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












2















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



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




I have this:



For any $hin langle f,grangle, h(x)=2^nx+frac{m}{2^k}$ 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 $varphi:F(S)=left{f,g,f^{-1},g^{-1}right}^{ast}to langle f,grangle$ epimorphism.



I want to prove that $kervarphi=langle langle Tranglerangle$ with $T=left{fgf^{-1}g^{-2}right}$.



Obviously $langle langle Trangleranglesubset kervarphi$ because $fgf^{-1}g^{-2}(x)=Id(x)$ then $varphi(fgf^{-1}g^{-2})=Id_{langle f,grangle }$.




But, how to prove that $kervarphisubset langle langle Tranglerangle$?







group-theory affine-geometry group-presentation







share|cite|improve this question




















  • 1




    Isn't it enough for you to prove that both functions have infinite order and they fulfill $;fcirc gcirc f^{-1}=g^{2};$ ?
    – DonAntonio
    Dec 9 '18 at 21:56












  • $fgg=g^{-1} $ is $f=g^{-3}$ but $2xneq x-3$...
    – eraldcoil
    Dec 9 '18 at 21:59










  • Read again the comment...
    – DonAntonio
    Dec 9 '18 at 22:01






  • 1




    @DonAntonio: or perhaps you are relying on this?
    – Hempelicious
    Dec 10 '18 at 4:34






  • 1




    Note that DonAntonio's comments together with Hempelicious's link solve the problem.
    – Derek Holt
    Dec 10 '18 at 8:56
















2















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



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




I have this:



For any $hin langle f,grangle, h(x)=2^nx+frac{m}{2^k}$ 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 $varphi:F(S)=left{f,g,f^{-1},g^{-1}right}^{ast}to langle f,grangle$ epimorphism.



I want to prove that $kervarphi=langle langle Tranglerangle$ with $T=left{fgf^{-1}g^{-2}right}$.



Obviously $langle langle Trangleranglesubset kervarphi$ because $fgf^{-1}g^{-2}(x)=Id(x)$ then $varphi(fgf^{-1}g^{-2})=Id_{langle f,grangle }$.




But, how to prove that $kervarphisubset langle langle Tranglerangle$?







group-theory affine-geometry group-presentation







share|cite|improve this question




















  • 1




    Isn't it enough for you to prove that both functions have infinite order and they fulfill $;fcirc gcirc f^{-1}=g^{2};$ ?
    – DonAntonio
    Dec 9 '18 at 21:56












  • $fgg=g^{-1} $ is $f=g^{-3}$ but $2xneq x-3$...
    – eraldcoil
    Dec 9 '18 at 21:59










  • Read again the comment...
    – DonAntonio
    Dec 9 '18 at 22:01






  • 1




    @DonAntonio: or perhaps you are relying on this?
    – Hempelicious
    Dec 10 '18 at 4:34






  • 1




    Note that DonAntonio's comments together with Hempelicious's link solve the problem.
    – Derek Holt
    Dec 10 '18 at 8:56














2












2








2


2






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



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




I have this:



For any $hin langle f,grangle, h(x)=2^nx+frac{m}{2^k}$ 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 $varphi:F(S)=left{f,g,f^{-1},g^{-1}right}^{ast}to langle f,grangle$ epimorphism.



I want to prove that $kervarphi=langle langle Tranglerangle$ with $T=left{fgf^{-1}g^{-2}right}$.



Obviously $langle langle Trangleranglesubset kervarphi$ because $fgf^{-1}g^{-2}(x)=Id(x)$ then $varphi(fgf^{-1}g^{-2})=Id_{langle f,grangle }$.




But, how to prove that $kervarphisubset langle langle Tranglerangle$?







group-theory affine-geometry group-presentation







share|cite|improve this question
















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



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




I have this:



For any $hin langle f,grangle, h(x)=2^nx+frac{m}{2^k}$ 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 $varphi:F(S)=left{f,g,f^{-1},g^{-1}right}^{ast}to langle f,grangle$ epimorphism.



I want to prove that $kervarphi=langle langle Tranglerangle$ with $T=left{fgf^{-1}g^{-2}right}$.



Obviously $langle langle Trangleranglesubset kervarphi$ because $fgf^{-1}g^{-2}(x)=Id(x)$ then $varphi(fgf^{-1}g^{-2})=Id_{langle f,grangle }$.




