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
9,620113684
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
$endgroup$
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
9,620113684
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
$endgroup$
$begingroup$
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
9,620113684
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
$endgroup$
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
9,620113684
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
edited Dec 29 '18 at 13:16
Shaun
9,620113684
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
edited Dec 29 '18 at 13:16
Shaun
9,620113684
edited Dec 29 '18 at 13:16
Shaun
9,620113684
edited Dec 29 '18 at 13:16
Shaun
9,620113684
9,620113684
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
asked Dec 29 '18 at 13:10
EklavyaEklavya
972515
972515
$begingroup$
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
7,9651830
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055832%2frelation-between-set-of-normal-subgroups-and-set-of-homomorphism-images-of-a-gro%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
7,9651830
$endgroup$
$begingroup$
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
7,9651830
$endgroup$
$begingroup$
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
7,9651830
$endgroup$
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
7,9651830
answered Dec 29 '18 at 13:19
YankoYanko
7,9651830
answered Dec 29 '18 at 13:19
YankoYanko
7,9651830
answered Dec 29 '18 at 13:19
YankoYanko
7,9651830
7,9651830
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055832%2frelation-between-set-of-normal-subgroups-and-set-of-homomorphism-images-of-a-gro%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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