and the result follows.

If you don't mind me asking: is this a question out of curiosity, or one that you've run into during your studies or research?

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#Pete: done. Hope that reads better. – Yemon Choi Feb 1 '10 at 0:18

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