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

Question

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 $\ker\varphi\subset \langle \langle T\rangle\rangle$?

Answer 1

Now, i have this:

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

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

further, $F(a,b)/\ker\varphi\simeq <f,g>$.

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

Therefore $\ker\varphi=<<aba^{-1}b^{-2}>>$

It is correct?

Answer 2

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.