Question about the proof of Second Isomorphism TheoremIsomorphism theorem and proving $f:Gto G'$ onto,...
How can we generalize the fact of finite dimensional vector space to an infinte dimensional case?
Drawing ramified coverings with tikz
Can someone explain how this makes sense electrically?
How can Trident be so inexpensive? Will it orbit Triton or just do a (slow) flyby?
Non-trope happy ending?
Question about the proof of Second Isomorphism Theorem
Longest common substring in linear time
What if a revenant (monster) gains fire resistance?
Loading commands from file
Biological Blimps: Propulsion
Are the IPv6 address space and IPv4 address space completely disjoint?
Creature in Shazam mid-credits scene?
Is "staff" singular or plural?
I am looking for the correct translation of love for the phrase "in this sign love"
why `nmap 192.168.1.97` returns less services than `nmap 127.0.0.1`?
Count the occurrence of each unique word in the file
Lowest total scrabble score
How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?
Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?
Problem with TransformedDistribution
The IT department bottlenecks progress. How should I handle this?
How to explain what's wrong with this application of the chain rule?
What should you do if you miss a job interview (deliberately)?
Is there a single word describing earning money through any means?
Question about the proof of Second Isomorphism Theorem
Isomorphism theorem and proving $f:Gto G'$ onto, $K'triangleleft G'Rightarrow G/f^{-1}(K')cong G'/K'$Interpretation of Second isomorphism theoremQuestion about second Isomorphism TheoremNeed isomorphism theorem intuitionWhy $phi(H) cong H/ kerphi$ in the Second Isomorphism Theorem?Intuition behind the first isomorphism theoremIntuition about the first isomorphism theoremIntuition about the second isomorphism theoremFundamental Isomorphism TheoremFinding the kernel of $phi$ of applying the First Isomorphism Theorem
$begingroup$
The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$
There is the proof of Abstract Algebra Thomas by W. Judson:
Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi={hin H:hin N}=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$
My question:
Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.
Thank you.
abstract-algebra group-theory group-isomorphism group-homomorphism
edited 4 hours ago
Andrews
1,2761421
asked 4 hours ago
NiaBieNiaBie
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$
There is the proof of Abstract Algebra Thomas by W. Judson:
Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi={hin H:hin N}=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$
My question:
Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.
Thank you.
abstract-algebra group-theory group-isomorphism group-homomorphism
edited 4 hours ago
Andrews
1,2761421
asked 4 hours ago
NiaBieNiaBie
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$
There is the proof of Abstract Algebra Thomas by W. Judson:
Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi={hin H:hin N}=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$
My question:
Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.
Thank you.
abstract-algebra group-theory group-isomorphism group-homomorphism
edited 4 hours ago
Andrews
1,2761421
asked 4 hours ago
NiaBieNiaBie
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$
There is the proof of Abstract Algebra Thomas by W. Judson:
Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi={hin H:hin N}=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$
My question:
Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.
Thank you.
abstract-algebra group-theory group-isomorphism group-homomorphism
abstract-algebra group-theory group-isomorphism group-homomorphism
edited 4 hours ago
Andrews
1,2761421
asked 4 hours ago
NiaBieNiaBie
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Andrews
1,2761421
asked 4 hours ago
NiaBieNiaBie
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Andrews
1,2761421
edited 4 hours ago
Andrews
1,2761421
edited 4 hours ago
Andrews
1,2761421
1,2761421
asked 4 hours ago
NiaBieNiaBie
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
NiaBieNiaBie
232
asked 4 hours ago
NiaBieNiaBie
232
232
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
$endgroup$
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
});
}
});
NiaBie is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3160013%2fquestion-about-the-proof-of-second-isomorphism-theorem%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
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
$endgroup$
add a comment |
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
$endgroup$
add a comment |
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
$endgroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
answered 4 hours ago
Joshua MundingerJoshua Mundinger
2,7621028
2,7621028
add a comment |
add a comment |
NiaBie is a new contributor. Be nice, and check out our Code of Conduct.
NiaBie is a new contributor. Be nice, and check out our Code of Conduct.
NiaBie is a new contributor. Be nice, and check out our Code of Conduct.
NiaBie is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3160013%2fquestion-about-the-proof-of-second-isomorphism-theorem%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
