Is studying a free group (or other free object) equivalent to considering only the consequences of the basic...
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I'm trying to get a better understanding of the rationale behind free groups, and more generally free objects.
This answer does a great job at explaining how various free objects are built, and I understand that one builds a set of "words", and defines an operation over this set, imposing only that a specific set of rules are satisfied.
This makes me wonder: is this construction with words really necessary, or is its only purpose to have a "concrete" object to reason with?
In other words, is studying a free (say) group equivalent to analysing what exactly can be said about a group, without attaching any specific meaning/interpretation to the group elements, so that the elements of the group are effectively only arbitrary symbols, and only the number of such symbols matters?
Along the same lines, when people say that $mathbb Z$ is a free Abelian group, is this statement effectively equivalent to be saying that $mathbb Z$ is entirely defined by its property of being an Abelian group with a single generator?
abstract-algebra group-theory free-groups free-abelian-group
asked Dec 3 at 22:33
glS
709520
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show 6 more comments
up vote
3
down vote
favorite
I'm trying to get a better understanding of the rationale behind free groups, and more generally free objects.
This answer does a great job at explaining how various free objects are built, and I understand that one builds a set of "words", and defines an operation over this set, imposing only that a specific set of rules are satisfied.
This makes me wonder: is this construction with words really necessary, or is its only purpose to have a "concrete" object to reason with?
In other words, is studying a free (say) group equivalent to analysing what exactly can be said about a group, without attaching any specific meaning/interpretation to the group elements, so that the elements of the group are effectively only arbitrary symbols, and only the number of such symbols matters?
Along the same lines, when people say that $mathbb Z$ is a free Abelian group, is this statement effectively equivalent to be saying that $mathbb Z$ is entirely defined by its property of being an Abelian group with a single generator?
abstract-algebra group-theory free-groups free-abelian-group
asked Dec 3 at 22:33
glS
709520
You got the general idea. There may be subtleties having to do with constructivity and infinities.
– Somos
Dec 3 at 22:36
I don't know what you mean by "really necessary."
– Qiaochu Yuan
Dec 3 at 22:46
I would say "characterized" (and then only up to isomorphism) rather than "defined".
– Derek Elkins
Dec 3 at 23:20
@QiaochuYuan I mean something on the lines of: can any result about free groups be derived via solely their "equational properties", without any reference to the model with words, or does defining a free group with "words" bring in more structure than that given by only saying "we are dealing with a group with a given number of generators", without making any statement about what the group or the generators "actually are"?
– glS
Dec 4 at 0:03
@glS: I don't know what you mean by "more structure." I try to avoid working with words as much as possible myself, although lots of people prove facts about free groups with them. Here's an example of a result about free groups I don't know how to prove via solely "equational properties": every subgroup of a free group is free. Note that the corresponding result for free objects in other settings is false, e.g. subgroups of free commutative rings are usually not free. So there is something special about groups here.
– Qiaochu Yuan
Dec 4 at 0:11
|
show 6 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm trying to get a better understanding of the rationale behind free groups, and more generally free objects.
This answer does a great job at explaining how various free objects are built, and I understand that one builds a set of "words", and defines an operation over this set, imposing only that a specific set of rules are satisfied.
This makes me wonder: is this construction with words really necessary, or is its only purpose to have a "concrete" object to reason with?
In other words, is studying a free (say) group equivalent to analysing what exactly can be said about a group, without attaching any specific meaning/interpretation to the group elements, so that the elements of the group are effectively only arbitrary symbols, and only the number of such symbols matters?
Along the same lines, when people say that $mathbb Z$ is a free Abelian group, is this statement effectively equivalent to be saying that $mathbb Z$ is entirely defined by its property of being an Abelian group with a single generator?
abstract-algebra group-theory free-groups free-abelian-group
asked Dec 3 at 22:33
glS
709520
I'm trying to get a better understanding of the rationale behind free groups, and more generally free objects.
This answer does a great job at explaining how various free objects are built, and I understand that one builds a set of "words", and defines an operation over this set, imposing only that a specific set of rules are satisfied.
This makes me wonder: is this construction with words really necessary, or is its only purpose to have a "concrete" object to reason with?
In other words, is studying a free (say) group equivalent to analysing what exactly can be said about a group, without attaching any specific meaning/interpretation to the group elements, so that the elements of the group are effectively only arbitrary symbols, and only the number of such symbols matters?
