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Aug 22, 2011 at 13:19 vote accept Mustafa Gokhan Benli
Aug 20, 2011 at 2:17 comment added Ben Wieland The group is $\mathbb Z^2\ltimes \mathbb Z[\frac12]^2$. The subgroup is $\mathbb Z\ltimes \mathbb Z[\frac12]^2$. The subgroup consists of those elements whose image in $\mathbb Z^2$ is anti-diagonal: $(n,-n)$. The group is fp because it's the square of the fp $BS(1,2)$. The subgroup is fg by 3 elements, the $1\in \mathbb Z$ and the two $1\in\mathbb Z[\frac12]$.
Aug 19, 2011 at 17:55 comment added Mustafa Gokhan Benli @Ben. I am sorry but I have difficulty following the argument . Can you explicitly tell me what is the group and what is the subgroup? Thanks alot.
Aug 19, 2011 at 3:43 comment added Autumn Kent That's also good.
Aug 19, 2011 at 3:27 history answered Ben Wieland CC BY-SA 3.0