How to find a homomorphic map in following question?
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Let $S1$ and $S2$ be two sets. Suppose that there exists a one-to-one
mapping $J$ of $S1$ into $S2$ . Show that there exists an isomorphism of
$A(S1)$ into $A(S2)$, where $A(S)$ means the set of all one-to-one mappings
of $S$ onto itself.
I am not able to find the homomorphic map because $J$ is not necessarily onto.If $J$ was onto we have define a map in which each symbol in an element $x$ belonging to $A(S1)$ could be replaced by corresponding in S2 by using the map $J$.
This Question is from Herstein 2.7.21 .
group-theory group-isomorphism group-homomorphism
asked Nov 13 at 16:36
Amit
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Let $S1$ and $S2$ be two sets. Suppose that there exists a one-to-one
mapping $J$ of $S1$ into $S2$ . Show that there exists an isomorphism of
$A(S1)$ into $A(S2)$, where $A(S)$ means the set of all one-to-one mappings
of $S$ onto itself.
I am not able to find the homomorphic map because $J$ is not necessarily onto.If $J$ was onto we have define a map in which each symbol in an element $x$ belonging to $A(S1)$ could be replaced by corresponding in S2 by using the map $J$.
This Question is from Herstein 2.7.21 .
group-theory group-isomorphism group-homomorphism
asked Nov 13 at 16:36
Amit
297
New contributor
Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think you are right that it is not true, even with finite sets, if J is not onto. I would assume that J is onto and then do it. Also, sometimes people write "a one-to-one mapping between sets" to mean a bijective map, rather than just an injective map.
– Keshav
Nov 13 at 17:19
@Keshav This Question is from Herstein 2.7.21 . No i am sure it's not bijective
– Amit
Nov 13 at 17:26
You need to check Herstein's definition of isomorphism. I think "isomorphism of $A(S_1)$ into $A(S_2)$" mean an injective homomorphism from $A(S_1)$ to $A(S_2)$. There is a bijection from $S_1$ to a subset of $T$ of $S_2$, which enables you to construct a bijective isomorphism from $A(S_1)$ to $A(T)$. Then you can embed $A(T)$ into $A(S_2)$.
– Derek Holt
Nov 13 at 17:50
@DerekHolt I know .But i am not able to do so.Still trying
– Amit
Nov 13 at 18:00
@DerekHolt Can you tell me how to embed .
– Amit
Nov 13 at 18:02
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $S1$ and $S2$ be two sets. Suppose that there exists a one-to-one
mapping $J$ of $S1$ into $S2$ . Show that there exists an isomorphism of
$A(S1)$ into $A(S2)$, where $A(S)$ means the set of all one-to-one mappings
of $S$ onto itself.
I am not able to find the homomorphic map because $J$ is not necessarily onto.If $J$ was onto we have define a map in which each symbol in an element $x$ belonging to $A(S1)$ could be replaced by corresponding in S2 by using the map $J$.
This Question is from Herstein 2.7.21 .
group-theory group-isomorphism group-homomorphism
asked Nov 13 at 16:36
Amit
297
New contributor
Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $S1$ and $S2$ be two sets. Suppose that there exists a one-to-one
mapping $J$ of $S1$ into $S2$ . Show that there exists an isomorphism of
$A(S1)$ into $A(S2)$, where $A(S)$ means the set of all one-to-one mappings
of $S$ onto itself.
I am not able to find the homomorphic map because $J$ is not necessarily onto.If $J$ was onto we have define a map in which each symbol in an element $x$ belonging to $A(S1)$ could be replaced by corresponding in S2 by using the map $J$.
This Question is from Herstein 2.7.21 .
group-theory group-isomorphism group-homomorphism
group-theory group-isomorphism group-homomorphism
asked Nov 13 at 16:36
Amit
297
New contributor
Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 13 at 16:36
Amit
297
New contributor
Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 13 at 17:26
asked Nov 13 at 16:36
Amit
297
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Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 13 at 16:36
Amit
297
asked Nov 13 at 16:36
Amit
297
297
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Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Amit is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I think you are right that it is not true, even with finite sets, if J is not onto. I would assume that J is onto and then do it. Also, sometimes people write "a one-to-one mapping between sets" to mean a bijective map, rather than just an injective map.
– Keshav
Nov 13 at 17:19
@Keshav This Question is from Herstein 2.7.21 . No i am sure it's not bijective
– Amit
Nov 13 at 17:26
You need to check Herstein's definition of isomorphism. I think "isomorphism of $A(S_1)$ into $A(S_2)$" mean an injective homomorphism from $A(S_1)$ to $A(S_2)$. There is a bijection from $S_1$ to a subset of $T$ of $S_2$, which enables you to construct a bijective isomorphism from $A(S_1)$ to $A(T)$. Then you can embed $A(T)$ into $A(S_2)$.
