# Showing affine transformations group generated by $2x$ and $x+1$ is the Baumslag-Solitar group.

I want to compute the presentation groups of $$\langle f,g\rangle$$ the generated group of affine transformations with $$f(x)=2x$$ and $$g(x)=x+1.$$

The affirmation is $$\langle f,g\rangle=\langle a,b\mid aba^{-1}=b^2\rangle$$ the Baumslag-Solitar group.

I have this:

For any $$h\in \langle f,g\rangle, h(x)=2^nx+\frac{m}{2^k}$$ with $$n,m,k$$ integers. And, the word $$f^{-k}g^{m}f^{k+n}$$ is associated with $$2^nx+\frac{m}{2^k}$$, because $$f^{-k}\circ g^{m}\circ f^{k+n}(x)=2^nx+\frac{m}{2^k}.$$

I know that exists $$\varphi:F(S)=\left\{f,g,f^{-1},g^{-1}\right\}^{\ast}\to \langle f,g\rangle$$ epimorphism.

I want to prove that $$\ker\varphi=\langle \langle T\rangle\rangle$$ with $$T=\left\{fgf^{-1}g^{-2}\right\}$$.

Obviously $$\langle \langle T\rangle\rangle\subset \ker\varphi$$ because $$fgf^{-1}g^{-2}(x)=Id(x)$$ then $$\varphi(fgf^{-1}g^{-2})=Id_{\langle f,g\rangle }$$.

But, how to prove that $$\ker\varphi\subset \langle \langle T\rangle\rangle$$?

• Isn't it enough for you to prove that both functions have infinite order and they fulfill $\;f\circ g\circ f^{-1}=g^{2}\;$ ? – DonAntonio Dec 9 '18 at 21:56
• $fgg=g^{-1}$ is $f=g^{-3}$ but $2x\neq x-3$... – eraldcoil Dec 9 '18 at 21:59
• Read again the comment... – DonAntonio Dec 9 '18 at 22:01
• @DonAntonio: or perhaps you are relying on this? – Hempelicious Dec 10 '18 at 4:34
• Note that DonAntonio's comments together with Hempelicious's link solve the problem. – Derek Holt Dec 10 '18 at 8:56

Now, i have this:

$$\varphi:\left\{a,b\right\}\to $$ with $$\varphi(a)=f$$ and $$\varphi(b)=g$$ homomorphism.

exists unique epimorphism $$\varphi F(a,b)\to $$ such that $$\varphi(w)=w$$ with $$w$$ word in Domain, and $$w$$ group element in Codomain.

further, $$F(a,b)/\ker\varphi\simeq $$.

Afirmation. $$\ker\varphi=<< aba^{-1}b^{-2}>>$$. Obviously $$<< aba^{-1}b^{-2}>>\subset \ker\varphi$$.

Now, let $$w\in \ker\varphi$$, then $$w=a^{-k}b^{m}a^{k+n}$$ with $$\varphi(a^{-k}b^{m}a^{k+n})=Id$$, or, equivalent, $$2^nx+\frac{m}{2^k}=x$$, and this implies $$n=m=0$$.

Therefore, $$w\sim a^{-k}a^{k}\sim \epsilon\sim aba^{-1}b^{-2}\in <>$$.

Therefore $$\ker\varphi=<>$$

It is correct?

• Why are you sure $w$ has that form?? – Hempelicious Dec 14 '18 at 18:52
• That is because all element inf $<f,g>$ is of the form $2^nx+\frac{m}{2^k}$ and $\varphi(a^{-k}b^ma^{k+n})=2^nx+\frac{m}{2^k}$ – eraldcoil Dec 14 '18 at 21:59
• But why does any $w\in\ker\phi$ have that form? You need to prove that! – Hempelicious Dec 15 '18 at 0:24
• I tried that any $h\in <f,g>,\ h=2^nx+\frac{m}{2^k}$ and $f^{-k}g^{m}f^{k+n}(x)=2^nx+\frac{m}{2^k}.$ Identifying $a$ by $f$ and $b$ by $g$ I have, and $\varphi:F(a,b)\to <f,g>$ epimorphism. For any, $h\in <f,g>$ exists $a^{-k}b^{m}a^{k+n}\in F(a,b) : \varphi(a^{-k}b^{m}a^{k+n})=h$. This requires that any word of $F(a,b)$, is $w=a^{-k}b^{m}a^{k+n}$ because $\varphi(w)\in <f,g>$ then $\varphi(w)=2^nx+\frac{m}{2^k}$ In particular, any word in $\ker\varphi$ is of the form $a^{-k}b^{m}a^{k+n}$. or not? – eraldcoil Dec 15 '18 at 2:13
• In the event that my previous comment was wrong. How can I prove that every word in $<a, b | aba = b^2>$ is of the form $a^{- k} b^{m} a^{k + n}$? – eraldcoil Dec 15 '18 at 2:17

Here is a different solution. From the set mapping \begin{align*} a&\mapsto f\\ b&\mapsto g \end{align*}

There is a group homomorphism $$F=\langle a,b\mid\rangle\rightarrow G=\langle f,g\rangle$$. Let $$K$$ be the kernel of this map, so that $$F/K\cong G$$.

Since $$aba^{-1}b^{-2}\in K$$, we can let $$N$$ be the normal subgroup of $$F$$ generated by it. Then $$(F/N)/(K/N)\cong G$$.

In $$F/N$$, we have $$\bar{a}\bar{b}=\bar{b}^2\bar{a}$$, from the relation $$aba^{-1}b^{-2}\in N$$. Thus every element of $$F/N$$ can be written as $$\bar{b}^n\bar{a}^m$$.

If $$\bar{b}^n\bar{a}^m\in K/N$$, then $$g^n\circ f^m$$ is the identity map. But $$g^n\circ f^m(x) = 2^mx+n$$ which is only the identity map if $$m=n=0$$. So $$K/N$$ only has the trivial element, so that $$K=N$$. This means $$F/N\cong G$$, which is what you wanted to prove.

• It is not true that every element of $F/N$ can be written as $\bar{b}^n\bar{a}^m$ for integers $m$ and $n$. For example $\bar{a}^{-1}\bar{ b}\bar{ a}$ cannot be written in that form (unless you are allowing $n$ to be a fraction). – Derek Holt Dec 14 '18 at 20:02
• @DerekHolt: yes you're right! I was thinking of $F/N$ as the semidirect product with the dyadic rational, but that didn't come through in my write-up. I'll update when I get a chance, thanks. – Hempelicious Dec 15 '18 at 0:25
• It is true that every element can be written as $a^{-k}b^ma^{k+n}$ (as in eradcoil's proof) but that needs justifying. You can collect all the negative powers of $a$ to the left. – Derek Holt Dec 15 '18 at 8:52
• How can I prove that every element of $BS(1,2)$ is of the form $a^{-p}a^{s}b^{q}$ with $s \in\mathbb{Z} , p,q\in \mathbb{N}_{0}$? I tried induction in the large of the word. With the relations $ab=b^2a$ and $ba^{-1}=a^{-1}b^2$ For $a,b, ab,ab^{-1}, a^{-1}b, a^{-1}b^{-1}$ etc etc this is true Let $w$ word in $BS(1,2)$ with the form $w=a^{-p}b^{s}a^{q}$ Let's prove that $wa, wb, wa^{-1}, wb^{-1}$ It has the same form. Por example, $wb=(a^{-p}b^{s}a^{q})b$ Here I have a problem. For more than trying to play with relationships I can not "move" that $b$ to the "center" – eraldcoil Dec 16 '18 at 5:24