Along the same lines, when people say that $mathbb Z$ is a free Abelian group, is this statement effectively equivalent to be saying that $mathbb Z$ is entirely defined by its property of being an Abelian group with a single generator?
abstract-algebra group-theory free-groups free-abelian-group
abstract-algebra group-theory free-groups free-abelian-group
asked Dec 3 at 22:33
glS
709520
asked Dec 3 at 22:33
glS
709520
edited Dec 4 at 0:07
asked Dec 3 at 22:33
glS
709520
asked Dec 3 at 22:33
glS
709520
asked Dec 3 at 22:33
glS
709520
709520
You got the general idea. There may be subtleties having to do with constructivity and infinities.
– Somos
Dec 3 at 22:36
I don't know what you mean by "really necessary."
– Qiaochu Yuan
Dec 3 at 22:46
I would say "characterized" (and then only up to isomorphism) rather than "defined".
– Derek Elkins
Dec 3 at 23:20
@QiaochuYuan I mean something on the lines of: can any result about free groups be derived via solely their "equational properties", without any reference to the model with words, or does defining a free group with "words" bring in more structure than that given by only saying "we are dealing with a group with a given number of generators", without making any statement about what the group or the generators "actually are"?
– glS
Dec 4 at 0:03
@glS: I don't know what you mean by "more structure." I try to avoid working with words as much as possible myself, although lots of people prove facts about free groups with them. Here's an example of a result about free groups I don't know how to prove via solely "equational properties": every subgroup of a free group is free. Note that the corresponding result for free objects in other settings is false, e.g. subgroups of free commutative rings are usually not free. So there is something special about groups here.
– Qiaochu Yuan
Dec 4 at 0:11
|
show 6 more comments
You got the general idea. There may be subtleties having to do with constructivity and infinities.
– Somos
Dec 3 at 22:36
I don't know what you mean by "really necessary."
– Qiaochu Yuan
Dec 3 at 22:46
I would say "characterized" (and then only up to isomorphism) rather than "defined".
– Derek Elkins
Dec 3 at 23:20
@QiaochuYuan I mean something on the lines of: can any result about free groups be derived via solely their "equational properties", without any reference to the model with words, or does defining a free group with "words" bring in more structure than that given by only saying "we are dealing with a group with a given number of generators", without making any statement about what the group or the generators "actually are"?
– glS
Dec 4 at 0:03
@glS: I don't know what you mean by "more structure." I try to avoid working with words as much as possible myself, although lots of people prove facts about free groups with them. Here's an example of a result about free groups I don't know how to prove via solely "equational properties": every subgroup of a free group is free. Note that the corresponding result for free objects in other settings is false, e.g. subgroups of free commutative rings are usually not free. So there is something special about groups here.
– Qiaochu Yuan
Dec 4 at 0:11
You got the general idea. There may be subtleties having to do with constructivity and infinities.
– Somos
Dec 3 at 22:36
You got the general idea. There may be subtleties having to do with constructivity and infinities.
– Somos
Dec 3 at 22:36
I don't know what you mean by "really necessary."
– Qiaochu Yuan
Dec 3 at 22:46
I don't know what you mean by "really necessary."
– Qiaochu Yuan
Dec 3 at 22:46
I would say "characterized" (and then only up to isomorphism) rather than "defined".
– Derek Elkins
Dec 3 at 23:20
I would say "characterized" (and then only up to isomorphism) rather than "defined".
– Derek Elkins
Dec 3 at 23:20
@QiaochuYuan I mean something on the lines of: can any result about free groups be derived via solely their "equational properties", without any reference to the model with words, or does defining a free group with "words" bring in more structure than that given by only saying "we are dealing with a group with a given number of generators", without making any statement about what the group or the generators "actually are"?
– glS
Dec 4 at 0:03
@QiaochuYuan I mean something on the lines of: can any result about free groups be derived via solely their "equational properties", without any reference to the model with words, or does defining a free group with "words" bring in more structure than that given by only saying "we are dealing with a group with a given number of generators", without making any statement about what the group or the generators "actually are"?
– glS
Dec 4 at 0:03
@glS: I don't know what you mean by "more structure." I try to avoid working with words as much as possible myself, although lots of people prove facts about free groups with them. Here's an example of a result about free groups I don't know how to prove via solely "equational properties": every subgroup of a free group is free. Note that the corresponding result for free objects in other settings is false, e.g. subgroups of free commutative rings are usually not free. So there is something special about groups here.
