How to prove that two groups with different presentations are isomorphic in a naive way?
2
$begingroup$
One can define a presentation of a group naively (ala Dummit-Foote in Chapter 1.2), i.e., as a group generated by certain elements with certain relations such that all other relations follow from the given ones. (By "naively" I mean not formally (as being an appropriate quotient of the free group on some letters).) I was wondering, what exactly does one need to show in order to prove that two "naively presented groups" (with different presentations) are isomorphic?
abstract-algebra group-theory group-presentation combinatorial-group-theory
edited Dec 3 '18 at 0:41
Shaun
8,893113681
asked Aug 25 '18 at 17:17
user437309user437309
659212
$endgroup$
|
show 4 more comments
2
$begingroup$
One can define a presentation of a group naively (ala Dummit-Foote in Chapter 1.2), i.e., as a group generated by certain elements with certain relations such that all other relations follow from the given ones. (By "naively" I mean not formally (as being an appropriate quotient of the free group on some letters).) I was wondering, what exactly does one need to show in order to prove that two "naively presented groups" (with different presentations) are isomorphic?
abstract-algebra group-theory group-presentation combinatorial-group-theory
edited Dec 3 '18 at 0:41
Shaun
8,893113681
asked Aug 25 '18 at 17:17
user437309user437309
659212
$endgroup$
5
$begingroup$
That's a very hard problem and there is (and there can be) no formal algorithm to do that in general, let alone a user-friendly one.
$endgroup$
– Arnaud Mortier
Aug 25 '18 at 17:18
2
$begingroup$
In general, it is an almost hopeless problem. In some particular cases, it can be a rather challenging problem
$endgroup$
– DonAntonio
Aug 25 '18 at 17:26
1
$begingroup$
The trivial group can have impressively complex presentations...and ti can be very, very tough to prove the group is the trivial one.
$endgroup$
– DonAntonio
Aug 25 '18 at 17:40
1
$begingroup$
@user437309 If I'm interprating what you mean correctly, it would mean to show that the relations are basically the same on a basically same set of generators. If we restrict ourselves to finite sets of generators and relations, it could, in principle, be possible to prove the minimal set of (irredundant) generators in both groups has the same cardinality and the relations in one imply the relations in the other, and the other way around
$endgroup$
– DonAntonio
Aug 25 '18 at 17:53
1
$begingroup$
You use Tietze transformations. It can be proved that if the groups are isomorphic then it can be proved in this way.
$endgroup$
– Derek Holt
Aug 25 '18 at 17:58
|
show 4 more comments
2
2
2
1
$begingroup$
One can define a presentation of a group naively (ala Dummit-Foote in Chapter 1.2), i.e., as a group generated by certain elements with certain relations such that all other relations follow from the given ones. (By "naively" I mean not formally (as being an appropriate quotient of the free group on some letters).) I was wondering, what exactly does one need to show in order to prove that two "naively presented groups" (with different presentations) are isomorphic?
abstract-algebra group-theory group-presentation combinatorial-group-theory
edited Dec 3 '18 at 0:41
Shaun
8,893113681
asked Aug 25 '18 at 17:17
user437309user437309
659212
$endgroup$
One can define a presentation of a group naively (ala Dummit-Foote in Chapter 1.2), i.e., as a group generated by certain elements with certain relations such that all other relations follow from the given ones. (By "naively" I mean not formally (as being an appropriate quotient of the free group on some letters).) I was wondering, what exactly does one need to show in order to prove that two "naively presented groups" (with different presentations) are isomorphic?
abstract-algebra group-theory group-presentation combinatorial-group-theory
abstract-algebra group-theory group-presentation combinatorial-group-theory
edited Dec 3 '18 at 0:41
Shaun
8,893113681
asked Aug 25 '18 at 17:17
user437309user437309
659212
edited Dec 3 '18 at 0:41
Shaun
8,893113681
asked Aug 25 '18 at 17:17
user437309user437309
659212
edited Dec 3 '18 at 0:41
Shaun
8,893113681
edited Dec 3 '18 at 0:41
Shaun
8,893113681
edited Dec 3 '18 at 0:41
Shaun
8,893113681
8,893113681
asked Aug 25 '18 at 17:17
user437309user437309
659212
asked Aug 25 '18 at 17:17
user437309user437309
659212
asked Aug 25 '18 at 17:17
user437309user437309
659212
659212
5
$begingroup$
That's a very hard problem and there is (and there can be) no formal algorithm to do that in general, let alone a user-friendly one.
$endgroup$
– Arnaud Mortier
Aug 25 '18 at 17:18
2
$begingroup$
In general, it is an almost hopeless problem. In some particular cases, it can be a rather challenging problem
$endgroup$
– DonAntonio
Aug 25 '18 at 17:26
1
$begingroup$
The trivial group can have impressively complex presentations...and ti can be very, very tough to prove the group is the trivial one.
$endgroup$
– DonAntonio
Aug 25 '18 at 17:40
1
$begingroup$
@user437309 If I'm interprating what you mean correctly, it would mean to show that the relations are basically the same on a basically same set of generators. If we restrict ourselves to finite sets of generators and relations, it could, in principle, be possible to prove the minimal set of (irredundant) generators in both groups has the same cardinality and the relations in one imply the relations in the other, and the other way around
$endgroup$
– DonAntonio
Aug 25 '18 at 17:53
1
$begingroup$
You use Tietze transformations. It can be proved that if the groups are isomorphic then it can be proved in this way.
