Mixing
Prove that $langle x,y,|x^2,y^2rangle$ is an infinite group
Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?
finitely-generated
asked Nov 21 '18 at 13:44
mathnoob
1,797422
add a comment |
Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?
finitely-generated
asked Nov 21 '18 at 13:44
mathnoob
1,797422
1
Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
Nov 21 '18 at 13:47
1
You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
Nov 21 '18 at 13:49
Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
Nov 21 '18 at 13:55
add a comment |
Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?
finitely-generated
asked Nov 21 '18 at 13:44
mathnoob
1,797422
Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?
finitely-generated
finitely-generated
asked Nov 21 '18 at 13:44
mathnoob
1,797422
asked Nov 21 '18 at 13:44
mathnoob
1,797422
asked Nov 21 '18 at 13:44
mathnoob
1,797422
asked Nov 21 '18 at 13:44
mathnoob
1,797422
asked Nov 21 '18 at 13:44
mathnoob
1,797422
1,797422
1
Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
Nov 21 '18 at 13:47
1
You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
Nov 21 '18 at 13:49
Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
Nov 21 '18 at 13:55
add a comment |
1
Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
Nov 21 '18 at 13:47
1
You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
Nov 21 '18 at 13:49
Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
Nov 21 '18 at 13:55
1
1
Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
Nov 21 '18 at 13:47
Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
Nov 21 '18 at 13:47
1
1
You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
Nov 21 '18 at 13:49
You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
Nov 21 '18 at 13:49
Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
Nov 21 '18 at 13:55
Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
Nov 21 '18 at 13:55
add a comment |
1 Answer
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You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.
We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.
Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.
Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.
To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?
It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
add a comment |
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You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.
We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.
Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.
Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.
To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?
It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
add a comment |
You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.
We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.
Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.
Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.
To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?
It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
add a comment |
You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.
We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.
Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.
Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.
To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?
It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.
We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.
Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.
Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.
To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?
It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
answered Nov 21 '18 at 22:03
Rafay Ashary
83618
83618
add a comment |
add a comment |
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1
Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
Nov 21 '18 at 13:47
1
You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
Nov 21 '18 at 13:49
Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
Nov 21 '18 at 13:55