Mixing
Normal subgroup and corresponding homomorphism
$begingroup$
Topics in Algebra, a book written by Herstein, said the following thing on the first isomorphism theory:
[The book says] Theorem 2.7.1 is important, for it tells us precisely what groups can
be expected to arise as homomorphic images of a given group. These
must be expressible in the form $G/K$, where $K$ is normal in $G$.
But, by Lemma 2.7.1, for any normal subgroup $N$ of $G$, $G/N$ is a
homomorphic image of $G$. Thus there is a one-to-one correspondence
between homomorphic images of $G$ and normal subgroups of $G$.
(In that book, Theorem 2.7.1 refers to the first isomorphism theorem, and Lemma 2.7.1 says for any normal subgroup N of G, G/N is a homomorphic image of G)
Would you check my following interpretation of Herstein's sentence?
[My interpretation] Let $Omega={N|Ntriangleleft G}$ and $Pi=mathrm{Hom}(G)/cong$,
where $mathrm{Hom}(G)$ denotes collection of homomorphism from $G$ to
somewhere and $cong$ denotes isomorphism relation of groups. Then,
there exists one-to-one correspondence $phi:Omegani Nmapsto {finmathrm{Hom}(G),|,f(G)cong G/N}inPi$.
I found a similar question but I couldn't understand this. I am just a beginner of abstract math, so I'm sorry in advance if I use some symbols and notations strangely.
abstract-algebra group-theory normal-subgroups group-isomorphism group-homomorphism
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
$endgroup$
add a comment |
$begingroup$
Topics in Algebra, a book written by Herstein, said the following thing on the first isomorphism theory:
[The book says] Theorem 2.7.1 is important, for it tells us precisely what groups can
be expected to arise as homomorphic images of a given group. These
must be expressible in the form $G/K$, where $K$ is normal in $G$.
But, by Lemma 2.7.1, for any normal subgroup $N$ of $G$, $G/N$ is a
homomorphic image of $G$. Thus there is a one-to-one correspondence
between homomorphic images of $G$ and normal subgroups of $G$.
(In that book, Theorem 2.7.1 refers to the first isomorphism theorem, and Lemma 2.7.1 says for any normal subgroup N of G, G/N is a homomorphic image of G)
Would you check my following interpretation of Herstein's sentence?
[My interpretation] Let $Omega={N|Ntriangleleft G}$ and $Pi=mathrm{Hom}(G)/cong$,
where $mathrm{Hom}(G)$ denotes collection of homomorphism from $G$ to
somewhere and $cong$ denotes isomorphism relation of groups. Then,
there exists one-to-one correspondence $phi:Omegani Nmapsto {finmathrm{Hom}(G),|,f(G)cong G/N}inPi$.
I found a similar question but I couldn't understand this. I am just a beginner of abstract math, so I'm sorry in advance if I use some symbols and notations strangely.
abstract-algebra group-theory normal-subgroups group-isomorphism group-homomorphism
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
$endgroup$
4
$begingroup$
Nothing against your interpretation, but I think the sentence "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$" is just short and clear. Yours is a bit more difficult.
$endgroup$
– Dietrich Burde
Jan 19 at 20:05
$begingroup$
@DietrichBurde Thanks for your comment. After thinking bit more, I came to suspect that my interpretaion was actually wrong. For example, let $A={(x, y)in R^2|y=0}$ and $B={(x, y)in R^2| x=0}$ be groups about usual vector's addition. In this case, $Atriangleleft R^2$ and $Btriangleleft R^2$, and $R^2/Acong R^2/B$. Thus, by the map $phi$ that I defined earlier, $A$ and $B$ send to the equal element of $Pi$, which means $phi$ is no longer be one-to-one.
$endgroup$
– orematasaburou
Jan 20 at 14:09
$begingroup$
@DietrichBurde Instead, I think the follwing is correct. Let $Omega={N|Ntriangleleft G}$, $Omega^{prime}=Omega/cong$, and $Pi=mathrm{Hom}(G)/cong$. Then, there exists one-to-one correspondence $phi:Omega^{prime}toPi$. I would be grateful if you give me comment on this.
$endgroup$
– orematasaburou
Jan 20 at 14:10
add a comment |
$begingroup$
Topics in Algebra, a book written by Herstein, said the following thing on the first isomorphism theory:
[The book says] Theorem 2.7.1 is important, for it tells us precisely what groups can
be expected to arise as homomorphic images of a given group. These
must be expressible in the form $G/K$, where $K$ is normal in $G$.
