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The title pretty much sums it up : I'm wondering if BS(2,1) contains the Thompson Group F ?

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    BS(2,1) is metabelian, yet F is not.2011-04-25

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I think the answer is no. This probably isn't the easiest argument, but it's the first one that came to my mind:

  1. Subgroups of residually finite groups are residually finite.
  2. $BS(2,1)$ is residually finite (as every $B(m,n)$ with $m=1$ or $n=1$ or $m=n$).
  3. Thompson's group $F$ isn't residually finite.