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