Relation between set of Normal subgroups and set of Homomorphism images of a group $G$
$begingroup$
let $G$ be any group.let $A$ is set of all normal subgroups of $G$, and $B$ is set of all homomorphic images of $G$.
Is there any type of relation between these sets.Mean some type of bijection like that.
For any normal subgroup $N in A$ there is a homomorphic image $G/N in B$ and conversely for any homomorphic image $G' in B$ there is normal subgroup $N=kerphi in A$ where $phi$ is homomorphism from $G$ to $G'$
Is this a bijection between A and B ?? or i am doing something wrong . please help.
abstract-algebra group-theory finite-groups group-homomorphism
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
$endgroup$
add a comment |
$begingroup$
let $G$ be any group.let $A$ is set of all normal subgroups of $G$, and $B$ is set of all homomorphic images of $G$.
Is there any type of relation between these sets.Mean some type of bijection like that.
For any normal subgroup $N in A$ there is a homomorphic image $G/N in B$ and conversely for any homomorphic image $G' in B$ there is normal subgroup $N=kerphi in A$ where $phi$ is homomorphism from $G$ to $G'$
Is this a bijection between A and B ?? or i am doing something wrong . please help.
abstract-algebra group-theory finite-groups group-homomorphism
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
$endgroup$
$begingroup$
I think you need to give a more precise definition of "the set of all homomorphic images of $G$". Clearly it is possible for there to exist distinct normal subgroups $M$ and $N$ of $G$ with $G/M cong G/N$.
$endgroup$
– Derek Holt
Dec 29 '18 at 13:59
add a comment |
$begingroup$
let $G$ be any group.let $A$ is set of all normal subgroups of $G$, and $B$ is set of all homomorphic images of $G$.
Is there any type of relation between these sets.Mean some type of bijection like that.
For any normal subgroup $N in A$ there is a homomorphic image $G/N in B$ and conversely for any homomorphic image $G' in B$ there is normal subgroup $N=kerphi in A$ where $phi$ is homomorphism from $G$ to $G'$
Is this a bijection between A and B ?? or i am doing something wrong . please help.
abstract-algebra group-theory finite-groups group-homomorphism
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
$endgroup$
let $G$ be any group.let $A$ is set of all normal subgroups of $G$, and $B$ is set of all homomorphic images of $G$.
Is there any type of relation between these sets.Mean some type of bijection like that.
For any normal subgroup $N in A$ there is a homomorphic image $G/N in B$ and conversely for any homomorphic image $G' in B$ there is normal subgroup $N=kerphi in A$ where $phi$ is homomorphism from $G$ to $G'$
Is this a bijection between A and B ?? or i am doing something wrong . please help.
abstract-algebra group-theory finite-groups group-homomorphism
abstract-algebra group-theory finite-groups group-homomorphism
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
edited Dec 29 '18 at 13:16
Shaun
10.6k113687
10.6k113687
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
asked Dec 29 '18 at 13:10
EklavyaEklavya
987515
987515
$begingroup$
I think you need to give a more precise definition of "the set of all homomorphic images of $G$". Clearly it is possible for there to exist distinct normal subgroups $M$ and $N$ of $G$ with $G/M cong G/N$.
$endgroup$
– Derek Holt
Dec 29 '18 at 13:59
add a comment |
$begingroup$
I think you need to give a more precise definition of "the set of all homomorphic images of $G$". Clearly it is possible for there to exist distinct normal subgroups $M$ and $N$ of $G$ with $G/M cong G/N$.
$endgroup$
– Derek Holt
Dec 29 '18 at 13:59
$begingroup$
I think you need to give a more precise definition of "the set of all homomorphic images of $G$". Clearly it is possible for there to exist distinct normal subgroups $M$ and $N$ of $G$ with $G/M cong G/N$.
$endgroup$
– Derek Holt
Dec 29 '18 at 13:59
$begingroup$
I think you need to give a more precise definition of "the set of all homomorphic images of $G$". Clearly it is possible for there to exist distinct normal subgroups $M$ and $N$ of $G$ with $G/M cong G/N$.