– Qiaochu Yuan
Dec 4 at 0:11
@glS: I don't know what you mean by "more structure." I try to avoid working with words as much as possible myself, although lots of people prove facts about free groups with them. Here's an example of a result about free groups I don't know how to prove via solely "equational properties": every subgroup of a free group is free. Note that the corresponding result for free objects in other settings is false, e.g. subgroups of free commutative rings are usually not free. So there is something special about groups here.
– Qiaochu Yuan
Dec 4 at 0:11
|
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
In a sense yes, studying a free object is very similar to studying the underlying equational theory : as you mentioned, a free group on one generator ($mathbb{Z}$) is the most general thing you get when you think of a group generated by one element.
But that sense is very limited, in that it seems like you want to restrict the study of an object to its equational theory. Free objects are much richer than that, and from time to time, having a concrete model for a free object (say reduced words for the free group) can sbe vert useful, even if most of the time the universal property is enough to get by.
An example that comes up way more often than one might think at first sight is the free (commutative, unital) ring on $n$ generators. One model for it is $mathbb{Z}[X_1,...,X_n]$, and this ring has (non equational) properties that are really interesting (its equational properties "aren't interesting", in that they're just the equational properties of any ring with $n$ fixed elements), for instance it's an integral domain, which allows us to use its fraction field in many arguments concerning general rings, and this comes in quite handy.
I don't know how "useful" you think that can be, but the "reduced words" model for the free group on $n$ generators allows us to prove that $F_2$ (free group on $2$ generators) contains a free group on $3$, or even infinitely many generators !
So yes, the free object on $n$ generators is "entirely defined by its property of being an object and having $n$ generators", and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.
answered Dec 3 at 23:14
Max
12.5k11040
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
add a comment |
up vote
2
down vote
I am not a hundred percent sure this answers your question, but the Tarski problem asks whether or not the first order theory of nonabelian free groups are equivalent. This was answered in the affirmative by two groups independently: Kharlampovich-Myasnikov and Sela (spanning hundreds of pages). While proving this they also showed that there are groups with the same first order theory as free groups, but not free! As an example surface groups also satisfy the same first order theory.
So from the perspective of elementary theories you can not tell $F_2,F_3$ or $pi_1(Sigma_2)=langle a_1,b_1,a_2,b_2 mid a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1} rangle$ apart! This should tell you just first order equations is very restrictive, and there are certainly a lot more to these groups than that.
You may be interested in this mathoverflow question too.
answered Dec 4 at 2:10
Paul Plummer
5,20221950
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
In a sense yes, studying a free object is very similar to studying the underlying equational theory : as you mentioned, a free group on one generator ($mathbb{Z}$) is the most general thing you get when you think of a group generated by one element.
But that sense is very limited, in that it seems like you want to restrict the study of an object to its equational theory. Free objects are much richer than that, and from time to time, having a concrete model for a free object (say reduced words for the free group) can sbe vert useful, even if most of the time the universal property is enough to get by.
An example that comes up way more often than one might think at first sight is the free (commutative, unital) ring on $n$ generators. One model for it is $mathbb{Z}[X_1,...,X_n]$, and this ring has (non equational) properties that are really interesting (its equational properties "aren't interesting", in that they're just the equational properties of any ring with $n$ fixed elements), for instance it's an integral domain, which allows us to use its fraction field in many arguments concerning general rings, and this comes in quite handy.
I don't know how "useful" you think that can be, but the "reduced words" model for the free group on $n$ generators allows us to prove that $F_2$ (free group on $2$ generators) contains a free group on $3$, or even infinitely many generators !
So yes, the free object on $n$ generators is "entirely defined by its property of being an object and having $n$ generators", and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.
answered Dec 3 at 23:14
Max
12.5k11040
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
add a comment |
up vote
5
down vote
accepted
In a sense yes, studying a free object is very similar to studying the underlying equational theory : as you mentioned, a free group on one generator ($mathbb{Z}$) is the most general thing you get when you think of a group generated by one element.
But that sense is very limited, in that it seems like you want to restrict the study of an object to its equational theory. Free objects are much richer than that, and from time to time, having a concrete model for a free object (say reduced words for the free group) can sbe vert useful, even if most of the time the universal property is enough to get by.
