Is there a surjective morphism from an infinite direct product of copies of $mathbbZ$ to an infinite direct sum of copies of $mathbbZ$?Countable infinite direct product of $mathbbZ$ modulo countable direct sumAn example of a morphism which does not preserve normality.Unique morphism from the additive group $mathbb Q$ to $mathbb Z$From semidirect to direct product of groupsThere is no injective morphism from $mathbb S_7$ to $mathbb A_8$Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?Proof/Disproof is there a surjective homomorphism $f:mathbbZ_20rightarrow mathbbZ_2 oplusmathbbZ_2$isomorphic direct product to $mathbbC_0, cdot$Show that the direct sum of countably many $mathbbZ$ copies is not finitely generatedIf $G$ is a $p$-group, show that there is an epimorphism from $G$ to $mathbbZ/mathbbZp$Is there such a thing as direct product of an infinite number of groups?
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Is there a surjective morphism from an infinite direct product of copies of $mathbbZ$ to an infinite direct sum of copies of $mathbbZ$?
Countable infinite direct product of $mathbbZ$ modulo countable direct sumAn example of a morphism which does not preserve normality.Unique morphism from the additive group $mathbb Q$ to $mathbb Z$From semidirect to direct product of groupsThere is no injective morphism from $mathbb S_7$ to $mathbb A_8$Does there exist a surjective homomorphism from $(mathbb R,+)$ to $(mathbb Q,+)$ ?Proof/Disproof is there a surjective homomorphism $f:mathbbZ_20rightarrow mathbbZ_2 oplusmathbbZ_2$isomorphic direct product to $mathbbC_0, cdot$Show that the direct sum of countably many $mathbbZ$ copies is not finitely generatedIf $G$ is a $p$-group, show that there is an epimorphism from $G$ to $mathbbZ/mathbbZp$Is there such a thing as direct product of an infinite number of groups?
$begingroup$
Is there a surjective morphism $mathbbZ^Ito mathbbZ^(J)$ for some $I,J$?
i) I'm asking about group morphisms
ii) $mathbbZ^(J)$ denotes the direct sum of $J$ copies of $mathbb Z$
iii) Of course, I want $J$, and thus $I$ to be infinite, otherwise it's trivial.
I want to know if there is such a surjection for any $J$, so it suffices to find one $J$ with no such surjection and it'll be done. However it seems that the answer won't really depend on $J$.
Thoughts : I think there is no surjective morphism, and so was trying to find a contradiction. Of course such a morphism is split, so I was trying to analyze summands of $mathbbZ^I$ but got essentially nowhere.
I also tried studying maps $mathbbZ^I to mathbbZ$, to study the projections. I know that if such a map vanishes on almost zero sequences, then it vanishes, so I was trying to prove that it would do that here, but with no success.
I don't really know how to proceed further.
EDIT: the link that was given in the comments can help. Indeed, if there were such a surjection, then by passing to hom's into $mathbbZ$, we would have an injection $mathbbZ^J to hom (mathbbZ^I, mathbbZ)$. Now I don't know how general the quoted Baer's result is, but if it generalizes to any exponent, then this would provide an injection $mathbbZ^Jto mathbbZ^(I)$, which is clearly contradictory, by looking for instance at "almost" $2^infty$-divisible elements of $mathbbZ^J$ (if this isn't clear I can sketch the proof here, but it's not that complicated)
Alternatively, if Baer's result isn't that general, it might be possible to use some trickery to reduce to $I= mathbbN$, at least when assuming that $J=mathbbN$.
abstract-algebra group-theory abelian-groups
asked Apr 1 at 15:50
MaxMax
16.1k11144
$endgroup$
|
show 6 more comments
$begingroup$
Is there a surjective morphism $mathbbZ^Ito mathbbZ^(J)$ for some $I,J$?
i) I'm asking about group morphisms
ii) $mathbbZ^(J)$ denotes the direct sum of $J$ copies of $mathbb Z$
iii) Of course, I want $J$, and thus $I$ to be infinite, otherwise it's trivial.
I want to know if there is such a surjection for any $J$, so it suffices to find one $J$ with no such surjection and it'll be done. However it seems that the answer won't really depend on $J$.
Thoughts : I think there is no surjective morphism, and so was trying to find a contradiction. Of course such a morphism is split, so I was trying to analyze summands of $mathbbZ^I$ but got essentially nowhere.
I also tried studying maps $mathbbZ^I to mathbbZ$, to study the projections. I know that if such a map vanishes on almost zero sequences, then it vanishes, so I was trying to prove that it would do that here, but with no success.