– Derek Holt
Nov 13 at 17:50
@DerekHolt I know .But i am not able to do so.Still trying
– Amit
Nov 13 at 18:00
@DerekHolt Can you tell me how to embed .
– Amit
Nov 13 at 18:02
|
show 1 more comment
I think you are right that it is not true, even with finite sets, if J is not onto. I would assume that J is onto and then do it. Also, sometimes people write "a one-to-one mapping between sets" to mean a bijective map, rather than just an injective map.
– Keshav
Nov 13 at 17:19
@Keshav This Question is from Herstein 2.7.21 . No i am sure it's not bijective
– Amit
Nov 13 at 17:26
You need to check Herstein's definition of isomorphism. I think "isomorphism of $A(S_1)$ into $A(S_2)$" mean an injective homomorphism from $A(S_1)$ to $A(S_2)$. There is a bijection from $S_1$ to a subset of $T$ of $S_2$, which enables you to construct a bijective isomorphism from $A(S_1)$ to $A(T)$. Then you can embed $A(T)$ into $A(S_2)$.
– Derek Holt
Nov 13 at 17:50
@DerekHolt I know .But i am not able to do so.Still trying
– Amit
Nov 13 at 18:00
@DerekHolt Can you tell me how to embed .
– Amit
Nov 13 at 18:02
I think you are right that it is not true, even with finite sets, if J is not onto. I would assume that J is onto and then do it. Also, sometimes people write "a one-to-one mapping between sets" to mean a bijective map, rather than just an injective map.
– Keshav
Nov 13 at 17:19
I think you are right that it is not true, even with finite sets, if J is not onto. I would assume that J is onto and then do it. Also, sometimes people write "a one-to-one mapping between sets" to mean a bijective map, rather than just an injective map.
– Keshav
Nov 13 at 17:19
@Keshav This Question is from Herstein 2.7.21 . No i am sure it's not bijective
– Amit
Nov 13 at 17:26
@Keshav This Question is from Herstein 2.7.21 . No i am sure it's not bijective
– Amit
Nov 13 at 17:26
You need to check Herstein's definition of isomorphism. I think "isomorphism of $A(S_1)$ into $A(S_2)$" mean an injective homomorphism from $A(S_1)$ to $A(S_2)$. There is a bijection from $S_1$ to a subset of $T$ of $S_2$, which enables you to construct a bijective isomorphism from $A(S_1)$ to $A(T)$. Then you can embed $A(T)$ into $A(S_2)$.
– Derek Holt
Nov 13 at 17:50
You need to check Herstein's definition of isomorphism. I think "isomorphism of $A(S_1)$ into $A(S_2)$" mean an injective homomorphism from $A(S_1)$ to $A(S_2)$. There is a bijection from $S_1$ to a subset of $T$ of $S_2$, which enables you to construct a bijective isomorphism from $A(S_1)$ to $A(T)$. Then you can embed $A(T)$ into $A(S_2)$.
– Derek Holt
Nov 13 at 17:50
@DerekHolt I know .But i am not able to do so.Still trying
– Amit
Nov 13 at 18:00
@DerekHolt I know .But i am not able to do so.Still trying
– Amit
Nov 13 at 18:00
@DerekHolt Can you tell me how to embed .
– Amit
Nov 13 at 18:02
@DerekHolt Can you tell me how to embed .
– Amit
Nov 13 at 18:02
|
show 1 more comment
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I think you are right that it is not true, even with finite sets, if J is not onto. I would assume that J is onto and then do it. Also, sometimes people write "a one-to-one mapping between sets" to mean a bijective map, rather than just an injective map.
– Keshav
Nov 13 at 17:19
@Keshav This Question is from Herstein 2.7.21 . No i am sure it's not bijective
– Amit
Nov 13 at 17:26
You need to check Herstein's definition of isomorphism. I think "isomorphism of $A(S_1)$ into $A(S_2)$" mean an injective homomorphism from $A(S_1)$ to $A(S_2)$. There is a bijection from $S_1$ to a subset of $T$ of $S_2$, which enables you to construct a bijective isomorphism from $A(S_1)$ to $A(T)$. Then you can embed $A(T)$ into $A(S_2)$.
– Derek Holt
Nov 13 at 17:50
@DerekHolt I know .But i am not able to do so.Still trying
– Amit
Nov 13 at 18:00
@DerekHolt Can you tell me how to embed .
– Amit
Nov 13 at 18:02