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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033073%2fshowing-affine-transformations-group-generated-by-2x-and-x1-is-the-baumslag%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033073%2fshowing-affine-transformations-group-generated-by-2x-and-x1-is-the-baumslag%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more