An example that comes up way more often than one might think at first sight is the free (commutative, unital) ring on $n$ generators. One model for it is $mathbb{Z}[X_1,...,X_n]$, and this ring has (non equational) properties that are really interesting (its equational properties "aren't interesting", in that they're just the equational properties of any ring with $n$ fixed elements), for instance it's an integral domain, which allows us to use its fraction field in many arguments concerning general rings, and this comes in quite handy.
I don't know how "useful" you think that can be, but the "reduced words" model for the free group on $n$ generators allows us to prove that $F_2$ (free group on $2$ generators) contains a free group on $3$, or even infinitely many generators !
So yes, the free object on $n$ generators is "entirely defined by its property of being an object and having $n$ generators", and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.
answered Dec 3 at 23:14
Max
12.5k11040
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
In a sense yes, studying a free object is very similar to studying the underlying equational theory : as you mentioned, a free group on one generator ($mathbb{Z}$) is the most general thing you get when you think of a group generated by one element.
But that sense is very limited, in that it seems like you want to restrict the study of an object to its equational theory. Free objects are much richer than that, and from time to time, having a concrete model for a free object (say reduced words for the free group) can sbe vert useful, even if most of the time the universal property is enough to get by.
An example that comes up way more often than one might think at first sight is the free (commutative, unital) ring on $n$ generators. One model for it is $mathbb{Z}[X_1,...,X_n]$, and this ring has (non equational) properties that are really interesting (its equational properties "aren't interesting", in that they're just the equational properties of any ring with $n$ fixed elements), for instance it's an integral domain, which allows us to use its fraction field in many arguments concerning general rings, and this comes in quite handy.
I don't know how "useful" you think that can be, but the "reduced words" model for the free group on $n$ generators allows us to prove that $F_2$ (free group on $2$ generators) contains a free group on $3$, or even infinitely many generators !
So yes, the free object on $n$ generators is "entirely defined by its property of being an object and having $n$ generators", and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.
answered Dec 3 at 23:14
Max
12.5k11040
In a sense yes, studying a free object is very similar to studying the underlying equational theory : as you mentioned, a free group on one generator ($mathbb{Z}$) is the most general thing you get when you think of a group generated by one element.
But that sense is very limited, in that it seems like you want to restrict the study of an object to its equational theory. Free objects are much richer than that, and from time to time, having a concrete model for a free object (say reduced words for the free group) can sbe vert useful, even if most of the time the universal property is enough to get by.
An example that comes up way more often than one might think at first sight is the free (commutative, unital) ring on $n$ generators. One model for it is $mathbb{Z}[X_1,...,X_n]$, and this ring has (non equational) properties that are really interesting (its equational properties "aren't interesting", in that they're just the equational properties of any ring with $n$ fixed elements), for instance it's an integral domain, which allows us to use its fraction field in many arguments concerning general rings, and this comes in quite handy.
I don't know how "useful" you think that can be, but the "reduced words" model for the free group on $n$ generators allows us to prove that $F_2$ (free group on $2$ generators) contains a free group on $3$, or even infinitely many generators !
So yes, the free object on $n$ generators is "entirely defined by its property of being an object and having $n$ generators", and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.
answered Dec 3 at 23:14
Max
12.5k11040
answered Dec 3 at 23:14
Max
12.5k11040
answered Dec 3 at 23:14
Max
12.5k11040
answered Dec 3 at 23:14
Max
12.5k11040
12.5k11040
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
add a comment |
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
you write "and yes its equational theory is not more interesting than simply the equational theory you're considering; but it can have some nontrivial/interesting non equational properties that can be very useful, or at the very least interesting.". Why isn't this a contradiction? I don't understand whether the "equational theory" is all there is with a free group or not. It looks like you are saying it is, but then you mention that there is more than that in the case e.g. of $mathbb Z[X_1,...,x_n]$. I'm confused
– glS
Dec 3 at 23:57
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
I'm saying that the equational properties of a free object are "boring", no more interesting than logical deductions from the definition of "object"; but that there are some non equational properties that arr interesting, I don't see how that would be a contradiction. Equations aren't all there is to life. Equationally, a free group is nothing more than "any group", but a free group is more than its equations. An example of a property that is not equational is "$ab=0 implies (a=0lor b=0)$ in the case of the free (commutative) ring, or "has exponential growth" in the case of the free group
– Max
Dec 4 at 11:13
add a comment |
up vote
2
down vote
I am not a hundred percent sure this answers your question, but the Tarski problem asks whether or not the first order theory of nonabelian free groups are equivalent. This was answered in the affirmative by two groups independently: Kharlampovich-Myasnikov and Sela (spanning hundreds of pages). While proving this they also showed that there are groups with the same first order theory as free groups, but not free! As an example surface groups also satisfy the same first order theory.
