Mixing
On isomorphisms between $G$ and $Htimes K$
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Here is a proposition from Artin:
Usually when one wants to prove that $Gsimeq Htimes K$, one verifies the conditions of $(d)$. But if $(d)$ holds, then $G$ is isomorphic to $Htimes K$ via the specific map $(h,k)mapsto hk$. But theoretically it may happen that for $G$, the conditions in $(d)$ do not hold (and hence $G$ is not isomorphic to the product $Htimes K$ via the product map), but $G$ is isomorphic to $Htimes K$ via $(h,k)mapsto h^17k^-1hkhk^2h^-28k$ or a similar map. Is my understanding correct? If so, what is an example when the above holds? And if $(d)$ does not hold, then how do I see whether there is an isomorphism between $G$ and $Htimes K$ different from $(h,k)mapsto hk$?
abstract-algebra group-theory group-isomorphism group-homomorphism
asked Jul 16 at 19:43
user437309
556212
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up vote
1
down vote
favorite
Here is a proposition from Artin:
Usually when one wants to prove that $Gsimeq Htimes K$, one verifies the conditions of $(d)$. But if $(d)$ holds, then $G$ is isomorphic to $Htimes K$ via the specific map $(h,k)mapsto hk$. But theoretically it may happen that for $G$, the conditions in $(d)$ do not hold (and hence $G$ is not isomorphic to the product $Htimes K$ via the product map), but $G$ is isomorphic to $Htimes K$ via $(h,k)mapsto h^17k^-1hkhk^2h^-28k$ or a similar map. Is my understanding correct? If so, what is an example when the above holds? And if $(d)$ does not hold, then how do I see whether there is an isomorphism between $G$ and $Htimes K$ different from $(h,k)mapsto hk$?
abstract-algebra group-theory group-isomorphism group-homomorphism
asked Jul 16 at 19:43
user437309
556212
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is a proposition from Artin:
Usually when one wants to prove that $Gsimeq Htimes K$, one verifies the conditions of $(d)$. But if $(d)$ holds, then $G$ is isomorphic to $Htimes K$ via the specific map $(h,k)mapsto hk$. But theoretically it may happen that for $G$, the conditions in $(d)$ do not hold (and hence $G$ is not isomorphic to the product $Htimes K$ via the product map), but $G$ is isomorphic to $Htimes K$ via $(h,k)mapsto h^17k^-1hkhk^2h^-28k$ or a similar map. Is my understanding correct? If so, what is an example when the above holds? And if $(d)$ does not hold, then how do I see whether there is an isomorphism between $G$ and $Htimes K$ different from $(h,k)mapsto hk$?
abstract-algebra group-theory group-isomorphism group-homomorphism
asked Jul 16 at 19:43
user437309
556212
Here is a proposition from Artin:
Usually when one wants to prove that $Gsimeq Htimes K$, one verifies the conditions of $(d)$. But if $(d)$ holds, then $G$ is isomorphic to $Htimes K$ via the specific map $(h,k)mapsto hk$. But theoretically it may happen that for $G$, the conditions in $(d)$ do not hold (and hence $G$ is not isomorphic to the product $Htimes K$ via the product map), but $G$ is isomorphic to $Htimes K$ via $(h,k)mapsto h^17k^-1hkhk^2h^-28k$ or a similar map. Is my understanding correct? If so, what is an example when the above holds? And if $(d)$ does not hold, then how do I see whether there is an isomorphism between $G$ and $Htimes K$ different from $(h,k)mapsto hk$?
abstract-algebra group-theory group-isomorphism group-homomorphism
asked Jul 16 at 19:43
user437309
556212
asked Jul 16 at 19:43
user437309
556212
asked Jul 16 at 19:43
user437309
556212
asked Jul 16 at 19:43
user437309
556212
556212
add a comment |Â
add a comment |Â
2 Answers
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3
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If you're not using the multiplication map from $H times K$ to $G$, then $H$ and $K$ aren't really specific subgroups of $G$ anymore. They're just abstract groups that each happen to be isomorphic to some subgroup of $G$, but that identification has nothing to do with the isomorphism from $H times K$ to $G$.