But, how to prove that $kervarphisubset langle langle Tranglerangle$?







group-theory affine-geometry group-presentation




group-theory affine-geometry group-presentation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 1:46









Shaun

8,721113680




8,721113680










asked Dec 9 '18 at 21:52









eraldcoil

358111




358111








  • 1




    Isn't it enough for you to prove that both functions have infinite order and they fulfill $;fcirc gcirc f^{-1}=g^{2};$ ?
    – DonAntonio
    Dec 9 '18 at 21:56












  • $fgg=g^{-1} $ is $f=g^{-3}$ but $2xneq x-3$...
    – eraldcoil
    Dec 9 '18 at 21:59










  • Read again the comment...
    – DonAntonio
    Dec 9 '18 at 22:01






  • 1




    @DonAntonio: or perhaps you are relying on this?
    – Hempelicious
    Dec 10 '18 at 4:34






  • 1




    Note that DonAntonio's comments together with Hempelicious's link solve the problem.
    – Derek Holt
    Dec 10 '18 at 8:56














  • 1




    Isn't it enough for you to prove that both functions have infinite order and they fulfill $;fcirc gcirc f^{-1}=g^{2};$ ?
    – DonAntonio
    Dec 9 '18 at 21:56












  • $fgg=g^{-1} $ is $f=g^{-3}$ but $2xneq x-3$...
    – eraldcoil
    Dec 9 '18 at 21:59










  • Read again the comment...
    – DonAntonio
    Dec 9 '18 at 22:01






  • 1




    @DonAntonio: or perhaps you are relying on this?
    – Hempelicious
    Dec 10 '18 at 4:34






  • 1




    Note that DonAntonio's comments together with Hempelicious's link solve the problem.
    – Derek Holt
    Dec 10 '18 at 8:56








1




1




Isn't it enough for you to prove that both functions have infinite order and they fulfill $;fcirc gcirc f^{-1}=g^{2};$ ?
– DonAntonio
Dec 9 '18 at 21:56






Isn't it enough for you to prove that both functions have infinite order and they fulfill $;fcirc gcirc f^{-1}=g^{2};$ ?
– DonAntonio
Dec 9 '18 at 21:56














$fgg=g^{-1} $ is $f=g^{-3}$ but $2xneq x-3$...
– eraldcoil
Dec 9 '18 at 21:59




$fgg=g^{-1} $ is $f=g^{-3}$ but $2xneq x-3$...
– eraldcoil
Dec 9 '18 at 21:59












Read again the comment...
– DonAntonio
Dec 9 '18 at 22:01




Read again the comment...
– DonAntonio
Dec 9 '18 at 22:01




1




1




@DonAntonio: or perhaps you are relying on this?
– Hempelicious
Dec 10 '18 at 4:34




@DonAntonio: or perhaps you are relying on this?
– Hempelicious
Dec 10 '18 at 4:34




1




1




Note that DonAntonio's comments together with Hempelicious's link solve the problem.
– Derek Holt
Dec 10 '18 at 8:56




Note that DonAntonio's comments together with Hempelicious's link solve the problem.
– Derek Holt
Dec 10 '18 at 8:56










2 Answers
2






active

oldest

votes


















3














Now, i have this:



$varphi:left{a,bright}to <f,g>$ 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, $F(a,b)/kervarphisimeq <f,g>$.



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



Now, let $win kervarphi$, 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, $wsim a^{-k}a^{k}sim epsilonsim aba^{-1}b^{-2}in <<aba^{-1}b^{-2}>>$.



Therefore $kervarphi=<<aba^{-1}b^{-2}>>$



It is correct?






share|cite|improve this answer





















  • 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 $winkerphi$ have that form? You need to prove that!
    – Hempelicious
    Dec 15 '18 at 0:24










  • I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ 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



















0














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



There is a group homomorphism $F=langle a,bmidranglerightarrow G=langle f,grangle$. Let $K$ be the kernel of this map, so that $F/Kcong G$.



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



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






share|cite|improve this answer





















  • It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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








  • 1




    @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 $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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













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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Now, i have this:



$varphi:left{a,bright}to <f,g>$ 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, $F(a,b)/kervarphisimeq <f,g>$.