I don't really know how to proceed further.
EDIT: the link that was given in the comments can help. Indeed, if there were such a surjection, then by passing to hom's into $mathbbZ$, we would have an injection $mathbbZ^J to hom (mathbbZ^I, mathbbZ)$. Now I don't know how general the quoted Baer's result is, but if it generalizes to any exponent, then this would provide an injection $mathbbZ^Jto mathbbZ^(I)$, which is clearly contradictory, by looking for instance at "almost" $2^infty$-divisible elements of $mathbbZ^J$ (if this isn't clear I can sketch the proof here, but it's not that complicated)
Alternatively, if Baer's result isn't that general, it might be possible to use some trickery to reduce to $I= mathbbN$, at least when assuming that $J=mathbbN$.
abstract-algebra group-theory abelian-groups
asked Apr 1 at 15:50
MaxMax
16.1k11144
$endgroup$
1
$begingroup$
Have you tried looking at the problem from a categorical perspective?
$endgroup$
– Shaun
Apr 1 at 16:03
1
$begingroup$
You defined $mathbbZ^(J)$, but not $mathbbZ^I$. Are they the same thing, or do your brackets have a meaning?
$endgroup$
– user1729
Apr 1 at 16:07
$begingroup$
@Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though
$endgroup$
– Max
Apr 1 at 16:09
1
$begingroup$
Okay (but call it "the direct product" or something, rather than just "the power" :-p)
$endgroup$
– user1729
Apr 1 at 16:13
2
$begingroup$
@user1729 : it is split because $mathbbZ^(I)$ is free
$endgroup$
– Max
Apr 1 at 16:27
|
show 6 more comments
$begingroup$
Is there a surjective morphism $mathbbZ^Ito mathbbZ^(J)$ for some $I,J$?
i) I'm asking about group morphisms
ii) $mathbbZ^(J)$ denotes the direct sum of $J$ copies of $mathbb Z$
iii) Of course, I want $J$, and thus $I$ to be infinite, otherwise it's trivial.
I want to know if there is such a surjection for any $J$, so it suffices to find one $J$ with no such surjection and it'll be done. However it seems that the answer won't really depend on $J$.
Thoughts : I think there is no surjective morphism, and so was trying to find a contradiction. Of course such a morphism is split, so I was trying to analyze summands of $mathbbZ^I$ but got essentially nowhere.
I also tried studying maps $mathbbZ^I to mathbbZ$, to study the projections. I know that if such a map vanishes on almost zero sequences, then it vanishes, so I was trying to prove that it would do that here, but with no success.
I don't really know how to proceed further.
EDIT: the link that was given in the comments can help. Indeed, if there were such a surjection, then by passing to hom's into $mathbbZ$, we would have an injection $mathbbZ^J to hom (mathbbZ^I, mathbbZ)$. Now I don't know how general the quoted Baer's result is, but if it generalizes to any exponent, then this would provide an injection $mathbbZ^Jto mathbbZ^(I)$, which is clearly contradictory, by looking for instance at "almost" $2^infty$-divisible elements of $mathbbZ^J$ (if this isn't clear I can sketch the proof here, but it's not that complicated)
Alternatively, if Baer's result isn't that general, it might be possible to use some trickery to reduce to $I= mathbbN$, at least when assuming that $J=mathbbN$.
abstract-algebra group-theory abelian-groups
asked Apr 1 at 15:50
MaxMax
16.1k11144
$endgroup$
Is there a surjective morphism $mathbbZ^Ito mathbbZ^(J)$ for some $I,J$?
i) I'm asking about group morphisms
ii) $mathbbZ^(J)$ denotes the direct sum of $J$ copies of $mathbb Z$
iii) Of course, I want $J$, and thus $I$ to be infinite, otherwise it's trivial.
I want to know if there is such a surjection for any $J$, so it suffices to find one $J$ with no such surjection and it'll be done. However it seems that the answer won't really depend on $J$.
Thoughts : I think there is no surjective morphism, and so was trying to find a contradiction. Of course such a morphism is split, so I was trying to analyze summands of $mathbbZ^I$ but got essentially nowhere.
I also tried studying maps $mathbbZ^I to mathbbZ$, to study the projections. I know that if such a map vanishes on almost zero sequences, then it vanishes, so I was trying to prove that it would do that here, but with no success.
I don't really know how to proceed further.
