Relation between set of Normal subgroups and set of Homomorphism images of a group $G$












1












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Dec 29 '18 at 13:16









Shaun

10.6k113687




10.6k113687










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $H=G/N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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1 Answer
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1 Answer
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0












$begingroup$

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 $H=G/N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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 $H=G/N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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 $H=G/N$.






share|cite|improve this answer









$endgroup$



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 $H=G/N$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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