But, by Lemma 2.7.1, for any normal subgroup $N$ of $G$, $G/N$ is a
homomorphic image of $G$. Thus there is a one-to-one correspondence
between homomorphic images of $G$ and normal subgroups of $G$.
(In that book, Theorem 2.7.1 refers to the first isomorphism theorem, and Lemma 2.7.1 says for any normal subgroup N of G, G/N is a homomorphic image of G)
Would you check my following interpretation of Herstein's sentence?
[My interpretation] Let $Omega={N|Ntriangleleft G}$ and $Pi=mathrm{Hom}(G)/cong$,
where $mathrm{Hom}(G)$ denotes collection of homomorphism from $G$ to
somewhere and $cong$ denotes isomorphism relation of groups. Then,
there exists one-to-one correspondence $phi:Omegani Nmapsto {finmathrm{Hom}(G),|,f(G)cong G/N}inPi$.
I found a similar question but I couldn't understand this. I am just a beginner of abstract math, so I'm sorry in advance if I use some symbols and notations strangely.
abstract-algebra group-theory normal-subgroups group-isomorphism group-homomorphism
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
$endgroup$
Topics in Algebra, a book written by Herstein, said the following thing on the first isomorphism theory:
[The book says] Theorem 2.7.1 is important, for it tells us precisely what groups can
be expected to arise as homomorphic images of a given group. These
must be expressible in the form $G/K$, where $K$ is normal in $G$.
But, by Lemma 2.7.1, for any normal subgroup $N$ of $G$, $G/N$ is a
homomorphic image of $G$. Thus there is a one-to-one correspondence
between homomorphic images of $G$ and normal subgroups of $G$.
(In that book, Theorem 2.7.1 refers to the first isomorphism theorem, and Lemma 2.7.1 says for any normal subgroup N of G, G/N is a homomorphic image of G)
Would you check my following interpretation of Herstein's sentence?
[My interpretation] Let $Omega={N|Ntriangleleft G}$ and $Pi=mathrm{Hom}(G)/cong$,
where $mathrm{Hom}(G)$ denotes collection of homomorphism from $G$ to
somewhere and $cong$ denotes isomorphism relation of groups. Then,
there exists one-to-one correspondence $phi:Omegani Nmapsto {finmathrm{Hom}(G),|,f(G)cong G/N}inPi$.
I found a similar question but I couldn't understand this. I am just a beginner of abstract math, so I'm sorry in advance if I use some symbols and notations strangely.
abstract-algebra group-theory normal-subgroups group-isomorphism group-homomorphism
abstract-algebra group-theory normal-subgroups group-isomorphism group-homomorphism
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
asked Jan 19 at 19:55
orematasaburouorematasaburou
447
447
4
$begingroup$
Nothing against your interpretation, but I think the sentence "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$" is just short and clear. Yours is a bit more difficult.
$endgroup$
– Dietrich Burde
Jan 19 at 20:05
$begingroup$
@DietrichBurde Thanks for your comment. After thinking bit more, I came to suspect that my interpretaion was actually wrong. For example, let $A={(x, y)in R^2|y=0}$ and $B={(x, y)in R^2| x=0}$ be groups about usual vector's addition. In this case, $Atriangleleft R^2$ and $Btriangleleft R^2$, and $R^2/Acong R^2/B$. Thus, by the map $phi$ that I defined earlier, $A$ and $B$ send to the equal element of $Pi$, which means $phi$ is no longer be one-to-one.
$endgroup$
– orematasaburou
Jan 20 at 14:09
$begingroup$
@DietrichBurde Instead, I think the follwing is correct. Let $Omega={N|Ntriangleleft G}$, $Omega^{prime}=Omega/cong$, and $Pi=mathrm{Hom}(G)/cong$. Then, there exists one-to-one correspondence $phi:Omega^{prime}toPi$. I would be grateful if you give me comment on this.
$endgroup$
– orematasaburou
Jan 20 at 14:10
add a comment |
4
$begingroup$
Nothing against your interpretation, but I think the sentence "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$" is just short and clear. Yours is a bit more difficult.
$endgroup$
– Dietrich Burde
Jan 19 at 20:05
$begingroup$
@DietrichBurde Thanks for your comment. After thinking bit more, I came to suspect that my interpretaion was actually wrong. For example, let $A={(x, y)in R^2|y=0}$ and $B={(x, y)in R^2| x=0}$ be groups about usual vector's addition. In this case, $Atriangleleft R^2$ and $Btriangleleft R^2$, and $R^2/Acong R^2/B$. Thus, by the map $phi$ that I defined earlier, $A$ and $B$ send to the equal element of $Pi$, which means $phi$ is no longer be one-to-one.