$endgroup$
– Derek Holt
Dec 29 '18 at 13:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes.
Let $G$ be a group. Given a normal subgroup $N$ it is well known that $G/N$ is a group and $varphi:Grightarrow G/N$ taking $g$ to $gN$ is a homomorphism whose kernel is $N$.
On the other hand if $varphi:Grightarrow H$ is a surjective homomorphism (so $H$ is an homomorphic image of $G$. Then by the first isomorphism theorem $N:=ker varphi$ is a normal subgroup of $G$ such that $H=G/N$.
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
$endgroup$
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
add a comment |
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1 Answer
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votes
$begingroup$
Yes.
Let $G$ be a group. Given a normal subgroup $N$ it is well known that $G/N$ is a group and $varphi:Grightarrow G/N$ taking $g$ to $gN$ is a homomorphism whose kernel is $N$.
On the other hand if $varphi:Grightarrow H$ is a surjective homomorphism (so $H$ is an homomorphic image of $G$. Then by the first isomorphism theorem $N:=ker varphi$ is a normal subgroup of $G$ such that $H=G/N$.
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
$endgroup$
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
add a comment |
$begingroup$
Yes.
Let $G$ be a group. Given a normal subgroup $N$ it is well known that $G/N$ is a group and $varphi:Grightarrow G/N$ taking $g$ to $gN$ is a homomorphism whose kernel is $N$.
On the other hand if $varphi:Grightarrow H$ is a surjective homomorphism (so $H$ is an homomorphic image of $G$. Then by the first isomorphism theorem $N:=ker varphi$ is a normal subgroup of $G$ such that $H=G/N$.
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
$endgroup$
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
add a comment |
$begingroup$
Yes.
Let $G$ be a group. Given a normal subgroup $N$ it is well known that $G/N$ is a group and $varphi:Grightarrow G/N$ taking $g$ to $gN$ is a homomorphism whose kernel is $N$.
On the other hand if $varphi:Grightarrow H$ is a surjective homomorphism (so $H$ is an homomorphic image of $G$. Then by the first isomorphism theorem $N:=ker varphi$ is a normal subgroup of $G$ such that $H=G/N$.
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
$endgroup$
Yes.
Let $G$ be a group. Given a normal subgroup $N$ it is well known that $G/N$ is a group and $varphi:Grightarrow G/N$ taking $g$ to $gN$ is a homomorphism whose kernel is $N$.
On the other hand if $varphi:Grightarrow H$ is a surjective homomorphism (so $H$ is an homomorphic image of $G$. Then by the first isomorphism theorem $N:=ker varphi$ is a normal subgroup of $G$ such that $H=G/N$.
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
answered Dec 29 '18 at 13:19
YankoYanko
8,4692830
8,4692830
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
add a comment |
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
how to prove this is bijection between $A$ and $B$ ? please explain
$endgroup$
– Eklavya
Dec 29 '18 at 13:30
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
@Eklavya Take $Nin A$ by the first two lines there exists $G/Nin B$. For the other direction take $Hin B$ then by the last two lines $H=G/N$ where $Nin A$. The first map is clearly the inverse of the second (and vice versa) so this is a bijection.
$endgroup$
– Yanko
Dec 29 '18 at 13:36
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
one more thing do we seeing element of set $A$ and $B$ unique up to isomorphism ? or we treat distinct elements of set $A$ and $B$ as distinct objects?
$endgroup$
– Eklavya
Dec 29 '18 at 13:42
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
$begingroup$
$H$ is isomorphic to $G/N$. So I think it's only up to isomorphism (I'm not sure though).
$endgroup$
– Yanko
Dec 29 '18 at 13:43
add a comment |
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$begingroup$
I think you need to give a more precise definition of "the set of all homomorphic images of $G$". Clearly it is possible for there to exist distinct normal subgroups $M$ and $N$ of $G$ with $G/M cong G/N$.
$endgroup$
– Derek Holt
Dec 29 '18 at 13:59