Is there a canonical “inverse” of Abelianization? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)The torsion submodule of $prod mathbbZ_p$ is not a direct summand of $prod mathbbZ_p$If $G, H, K$ are divisible abelian groups and $G oplus H cong G oplus K$ then $H cong K$Divisible Direct Sum or Direct ProductIf $G$ is a non-cyclic group of order $n^2$, then $G$ is isomorphic to $mathbbZ_n oplus mathbbZ_n$Commutative Monoid and Invertible ElementsHow many abelian groups are there of order $p^5q^4$Intuition on Rational Canonical FormNotation for multiple direct sumReference Request: Abelianization of free product is the direct sum of abelianizationsWhat is the abelianization of $langle x,y,zmid x^2=y^2z^2rangle?$
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Is there a canonical “inverse” of Abelianization?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)The torsion submodule of $prod mathbbZ_p$ is not a direct summand of $prod mathbbZ_p$If $G, H, K$ are divisible abelian groups and $G oplus H cong G oplus K$ then $H cong K$Divisible Direct Sum or Direct ProductIf $G$ is a non-cyclic group of order $n^2$, then $G$ is isomorphic to $mathbbZ_n oplus mathbbZ_n$Commutative Monoid and Invertible ElementsHow many abelian groups are there of order $p^5q^4$Intuition on Rational Canonical FormNotation for multiple direct sumReference Request: Abelianization of free product is the direct sum of abelianizationsWhat is the abelianization of $langle x,y,zmid x^2=y^2z^2rangle?$
$begingroup$
We know that the abelianization of a free product is the direct sum, for example $Ab(mathbbZ*mathbbZ)=mathbbZoplusmathbbZ$.
Is there a “canonical” (or even non-canonical) operator that reverses the process of Abelianization?
(For instance, it may goes from direct sum to free product?)
Thanks.
abstract-algebra group-theory
asked 3 hours ago
yoyosteinyoyostein
8,187104074
$endgroup$
add a comment |
$begingroup$
We know that the abelianization of a free product is the direct sum, for example $Ab(mathbbZ*mathbbZ)=mathbbZoplusmathbbZ$.
Is there a “canonical” (or even non-canonical) operator that reverses the process of Abelianization?
(For instance, it may goes from direct sum to free product?)
Thanks.
abstract-algebra group-theory
asked 3 hours ago
yoyosteinyoyostein
8,187104074
$endgroup$
$begingroup$
If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way.
$endgroup$
– KCd
2 hours ago
add a comment |
$begingroup$
We know that the abelianization of a free product is the direct sum, for example $Ab(mathbbZ*mathbbZ)=mathbbZoplusmathbbZ$.
Is there a “canonical” (or even non-canonical) operator that reverses the process of Abelianization?
(For instance, it may goes from direct sum to free product?)
Thanks.
abstract-algebra group-theory
asked 3 hours ago
yoyosteinyoyostein
8,187104074
$endgroup$
We know that the abelianization of a free product is the direct sum, for example $Ab(mathbbZ*mathbbZ)=mathbbZoplusmathbbZ$.
Is there a “canonical” (or even non-canonical) operator that reverses the process of Abelianization?
(For instance, it may goes from direct sum to free product?)
Thanks.
abstract-algebra group-theory
abstract-algebra group-theory
asked 3 hours ago
yoyosteinyoyostein
8,187104074
asked 3 hours ago
yoyosteinyoyostein
8,187104074
asked 3 hours ago
yoyosteinyoyostein
8,187104074
asked 3 hours ago
yoyosteinyoyostein
8,187104074
asked 3 hours ago
yoyosteinyoyostein
8,187104074
8,187104074
$begingroup$
If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way.
$endgroup$
– KCd
2 hours ago
add a comment |
$begingroup$
If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way.
$endgroup$
– KCd
2 hours ago
$begingroup$
If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way.
$endgroup$
– KCd
2 hours ago
$begingroup$
If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way.
$endgroup$
– KCd
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The abelianization functor, from the category of groups to the category of abelian groups, has a right adjoint, which is the forgetful functor, in other words the functor that sends an abelian group $A$ to the group $A$ itself. I guess this is as close to an "inverse" as you can reasonably hope.
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
$endgroup$
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The abelianization functor, from the category of groups to the category of abelian groups, has a right adjoint, which is the forgetful functor, in other words the functor that sends an abelian group $A$ to the group $A$ itself. I guess this is as close to an "inverse" as you can reasonably hope.
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
$endgroup$
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
add a comment |
$begingroup$
The abelianization functor, from the category of groups to the category of abelian groups, has a right adjoint, which is the forgetful functor, in other words the functor that sends an abelian group $A$ to the group $A$ itself. I guess this is as close to an "inverse" as you can reasonably hope.
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
$endgroup$
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
add a comment |
$begingroup$
The abelianization functor, from the category of groups to the category of abelian groups, has a right adjoint, which is the forgetful functor, in other words the functor that sends an abelian group $A$ to the group $A$ itself. I guess this is as close to an "inverse" as you can reasonably hope.
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
$endgroup$
The abelianization functor, from the category of groups to the category of abelian groups, has a right adjoint, which is the forgetful functor, in other words the functor that sends an abelian group $A$ to the group $A$ itself. I guess this is as close to an "inverse" as you can reasonably hope.
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
answered 2 hours ago
Captain LamaCaptain Lama
10.5k1030
10.5k1030
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
add a comment |
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
$begingroup$
Thanks for this answer.
$endgroup$
– yoyostein
2 hours ago
add a comment |
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$begingroup$
If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way.
$endgroup$
– KCd
2 hours ago