$endgroup$
– orematasaburou
Jan 20 at 14:09
$begingroup$
@DietrichBurde Instead, I think the follwing is correct. Let $Omega={N|Ntriangleleft G}$, $Omega^{prime}=Omega/cong$, and $Pi=mathrm{Hom}(G)/cong$. Then, there exists one-to-one correspondence $phi:Omega^{prime}toPi$. I would be grateful if you give me comment on this.
$endgroup$
– orematasaburou
Jan 20 at 14:10
4
4
$begingroup$
Nothing against your interpretation, but I think the sentence "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$" is just short and clear. Yours is a bit more difficult.
$endgroup$
– Dietrich Burde
Jan 19 at 20:05
$begingroup$
Nothing against your interpretation, but I think the sentence "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$" is just short and clear. Yours is a bit more difficult.
$endgroup$
– Dietrich Burde
Jan 19 at 20:05
$begingroup$
@DietrichBurde Thanks for your comment. After thinking bit more, I came to suspect that my interpretaion was actually wrong. For example, let $A={(x, y)in R^2|y=0}$ and $B={(x, y)in R^2| x=0}$ be groups about usual vector's addition. In this case, $Atriangleleft R^2$ and $Btriangleleft R^2$, and $R^2/Acong R^2/B$. Thus, by the map $phi$ that I defined earlier, $A$ and $B$ send to the equal element of $Pi$, which means $phi$ is no longer be one-to-one.
$endgroup$
– orematasaburou
Jan 20 at 14:09
$begingroup$
@DietrichBurde Thanks for your comment. After thinking bit more, I came to suspect that my interpretaion was actually wrong. For example, let $A={(x, y)in R^2|y=0}$ and $B={(x, y)in R^2| x=0}$ be groups about usual vector's addition. In this case, $Atriangleleft R^2$ and $Btriangleleft R^2$, and $R^2/Acong R^2/B$. Thus, by the map $phi$ that I defined earlier, $A$ and $B$ send to the equal element of $Pi$, which means $phi$ is no longer be one-to-one.
$endgroup$
– orematasaburou
Jan 20 at 14:09
$begingroup$
@DietrichBurde Instead, I think the follwing is correct. Let $Omega={N|Ntriangleleft G}$, $Omega^{prime}=Omega/cong$, and $Pi=mathrm{Hom}(G)/cong$. Then, there exists one-to-one correspondence $phi:Omega^{prime}toPi$. I would be grateful if you give me comment on this.
$endgroup$
– orematasaburou
Jan 20 at 14:10
$begingroup$
@DietrichBurde Instead, I think the follwing is correct. Let $Omega={N|Ntriangleleft G}$, $Omega^{prime}=Omega/cong$, and $Pi=mathrm{Hom}(G)/cong$. Then, there exists one-to-one correspondence $phi:Omega^{prime}toPi$. I would be grateful if you give me comment on this.
$endgroup$
– orematasaburou
Jan 20 at 14:10
add a comment |
0
active
oldest
votes
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%2f3079745%2fnormal-subgroup-and-corresponding-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3079745%2fnormal-subgroup-and-corresponding-homomorphism%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
4
$begingroup$
Nothing against your interpretation, but I think the sentence "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$" is just short and clear. Yours is a bit more difficult.
$endgroup$
– Dietrich Burde
Jan 19 at 20:05
$begingroup$
@DietrichBurde Thanks for your comment. After thinking bit more, I came to suspect that my interpretaion was actually wrong. For example, let $A={(x, y)in R^2|y=0}$ and $B={(x, y)in R^2| x=0}$ be groups about usual vector's addition. In this case, $Atriangleleft R^2$ and $Btriangleleft R^2$, and $R^2/Acong R^2/B$. Thus, by the map $phi$ that I defined earlier, $A$ and $B$ send to the equal element of $Pi$, which means $phi$ is no longer be one-to-one.
$endgroup$
– orematasaburou
Jan 20 at 14:09
$begingroup$
@DietrichBurde Instead, I think the follwing is correct. Let $Omega={N|Ntriangleleft G}$, $Omega^{prime}=Omega/cong$, and $Pi=mathrm{Hom}(G)/cong$. Then, there exists one-to-one correspondence $phi:Omega^{prime}toPi$. I would be grateful if you give me comment on this.
$endgroup$
– orematasaburou
Jan 20 at 14:10