Mixing
What are the completions of first-order group theory?
$begingroup$
A completion of some theory $T$ (i.e. set of first order statements $T$) is a consistient theory $T' supseteq T$ such that for every first order statement $phi$, either $phi in T'$ or $lnot phi in T'$.
For example, the completions of the theory of algebraically closed fields consist of:
- The theory of algebraically closed fields of characteristic $0$ (which the complex numbers are a model of).
- For every prime $p$, the theory of algebraically closed fields of characteristic $p$.
The first order theory of groups is expressed in the language with a single binary operator, and is axiomatized by the following statements:
- $forall a.b.c. (a ast b)ast c = a ast (b ast c)$
- $exists e. forall a. e ast a = a = a ast e$
- $forall a. exists z. forall b. (a ast z) ast b = b = b ast (a ast z) land (z ast a) ast b = b = b ast (z ast a)$
My question is, what are the completions of this theory?
Given a group $G$, we define $Th(G)$ (the theory of $G$) as the set of true statements about $G$.
Note that although every complete theory arises as the theory of some group, two groups might have the same theory. Since there are $aleph_0$ statements, there are at most $2^{aleph_0} = mathfrak c$ consistient and complete theories, and so many "collisions" will occur.
An obvious example is that two isomorphic groups will have the same theory. More generally two groups have the same theory iff they are elementarily equivalent (by definition).
group-theory first-order-logic model-theory
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
$endgroup$
add a comment |
$begingroup$
A completion of some theory $T$ (i.e. set of first order statements $T$) is a consistient theory $T' supseteq T$ such that for every first order statement $phi$, either $phi in T'$ or $lnot phi in T'$.
For example, the completions of the theory of algebraically closed fields consist of:
- The theory of algebraically closed fields of characteristic $0$ (which the complex numbers are a model of).
- For every prime $p$, the theory of algebraically closed fields of characteristic $p$.
The first order theory of groups is expressed in the language with a single binary operator, and is axiomatized by the following statements:
- $forall a.b.c. (a ast b)ast c = a ast (b ast c)$
- $exists e. forall a. e ast a = a = a ast e$
- $forall a. exists z. forall b. (a ast z) ast b = b = b ast (a ast z) land (z ast a) ast b = b = b ast (z ast a)$
My question is, what are the completions of this theory?
Given a group $G$, we define $Th(G)$ (the theory of $G$) as the set of true statements about $G$.
Note that although every complete theory arises as the theory of some group, two groups might have the same theory. Since there are $aleph_0$ statements, there are at most $2^{aleph_0} = mathfrak c$ consistient and complete theories, and so many "collisions" will occur.
An obvious example is that two isomorphic groups will have the same theory. More generally two groups have the same theory iff they are elementarily equivalent (by definition).
group-theory first-order-logic model-theory
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
$endgroup$
3
$begingroup$
Do you have a reason to expect there is a sensible description of all the completions? It's easy to see there indeed are $mathfrak c$ many of them - for any set of primes, consider the theory stating that there are elements of orders in this set but not of order outside this set. Those necessarily have distinct completions.
$endgroup$
– Wojowu
Jan 24 at 8:47
$begingroup$
@Wojowu Not necessarily. I was hoping there was.
$endgroup$
– PyRulez
Jan 24 at 8:48
add a comment |
$begingroup$
A completion of some theory $T$ (i.e. set of first order statements $T$) is a consistient theory $T' supseteq T$ such that for every first order statement $phi$, either $phi in T'$ or $lnot phi in T'$.
For example, the completions of the theory of algebraically closed fields consist of:
- The theory of algebraically closed fields of characteristic $0$ (which the complex numbers are a model of).
- For every prime $p$, the theory of algebraically closed fields of characteristic $p$.
The first order theory of groups is expressed in the language with a single binary operator, and is axiomatized by the following statements:
- $forall a.b.c. (a ast b)ast c = a ast (b ast c)$
- $exists e. forall a. e ast a = a = a ast e$
- $forall a. exists z. forall b. (a ast z) ast b = b = b ast (a ast z) land (z ast a) ast b = b = b ast (z ast a)$
My question is, what are the completions of this theory?