For example, let $G = mathbbZ times mathbbZ$, and let $H$ and $K$ both be the first factor of $mathbbZ$. Then the multiplication map isn't injective, so it's not an isomorphism, but abstractly $G$ is still isomorphic to $H times K$.
answered Jul 17 at 6:07
Charlie Cunningham
8114
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0
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I believe what you are referring to is a semidirect product. These have the same cardinality as $H times K$, but the actual group structure is different.
https://en.wikipedia.org/wiki/Semidirect_product
answered Jul 16 at 19:52
Allen O'Hara
1086
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If you're not using the multiplication map from $H times K$ to $G$, then $H$ and $K$ aren't really specific subgroups of $G$ anymore. They're just abstract groups that each happen to be isomorphic to some subgroup of $G$, but that identification has nothing to do with the isomorphism from $H times K$ to $G$.
For example, let $G = mathbbZ times mathbbZ$, and let $H$ and $K$ both be the first factor of $mathbbZ$. Then the multiplication map isn't injective, so it's not an isomorphism, but abstractly $G$ is still isomorphic to $H times K$.
answered Jul 17 at 6:07
Charlie Cunningham
8114
add a comment |Â
up vote
3
down vote
If you're not using the multiplication map from $H times K$ to $G$, then $H$ and $K$ aren't really specific subgroups of $G$ anymore. They're just abstract groups that each happen to be isomorphic to some subgroup of $G$, but that identification has nothing to do with the isomorphism from $H times K$ to $G$.
For example, let $G = mathbbZ times mathbbZ$, and let $H$ and $K$ both be the first factor of $mathbbZ$. Then the multiplication map isn't injective, so it's not an isomorphism, but abstractly $G$ is still isomorphic to $H times K$.
answered Jul 17 at 6:07
Charlie Cunningham
8114
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If you're not using the multiplication map from $H times K$ to $G$, then $H$ and $K$ aren't really specific subgroups of $G$ anymore. They're just abstract groups that each happen to be isomorphic to some subgroup of $G$, but that identification has nothing to do with the isomorphism from $H times K$ to $G$.
For example, let $G = mathbbZ times mathbbZ$, and let $H$ and $K$ both be the first factor of $mathbbZ$. Then the multiplication map isn't injective, so it's not an isomorphism, but abstractly $G$ is still isomorphic to $H times K$.
answered Jul 17 at 6:07
Charlie Cunningham
8114
If you're not using the multiplication map from $H times K$ to $G$, then $H$ and $K$ aren't really specific subgroups of $G$ anymore. They're just abstract groups that each happen to be isomorphic to some subgroup of $G$, but that identification has nothing to do with the isomorphism from $H times K$ to $G$.
For example, let $G = mathbbZ times mathbbZ$, and let $H$ and $K$ both be the first factor of $mathbbZ$. Then the multiplication map isn't injective, so it's not an isomorphism, but abstractly $G$ is still isomorphic to $H times K$.
answered Jul 17 at 6:07
Charlie Cunningham
8114
answered Jul 17 at 6:07
Charlie Cunningham
8114
answered Jul 17 at 6:07
Charlie Cunningham
8114
answered Jul 17 at 6:07
Charlie Cunningham
8114
8114
add a comment |Â
add a comment |Â
up vote
0
down vote
I believe what you are referring to is a semidirect product. These have the same cardinality as $H times K$, but the actual group structure is different.
https://en.wikipedia.org/wiki/Semidirect_product
answered Jul 16 at 19:52
Allen O'Hara
1086
add a comment |Â
up vote
0
down vote
I believe what you are referring to is a semidirect product. These have the same cardinality as $H times K$, but the actual group structure is different.
https://en.wikipedia.org/wiki/Semidirect_product
answered Jul 16 at 19:52
Allen O'Hara
1086
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I believe what you are referring to is a semidirect product. These have the same cardinality as $H times K$, but the actual group structure is different.
https://en.wikipedia.org/wiki/Semidirect_product
answered Jul 16 at 19:52
Allen O'Hara
1086
I believe what you are referring to is a semidirect product. These have the same cardinality as $H times K$, but the actual group structure is different.
https://en.wikipedia.org/wiki/Semidirect_product
answered Jul 16 at 19:52
Allen O'Hara
1086
answered Jul 16 at 19:52
Allen O'Hara
1086
answered Jul 16 at 19:52
Allen O'Hara
1086
answered Jul 16 at 19:52
Allen O'Hara
1086
1086
add a comment |Â
add a comment |Â
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