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, 2018 at 21:56
  • $\begingroup$ $fgg=g^{-1} $ is $f=g^{-3}$ but $2x\neq x-3$... $\endgroup$
    – eraldcoil
    Dec 9, 2018 at 21:59
  • $\begingroup$ Read again the comment... $\endgroup$
    – DonAntonio
    Dec 9, 2018 at 22:01
  • 1
    $\begingroup$ @DonAntonio: or perhaps you are relying on this? $\endgroup$ Dec 10, 2018 at 4:34
  • 2
    $\begingroup$ Note that DonAntonio's comments together with Hempelicious's link solve the problem. $\endgroup$
    – Derek Holt
    Dec 10, 2018 at 8:56

2 Answers 2


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?? $\endgroup$ Dec 14, 2018 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, 2018 at 21:59
  • $\begingroup$ But why does any $w\in\ker\phi$ have that form? You need to prove that! $\endgroup$ Dec 15, 2018 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, 2018 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, 2018 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, 2018 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. $\endgroup$ Dec 15, 2018 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, 2018 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, 2018 at 5:24

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