Given a group $G$, we define $Th(G)$ (the theory of $G$) as the set of true statements about $G$.
Note that although every complete theory arises as the theory of some group, two groups might have the same theory. Since there are $aleph_0$ statements, there are at most $2^{aleph_0} = mathfrak c$ consistient and complete theories, and so many "collisions" will occur.
An obvious example is that two isomorphic groups will have the same theory. More generally two groups have the same theory iff they are elementarily equivalent (by definition).
group-theory first-order-logic model-theory
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
$endgroup$
A completion of some theory $T$ (i.e. set of first order statements $T$) is a consistient theory $T' supseteq T$ such that for every first order statement $phi$, either $phi in T'$ or $lnot phi in T'$.
For example, the completions of the theory of algebraically closed fields consist of:
- The theory of algebraically closed fields of characteristic $0$ (which the complex numbers are a model of).
- For every prime $p$, the theory of algebraically closed fields of characteristic $p$.
The first order theory of groups is expressed in the language with a single binary operator, and is axiomatized by the following statements:
- $forall a.b.c. (a ast b)ast c = a ast (b ast c)$
- $exists e. forall a. e ast a = a = a ast e$
- $forall a. exists z. forall b. (a ast z) ast b = b = b ast (a ast z) land (z ast a) ast b = b = b ast (z ast a)$
My question is, what are the completions of this theory?
Given a group $G$, we define $Th(G)$ (the theory of $G$) as the set of true statements about $G$.
Note that although every complete theory arises as the theory of some group, two groups might have the same theory. Since there are $aleph_0$ statements, there are at most $2^{aleph_0} = mathfrak c$ consistient and complete theories, and so many "collisions" will occur.
An obvious example is that two isomorphic groups will have the same theory. More generally two groups have the same theory iff they are elementarily equivalent (by definition).
group-theory first-order-logic model-theory
group-theory first-order-logic model-theory
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
asked Jan 24 at 8:39
PyRulezPyRulez
4,99522471
4,99522471
3
$begingroup$
Do you have a reason to expect there is a sensible description of all the completions? It's easy to see there indeed are $mathfrak c$ many of them - for any set of primes, consider the theory stating that there are elements of orders in this set but not of order outside this set. Those necessarily have distinct completions.
$endgroup$
– Wojowu
Jan 24 at 8:47
$begingroup$
@Wojowu Not necessarily. I was hoping there was.
$endgroup$
– PyRulez
Jan 24 at 8:48
add a comment |
3
$begingroup$
Do you have a reason to expect there is a sensible description of all the completions? It's easy to see there indeed are $mathfrak c$ many of them - for any set of primes, consider the theory stating that there are elements of orders in this set but not of order outside this set. Those necessarily have distinct completions.
$endgroup$
– Wojowu
Jan 24 at 8:47
$begingroup$
@Wojowu Not necessarily. I was hoping there was.
$endgroup$
– PyRulez
Jan 24 at 8:48
3
3
$begingroup$
Do you have a reason to expect there is a sensible description of all the completions? It's easy to see there indeed are $mathfrak c$ many of them - for any set of primes, consider the theory stating that there are elements of orders in this set but not of order outside this set. Those necessarily have distinct completions.
$endgroup$
– Wojowu
Jan 24 at 8:47
$begingroup$
Do you have a reason to expect there is a sensible description of all the completions? It's easy to see there indeed are $mathfrak c$ many of them - for any set of primes, consider the theory stating that there are elements of orders in this set but not of order outside this set. Those necessarily have distinct completions.
$endgroup$
– Wojowu
Jan 24 at 8:47
$begingroup$
@Wojowu Not necessarily. I was hoping there was.
$endgroup$
– PyRulez
Jan 24 at 8:48
$begingroup$
@Wojowu Not necessarily. I was hoping there was.
