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

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

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?

  • $\begingroup$ Why are you sure $w$ has that form?? $\endgroup$ – Hempelicious Dec 14 '18 at 18:52
  • $\begingroup$ That is because all element inf $<f,g>$ is of the form $2^nx+\frac{m}{2^k}$ and $\varphi(a^{-k}b^ma^{k+n})=2^nx+\frac{m}{2^k}$ $\endgroup$ – eraldcoil Dec 14 '18 at 21:59
  • $\begingroup$ But why does any $w\in\ker\phi$ have that form? You need to prove that! $\endgroup$ – Hempelicious Dec 15 '18 at 0:24
  • $\begingroup$ 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? $\endgroup$ – eraldcoil Dec 15 '18 at 2:13
  • $\begingroup$ 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 $a^{- k} b^{m} a^{k + n}$? $\endgroup$ – 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.

  • $\begingroup$ 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). $\endgroup$ – Derek Holt Dec 14 '18 at 20:02
  • 1
    $\begingroup$ @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. $\endgroup$ – Hempelicious Dec 15 '18 at 0:25
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Dec 15 '18 at 8:52
  • $\begingroup$ 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" $\endgroup$ – eraldcoil Dec 16 '18 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.