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



Now, let $win kervarphi$, 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, $wsim a^{-k}a^{k}sim epsilonsim aba^{-1}b^{-2}in <<aba^{-1}b^{-2}>>$.



Therefore $kervarphi=<<aba^{-1}b^{-2}>>$



It is correct?






share|cite|improve this answer





















  • 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 $winkerphi$ have that form? You need to prove that!
    – Hempelicious
    Dec 15 '18 at 0:24










  • I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ 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
















3














Now, i have this:



$varphi:left{a,bright}to <f,g>$ 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, $F(a,b)/kervarphisimeq <f,g>$.



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



Now, let $win kervarphi$, 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, $wsim a^{-k}a^{k}sim epsilonsim aba^{-1}b^{-2}in <<aba^{-1}b^{-2}>>$.



Therefore $kervarphi=<<aba^{-1}b^{-2}>>$



It is correct?






share|cite|improve this answer





















  • 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 $winkerphi$ have that form? You need to prove that!
    – Hempelicious
    Dec 15 '18 at 0:24










  • I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ 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














3












3








3






Now, i have this:



$varphi:left{a,bright}to <f,g>$ 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, $F(a,b)/kervarphisimeq <f,g>$.



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



Now, let $win kervarphi$, 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, $wsim a^{-k}a^{k}sim epsilonsim aba^{-1}b^{-2}in <<aba^{-1}b^{-2}>>$.



Therefore $kervarphi=<<aba^{-1}b^{-2}>>$



It is correct?






share|cite|improve this answer












Now, i have this:



$varphi:left{a,bright}to <f,g>$ 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, $F(a,b)/kervarphisimeq <f,g>$.



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



Now, let $win kervarphi$, 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, $wsim a^{-k}a^{k}sim epsilonsim aba^{-1}b^{-2}in <<aba^{-1}b^{-2}>>$.



Therefore $kervarphi=<<aba^{-1}b^{-2}>>$



It is correct?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 6:19









eraldcoil

358111




358111












  • 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 $winkerphi$ have that form? You need to prove that!
    – Hempelicious
    Dec 15 '18 at 0:24










  • I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ 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


















  • 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 $winkerphi$ have that form? You need to prove that!
    – Hempelicious
    Dec 15 '18 at 0:24










  • I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ 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
















Why are you sure $w$ has that form??
– Hempelicious
Dec 14 '18 at 18:52




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






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 $winkerphi$ have that form? You need to prove that!
– Hempelicious
Dec 15 '18 at 0:24




But why does any $winkerphi$ have that form? You need to prove that!
– Hempelicious
Dec 15 '18 at 0:24












I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ is of the form $a^{-k}b^{m}a^{k+n}$. or not?
– eraldcoil
Dec 15 '18 at 2:13






I tried that any $hin <f,g>, h=2^nx+frac{m}{2^k}$ 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, $hin <f,g>$ 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 $kervarphi$ 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




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











0














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



There is a group homomorphism $F=langle a,bmidranglerightarrow G=langle f,grangle$. Let $K$ be the kernel of this map, so that $F/Kcong G$.



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



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






share|cite|improve this answer





















  • It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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








  • 1




    @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 $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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


















0














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



There is a group homomorphism $F=langle a,bmidranglerightarrow G=langle f,grangle$. Let $K$ be the kernel of this map, so that $F/Kcong G$.



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



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






share|cite|improve this answer





















  • It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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








  • 1




    @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 $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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
















0












0








0






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



There is a group homomorphism $F=langle a,bmidranglerightarrow G=langle f,grangle$. Let $K$ be the kernel of this map, so that $F/Kcong G$.



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



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






share|cite|improve this answer












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



There is a group homomorphism $F=langle a,bmidranglerightarrow G=langle f,grangle$. Let $K$ be the kernel of this map, so that $F/Kcong G$.



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



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 19:55









Hempelicious

127110




127110












  • It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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








  • 1




    @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 $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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




















  • It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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








  • 1




    @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 $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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


















It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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






It is not true that every element of $F/N$ can be written as $bar{b}^nbar{a}^m$ 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






1




1




@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




@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




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 $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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






How can I prove that every element of $BS(1,2)$ is of the form $a^{-p}a^{s}b^{q}$ with $s inmathbb{Z} , p,qin mathbb{N}_{0}$? 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




















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