So from the perspective of elementary theories you can not tell $F_2,F_3$ or $pi_1(Sigma_2)=langle a_1,b_1,a_2,b_2 mid a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1} rangle$ apart! This should tell you just first order equations is very restrictive, and there are certainly a lot more to these groups than that.
You may be interested in this mathoverflow question too.
answered Dec 4 at 2:10
Paul Plummer
5,20221950
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
add a comment |
up vote
2
down vote
I am not a hundred percent sure this answers your question, but the Tarski problem asks whether or not the first order theory of nonabelian free groups are equivalent. This was answered in the affirmative by two groups independently: Kharlampovich-Myasnikov and Sela (spanning hundreds of pages). While proving this they also showed that there are groups with the same first order theory as free groups, but not free! As an example surface groups also satisfy the same first order theory.
So from the perspective of elementary theories you can not tell $F_2,F_3$ or $pi_1(Sigma_2)=langle a_1,b_1,a_2,b_2 mid a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1} rangle$ apart! This should tell you just first order equations is very restrictive, and there are certainly a lot more to these groups than that.
You may be interested in this mathoverflow question too.
answered Dec 4 at 2:10
Paul Plummer
5,20221950
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
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I am not a hundred percent sure this answers your question, but the Tarski problem asks whether or not the first order theory of nonabelian free groups are equivalent. This was answered in the affirmative by two groups independently: Kharlampovich-Myasnikov and Sela (spanning hundreds of pages). While proving this they also showed that there are groups with the same first order theory as free groups, but not free! As an example surface groups also satisfy the same first order theory.
So from the perspective of elementary theories you can not tell $F_2,F_3$ or $pi_1(Sigma_2)=langle a_1,b_1,a_2,b_2 mid a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1} rangle$ apart! This should tell you just first order equations is very restrictive, and there are certainly a lot more to these groups than that.
You may be interested in this mathoverflow question too.
answered Dec 4 at 2:10
Paul Plummer
5,20221950
I am not a hundred percent sure this answers your question, but the Tarski problem asks whether or not the first order theory of nonabelian free groups are equivalent. This was answered in the affirmative by two groups independently: Kharlampovich-Myasnikov and Sela (spanning hundreds of pages). While proving this they also showed that there are groups with the same first order theory as free groups, but not free! As an example surface groups also satisfy the same first order theory.
So from the perspective of elementary theories you can not tell $F_2,F_3$ or $pi_1(Sigma_2)=langle a_1,b_1,a_2,b_2 mid a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1} rangle$ apart! This should tell you just first order equations is very restrictive, and there are certainly a lot more to these groups than that.
You may be interested in this mathoverflow question too.
answered Dec 4 at 2:10
Paul Plummer
5,20221950
edited Dec 4 at 2:25
answered Dec 4 at 2:10
Paul Plummer
5,20221950
answered Dec 4 at 2:10
Paul Plummer
5,20221950
answered Dec 4 at 2:10
Paul Plummer
5,20221950
5,20221950
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
add a comment |
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
A question and answer on real closed fields might be of interest too, since the first order theory stuff might be more intuitive
– Paul Plummer
Dec 4 at 2:59
add a comment |
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You got the general idea. There may be subtleties having to do with constructivity and infinities.
– Somos
Dec 3 at 22:36
I don't know what you mean by "really necessary."
– Qiaochu Yuan
Dec 3 at 22:46
I would say "characterized" (and then only up to isomorphism) rather than "defined".
– Derek Elkins
Dec 3 at 23:20
@QiaochuYuan I mean something on the lines of: can any result about free groups be derived via solely their "equational properties", without any reference to the model with words, or does defining a free group with "words" bring in more structure than that given by only saying "we are dealing with a group with a given number of generators", without making any statement about what the group or the generators "actually are"?
– glS
Dec 4 at 0:03
@glS: I don't know what you mean by "more structure." I try to avoid working with words as much as possible myself, although lots of people prove facts about free groups with them. Here's an example of a result about free groups I don't know how to prove via solely "equational properties": every subgroup of a free group is free. Note that the corresponding result for free objects in other settings is false, e.g. subgroups of free commutative rings are usually not free. So there is something special about groups here.
– Qiaochu Yuan
Dec 4 at 0:11