Quotient of a finitely presented group by a finitely presented group
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Is a quotient of a finitely presentable group $G$ by a subgroup $N$ necessarily finitely presentable? What about if the subgroup $N$ is also finitely presentable?
group-theory group-presentation
edited Dec 10 '18 at 23:35
Shaun
9,241113684
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
$endgroup$
add a comment |
$begingroup$
Is a quotient of a finitely presentable group $G$ by a subgroup $N$ necessarily finitely presentable? What about if the subgroup $N$ is also finitely presentable?
group-theory group-presentation
edited Dec 10 '18 at 23:35
Shaun
9,241113684
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
$endgroup$
add a comment |
$begingroup$
Is a quotient of a finitely presentable group $G$ by a subgroup $N$ necessarily finitely presentable? What about if the subgroup $N$ is also finitely presentable?
group-theory group-presentation
edited Dec 10 '18 at 23:35
Shaun
9,241113684
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
$endgroup$
Is a quotient of a finitely presentable group $G$ by a subgroup $N$ necessarily finitely presentable? What about if the subgroup $N$ is also finitely presentable?
group-theory group-presentation
group-theory group-presentation
edited Dec 10 '18 at 23:35
Shaun
9,241113684
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
edited Dec 10 '18 at 23:35
Shaun
9,241113684
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
edited Dec 10 '18 at 23:35
Shaun
9,241113684
edited Dec 10 '18 at 23:35
Shaun
9,241113684
edited Dec 10 '18 at 23:35
Shaun
9,241113684
9,241113684
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
asked Dec 2 '16 at 2:27
user101010user101010
1,943416
1,943416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $N$ is finitely generated, then you can just add the generators to the relation set of $G$ to get $G/N$, which will be a finite presentation. (after all a presentation is just a list of generators and things which normally generate some normal subgroup).
And Mariano's answer is right, although it isn't completely trivial to construct/show a group is infinitely presented, but finitely generated. Some well known groups are $B wr A$ (restricted wreath product) where $B$ is nontrivial and $A$ is infinite. Another way to construct infinitely presented groups is through small cancellation theory.
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
$endgroup$
2
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
add a comment |
$begingroup$
Since there exist finitely generated groups without finite presentations, the answer to your first question is clearly no, since free groups of finite rank are finitely presentable.
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
If $N$ is finitely generated, then you can just add the generators to the relation set of $G$ to get $G/N$, which will be a finite presentation. (after all a presentation is just a list of generators and things which normally generate some normal subgroup).
And Mariano's answer is right, although it isn't completely trivial to construct/show a group is infinitely presented, but finitely generated. Some well known groups are $B wr A$ (restricted wreath product) where $B$ is nontrivial and $A$ is infinite. Another way to construct infinitely presented groups is through small cancellation theory.
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
$endgroup$
2
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
add a comment |
$begingroup$
If $N$ is finitely generated, then you can just add the generators to the relation set of $G$ to get $G/N$, which will be a finite presentation. (after all a presentation is just a list of generators and things which normally generate some normal subgroup).
And Mariano's answer is right, although it isn't completely trivial to construct/show a group is infinitely presented, but finitely generated. Some well known groups are $B wr A$ (restricted wreath product) where $B$ is nontrivial and $A$ is infinite. Another way to construct infinitely presented groups is through small cancellation theory.
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
$endgroup$
2
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
add a comment |
$begingroup$
If $N$ is finitely generated, then you can just add the generators to the relation set of $G$ to get $G/N$, which will be a finite presentation. (after all a presentation is just a list of generators and things which normally generate some normal subgroup).
And Mariano's answer is right, although it isn't completely trivial to construct/show a group is infinitely presented, but finitely generated. Some well known groups are $B wr A$ (restricted wreath product) where $B$ is nontrivial and $A$ is infinite. Another way to construct infinitely presented groups is through small cancellation theory.
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
$endgroup$
If $N$ is finitely generated, then you can just add the generators to the relation set of $G$ to get $G/N$, which will be a finite presentation. (after all a presentation is just a list of generators and things which normally generate some normal subgroup).
And Mariano's answer is right, although it isn't completely trivial to construct/show a group is infinitely presented, but finitely generated. Some well known groups are $B wr A$ (restricted wreath product) where $B$ is nontrivial and $A$ is infinite. Another way to construct infinitely presented groups is through small cancellation theory.
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
answered Dec 2 '16 at 3:08
Paul PlummerPaul Plummer
5,26221950
5,26221950
2
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
add a comment |
2
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
2
2
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
$begingroup$
Examples were discussed at MSE at least twice: math.stackexchange.com/questions/547087/…, math.stackexchange.com/questions/1323510/…. Personally, the simplest example and a proof that I know is the amalgam $G=F_2 star_{F_infty} F_2$. Then $b_2(G)=infty$ by the Mayer-Vietoris sequence, hence, $G$ is not finitely presentable.
$endgroup$
– Moishe Cohen
Dec 3 '16 at 5:08
add a comment |
$begingroup$
Since there exist finitely generated groups without finite presentations, the answer to your first question is clearly no, since free groups of finite rank are finitely presentable.
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
$endgroup$
add a comment |
$begingroup$
Since there exist finitely generated groups without finite presentations, the answer to your first question is clearly no, since free groups of finite rank are finitely presentable.
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
$endgroup$
add a comment |
$begingroup$
Since there exist finitely generated groups without finite presentations, the answer to your first question is clearly no, since free groups of finite rank are finitely presentable.
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
$endgroup$
Since there exist finitely generated groups without finite presentations, the answer to your first question is clearly no, since free groups of finite rank are finitely presentable.
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
answered Dec 2 '16 at 2:32
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155286
111k7155286
add a comment |
add a comment |
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