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Let $G$ be the Baumslag-Solitar group $\langle a,t \mid tat^{-1}=a^k\rangle$ and $$\mathbb{Z}[1/k]:=\left\{\frac{x}{k^n}\mid x\in\mathbb{Z},n\in\mathbb{N}\cup\{0\}\right\}.$$ I'm searching for an isomorphism $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$. I only need to know the map $\Psi:\langle a\rangle^G\to\mathbb{Z}[1/k]$. I defined $$\Psi(t^na^xt^{-n}):=\frac{x}{k^n}.$$ But for one hour I'm calculating to prove that this map induces an homomorphism, but I always get for $n>m$:

$$\Psi(t^na^xt^{-n})\cdot \Psi(t^ma^yt^{-m})=\frac{x}{k^n}+\frac{y}{k^m}=\frac{x+y\cdot k^{n-m}}{k^n}$$ and $$\Psi(t^na^xt^{-n}\cdot t^ma^yt^{-m})=\Psi(t^ma^{(x\cdot k^{n-m})}t^{-m}t^myt^{-m})=\Psi(t^ma^{x\cdot k^{n-m}+y}t^{-m})=\frac{x\cdot k^{n-m}+y}{k^m}.$$ And these images are obvisually unequal. So what is my fault? Is this map the wrong one?

Thanks for your help.

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I think Baumslag–Solitar group is correct but I am not sure. – Babak Miraftab May 28 '12 at 17:04
Yes. Thank you. – Peter May 28 '12 at 17:30
$tat^{-1}=a^k$, so $\Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$. – Ted May 28 '12 at 21:11
Thank you very much. – Peter May 28 '12 at 21:24

You need to modify the definition of $\Psi$: put $$\Psi(t^na^xt^{-n})=xk^n$$ (instead of $\Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.

Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $\Psi$ as originally given is a homomorphism.

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Thank you very much. – Peter May 28 '12 at 21:24

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