$endgroup$
– PyRulez
Jan 24 at 8:48
add a comment |
1 Answer
1
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oldest
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$begingroup$
This is totally intractible. For instance, the complete theory of every finite group is a completion (and these completions are distinct for non-isomorphic finite groups since elementarily equivalent finite structures are isomorphic), so describing all the completions is at least as hard as classifying finite groups. It is also easy to see that there are $mathfrak{c}$ different completions: for instance, any subset of the statements "there exists an element of order $p$" where $p$ ranges over all primes can be true in a completion.
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
2
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
1
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
add a comment |
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$begingroup$
This is totally intractible. For instance, the complete theory of every finite group is a completion (and these completions are distinct for non-isomorphic finite groups since elementarily equivalent finite structures are isomorphic), so describing all the completions is at least as hard as classifying finite groups. It is also easy to see that there are $mathfrak{c}$ different completions: for instance, any subset of the statements "there exists an element of order $p$" where $p$ ranges over all primes can be true in a completion.
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
2
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
1
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
add a comment |
$begingroup$
This is totally intractible. For instance, the complete theory of every finite group is a completion (and these completions are distinct for non-isomorphic finite groups since elementarily equivalent finite structures are isomorphic), so describing all the completions is at least as hard as classifying finite groups. It is also easy to see that there are $mathfrak{c}$ different completions: for instance, any subset of the statements "there exists an element of order $p$" where $p$ ranges over all primes can be true in a completion.
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
2
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
1
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
add a comment |
$begingroup$
This is totally intractible. For instance, the complete theory of every finite group is a completion (and these completions are distinct for non-isomorphic finite groups since elementarily equivalent finite structures are isomorphic), so describing all the completions is at least as hard as classifying finite groups. It is also easy to see that there are $mathfrak{c}$ different completions: for instance, any subset of the statements "there exists an element of order $p$" where $p$ ranges over all primes can be true in a completion.
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
$endgroup$
This is totally intractible. For instance, the complete theory of every finite group is a completion (and these completions are distinct for non-isomorphic finite groups since elementarily equivalent finite structures are isomorphic), so describing all the completions is at least as hard as classifying finite groups. It is also easy to see that there are $mathfrak{c}$ different completions: for instance, any subset of the statements "there exists an element of order $p$" where $p$ ranges over all primes can be true in a completion.
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 8:49
Eric WofseyEric Wofsey
189k14216347
189k14216347
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
2
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
1
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
add a comment |
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
2
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
1
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
$begingroup$
Does it become more tractable if we mod out the finitely axiomatized theories (i.e. considering theories $T_1, T_2$ equivalent if there are sets of statements $t_1, t_2$ with finite symmetric difference such that $t_1 vdash T_1$ and $t_2 vdash T_2$)? Since finite groups are finitely axiomatizable, they will all become equivalent to each other.
$endgroup$
– PyRulez
Jan 24 at 9:11
2
2
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
$begingroup$
That equivalence relation makes all theories equivalent. Indeed, if $T_1$ and $T_2$ are complete theories, let $T_0=T_1cap T_2$. Then for any $varphiin T_1setminus T_2$, $T_0cup{varphi}vdash T_1$ and $T_0cup{negvarphi}vdash T_2$. (To prove this, note that for any $psiin T_1$, $psiveenegvarphiin T_0$, and for any $psiin T_2$, $psivee varphiin T_0$.)
$endgroup$
– Eric Wofsey
Jan 24 at 16:12
1
1
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
$begingroup$
I guess then that's a yes, but not in an interesting way.
$endgroup$
– PyRulez
Jan 24 at 16:43
add a comment |
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$begingroup$
Do you have a reason to expect there is a sensible description of all the completions? It's easy to see there indeed are $mathfrak c$ many of them - for any set of primes, consider the theory stating that there are elements of orders in this set but not of order outside this set. Those necessarily have distinct completions.
$endgroup$
– Wojowu
Jan 24 at 8:47
$begingroup$
@Wojowu Not necessarily. I was hoping there was.
$endgroup$
– PyRulez
Jan 24 at 8:48