The title pretty much sums it up : I'm wondering if BS(2,1) contains the Thompson Group F ?
Is the Thompson Group F a subgroup of the Baumslag-Solitar group BS(2,1)?
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group-theory
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2BS(2,1) is metabelian, yet F is not. – 2011-04-25
1 Answers
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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:
- Subgroups of residually finite groups are residually finite.
- $BS(2,1)$ is residually finite (as every $B(m,n)$ with $m=1$ or $n=1$ or $m=n$).
- Thompson's group $F$ isn't residually finite.