$endgroup$
– Derek Holt
Aug 25 '18 at 17:58
|
show 4 more comments
5
$begingroup$
That's a very hard problem and there is (and there can be) no formal algorithm to do that in general, let alone a user-friendly one.
$endgroup$
– Arnaud Mortier
Aug 25 '18 at 17:18
2
$begingroup$
In general, it is an almost hopeless problem. In some particular cases, it can be a rather challenging problem
$endgroup$
– DonAntonio
Aug 25 '18 at 17:26
1
$begingroup$
The trivial group can have impressively complex presentations...and ti can be very, very tough to prove the group is the trivial one.
$endgroup$
– DonAntonio
Aug 25 '18 at 17:40
1
$begingroup$
@user437309 If I'm interprating what you mean correctly, it would mean to show that the relations are basically the same on a basically same set of generators. If we restrict ourselves to finite sets of generators and relations, it could, in principle, be possible to prove the minimal set of (irredundant) generators in both groups has the same cardinality and the relations in one imply the relations in the other, and the other way around
$endgroup$
– DonAntonio
Aug 25 '18 at 17:53
1
$begingroup$
You use Tietze transformations. It can be proved that if the groups are isomorphic then it can be proved in this way.
$endgroup$
– Derek Holt
Aug 25 '18 at 17:58
5
5
$begingroup$
That's a very hard problem and there is (and there can be) no formal algorithm to do that in general, let alone a user-friendly one.
$endgroup$
– Arnaud Mortier
Aug 25 '18 at 17:18
$begingroup$
That's a very hard problem and there is (and there can be) no formal algorithm to do that in general, let alone a user-friendly one.
$endgroup$
– Arnaud Mortier
Aug 25 '18 at 17:18
2
2
$begingroup$
In general, it is an almost hopeless problem. In some particular cases, it can be a rather challenging problem
$endgroup$
– DonAntonio
Aug 25 '18 at 17:26
$begingroup$
In general, it is an almost hopeless problem. In some particular cases, it can be a rather challenging problem
$endgroup$
– DonAntonio
Aug 25 '18 at 17:26
1
1
$begingroup$
The trivial group can have impressively complex presentations...and ti can be very, very tough to prove the group is the trivial one.
$endgroup$
– DonAntonio
Aug 25 '18 at 17:40
$begingroup$
The trivial group can have impressively complex presentations...and ti can be very, very tough to prove the group is the trivial one.
$endgroup$
– DonAntonio
Aug 25 '18 at 17:40
1
1
$begingroup$
@user437309 If I'm interprating what you mean correctly, it would mean to show that the relations are basically the same on a basically same set of generators. If we restrict ourselves to finite sets of generators and relations, it could, in principle, be possible to prove the minimal set of (irredundant) generators in both groups has the same cardinality and the relations in one imply the relations in the other, and the other way around
$endgroup$
– DonAntonio
Aug 25 '18 at 17:53
$begingroup$
@user437309 If I'm interprating what you mean correctly, it would mean to show that the relations are basically the same on a basically same set of generators. If we restrict ourselves to finite sets of generators and relations, it could, in principle, be possible to prove the minimal set of (irredundant) generators in both groups has the same cardinality and the relations in one imply the relations in the other, and the other way around
$endgroup$
– DonAntonio
Aug 25 '18 at 17:53
1
1
$begingroup$
You use Tietze transformations. It can be proved that if the groups are isomorphic then it can be proved in this way.
$endgroup$
– Derek Holt
Aug 25 '18 at 17:58
$begingroup$
You use Tietze transformations. It can be proved that if the groups are isomorphic then it can be proved in this way.
$endgroup$
– Derek Holt
Aug 25 '18 at 17:58
|
show 4 more comments
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5
$begingroup$
That's a very hard problem and there is (and there can be) no formal algorithm to do that in general, let alone a user-friendly one.
$endgroup$
– Arnaud Mortier
Aug 25 '18 at 17:18
2
$begingroup$
In general, it is an almost hopeless problem. In some particular cases, it can be a rather challenging problem
$endgroup$
– DonAntonio
Aug 25 '18 at 17:26
1
$begingroup$
The trivial group can have impressively complex presentations...and ti can be very, very tough to prove the group is the trivial one.
$endgroup$
– DonAntonio
Aug 25 '18 at 17:40
1
$begingroup$
@user437309 If I'm interprating what you mean correctly, it would mean to show that the relations are basically the same on a basically same set of generators. If we restrict ourselves to finite sets of generators and relations, it could, in principle, be possible to prove the minimal set of (irredundant) generators in both groups has the same cardinality and the relations in one imply the relations in the other, and the other way around
$endgroup$
– DonAntonio
Aug 25 '18 at 17:53
1
$begingroup$
You use Tietze transformations. It can be proved that if the groups are isomorphic then it can be proved in this way.
$endgroup$
– Derek Holt
Aug 25 '18 at 17:58