EDIT: the link that was given in the comments can help. Indeed, if there were such a surjection, then by passing to hom's into $mathbbZ$, we would have an injection $mathbbZ^J to hom (mathbbZ^I, mathbbZ)$. Now I don't know how general the quoted Baer's result is, but if it generalizes to any exponent, then this would provide an injection $mathbbZ^Jto mathbbZ^(I)$, which is clearly contradictory, by looking for instance at "almost" $2^infty$-divisible elements of $mathbbZ^J$ (if this isn't clear I can sketch the proof here, but it's not that complicated)
Alternatively, if Baer's result isn't that general, it might be possible to use some trickery to reduce to $I= mathbbN$, at least when assuming that $J=mathbbN$.
abstract-algebra group-theory abelian-groups
abstract-algebra group-theory abelian-groups
asked Apr 1 at 15:50
MaxMax
16.1k11144
asked Apr 1 at 15:50
MaxMax
16.1k11144
edited Apr 1 at 20:16
Max
asked Apr 1 at 15:50
MaxMax
16.1k11144
asked Apr 1 at 15:50
MaxMax
16.1k11144
asked Apr 1 at 15:50
MaxMax
16.1k11144
16.1k11144
1
$begingroup$
Have you tried looking at the problem from a categorical perspective?
$endgroup$
– Shaun
Apr 1 at 16:03
1
$begingroup$
You defined $mathbbZ^(J)$, but not $mathbbZ^I$. Are they the same thing, or do your brackets have a meaning?
$endgroup$
– user1729
Apr 1 at 16:07
$begingroup$
@Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though
$endgroup$
– Max
Apr 1 at 16:09
1
$begingroup$
Okay (but call it "the direct product" or something, rather than just "the power" :-p)
$endgroup$
– user1729
Apr 1 at 16:13
2
$begingroup$
@user1729 : it is split because $mathbbZ^(I)$ is free
$endgroup$
– Max
Apr 1 at 16:27
|
show 6 more comments
1
$begingroup$
Have you tried looking at the problem from a categorical perspective?
$endgroup$
– Shaun
Apr 1 at 16:03
1
$begingroup$
You defined $mathbbZ^(J)$, but not $mathbbZ^I$. Are they the same thing, or do your brackets have a meaning?
$endgroup$
– user1729
Apr 1 at 16:07
$begingroup$
@Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though
$endgroup$
– Max
Apr 1 at 16:09
1
$begingroup$
Okay (but call it "the direct product" or something, rather than just "the power" :-p)
$endgroup$
– user1729
Apr 1 at 16:13
2
$begingroup$
@user1729 : it is split because $mathbbZ^(I)$ is free
$endgroup$
– Max
Apr 1 at 16:27
1
1
$begingroup$
Have you tried looking at the problem from a categorical perspective?
$endgroup$
– Shaun
Apr 1 at 16:03
$begingroup$
Have you tried looking at the problem from a categorical perspective?
$endgroup$
– Shaun
Apr 1 at 16:03
1
1
$begingroup$
You defined $mathbbZ^(J)$, but not $mathbbZ^I$. Are they the same thing, or do your brackets have a meaning?
$endgroup$
– user1729
Apr 1 at 16:07
$begingroup$
You defined $mathbbZ^(J)$, but not $mathbbZ^I$. Are they the same thing, or do your brackets have a meaning?
$endgroup$
– user1729
Apr 1 at 16:07
$begingroup$
@Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though
$endgroup$
– Max
Apr 1 at 16:09
$begingroup$
@Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though
$endgroup$
– Max
Apr 1 at 16:09
1
1
$begingroup$
Okay (but call it "the direct product" or something, rather than just "the power" :-p)
$endgroup$
– user1729
Apr 1 at 16:13
$begingroup$
Okay (but call it "the direct product" or something, rather than just "the power" :-p)
$endgroup$
– user1729
Apr 1 at 16:13
2
2
$begingroup$
@user1729 : it is split because $mathbbZ^(I)$ is free
$endgroup$
– Max
Apr 1 at 16:27
$begingroup$
@user1729 : it is split because $mathbbZ^(I)$ is free
$endgroup$
– Max
Apr 1 at 16:27
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
By the Łoś–Eda Theorem, $operatornameHom(mathbbZ^I,mathbbZ)$ is a free abelian group for any set $I$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $I$). If a surjective homomorphism $mathbbZ^ItomathbbZ^(J)$ existed, it would induce an injective homomorphism $operatornameHom(mathbbZ^(J),mathbbZ)tooperatornameHom(mathbbZ^I,mathbbZ)$, and so $operatornameHom(mathbbZ^(J),mathbbZ)$ would be free since $operatornameHom(mathbbZ^I,mathbbZ)$ is free. But $operatornameHom(mathbbZ^(J),mathbbZ)congmathbbZ^J$ is not free if $J$ is infinite, so this is a contradiction.
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
By the Łoś–Eda Theorem, $operatornameHom(mathbbZ^I,mathbbZ)$ is a free abelian group for any set $I$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $I$). If a surjective homomorphism $mathbbZ^ItomathbbZ^(J)$ existed, it would induce an injective homomorphism $operatornameHom(mathbbZ^(J),mathbbZ)tooperatornameHom(mathbbZ^I,mathbbZ)$, and so $operatornameHom(mathbbZ^(J),mathbbZ)$ would be free since $operatornameHom(mathbbZ^I,mathbbZ)$ is free. But $operatornameHom(mathbbZ^(J),mathbbZ)congmathbbZ^J$ is not free if $J$ is infinite, so this is a contradiction.
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
|
show 1 more comment
$begingroup$
By the Łoś–Eda Theorem, $operatornameHom(mathbbZ^I,mathbbZ)$ is a free abelian group for any set $I$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $I$). If a surjective homomorphism $mathbbZ^ItomathbbZ^(J)$ existed, it would induce an injective homomorphism $operatornameHom(mathbbZ^(J),mathbbZ)tooperatornameHom(mathbbZ^I,mathbbZ)$, and so $operatornameHom(mathbbZ^(J),mathbbZ)$ would be free since $operatornameHom(mathbbZ^I,mathbbZ)$ is free. But $operatornameHom(mathbbZ^(J),mathbbZ)congmathbbZ^J$ is not free if $J$ is infinite, so this is a contradiction.
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
|
show 1 more comment
$begingroup$
By the Łoś–Eda Theorem, $operatornameHom(mathbbZ^I,mathbbZ)$ is a free abelian group for any set $I$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $I$). If a surjective homomorphism $mathbbZ^ItomathbbZ^(J)$ existed, it would induce an injective homomorphism $operatornameHom(mathbbZ^(J),mathbbZ)tooperatornameHom(mathbbZ^I,mathbbZ)$, and so $operatornameHom(mathbbZ^(J),mathbbZ)$ would be free since $operatornameHom(mathbbZ^I,mathbbZ)$ is free. But $operatornameHom(mathbbZ^(J),mathbbZ)congmathbbZ^J$ is not free if $J$ is infinite, so this is a contradiction.
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
$endgroup$
By the Łoś–Eda Theorem, $operatornameHom(mathbbZ^I,mathbbZ)$ is a free abelian group for any set $I$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $I$). If a surjective homomorphism $mathbbZ^ItomathbbZ^(J)$ existed, it would induce an injective homomorphism $operatornameHom(mathbbZ^(J),mathbbZ)tooperatornameHom(mathbbZ^I,mathbbZ)$, and so $operatornameHom(mathbbZ^(J),mathbbZ)$ would be free since $operatornameHom(mathbbZ^I,mathbbZ)$ is free. But $operatornameHom(mathbbZ^(J),mathbbZ)congmathbbZ^J$ is not free if $J$ is infinite, so this is a contradiction.
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
answered Apr 1 at 21:14
Eric WofseyEric Wofsey
192k14220352
192k14220352
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
|
show 1 more comment
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
I have a problem : this paper seems to state that $hom (mathbbZ^I, mathbbZ) = mathbbZ^(I)$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well)
$endgroup$
– Max
Apr 1 at 21:19
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
What paper are you referring to?
$endgroup$
– Eric Wofsey
Apr 1 at 21:20
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/…
$endgroup$
– Max
Apr 1 at 21:25
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals.
$endgroup$
– Max
Apr 1 at 21:40
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
$begingroup$
But I thought I had a proof of the result stated there so I'll have to find where it goes wrong
$endgroup$
– Max
Apr 1 at 21:41
|
show 1 more comment
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1
$begingroup$
Have you tried looking at the problem from a categorical perspective?
$endgroup$
– Shaun
Apr 1 at 16:03
1
$begingroup$
You defined $mathbbZ^(J)$, but not $mathbbZ^I$. Are they the same thing, or do your brackets have a meaning?
$endgroup$
– user1729
Apr 1 at 16:07
$begingroup$
@Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though
$endgroup$
– Max
Apr 1 at 16:09
1
$begingroup$
Okay (but call it "the direct product" or something, rather than just "the power" :-p)
$endgroup$
– user1729
Apr 1 at 16:13
2
$begingroup$
@user1729 : it is split because $mathbbZ^(I)$ is free
$endgroup$
– Max
Apr 1 at 16:27