If we divide a group and a proper subgroup of this group by the same normal subgroup, can the quotients be...
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 at 23:30
Diglett
879520
add a comment |
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 at 23:30
Diglett
879520
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
Nov 20 at 0:43
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
Nov 20 at 2:37
add a comment |
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 at 23:30
Diglett
879520
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 at 23:30
Diglett
879520
asked Nov 19 at 23:30
Diglett
879520
asked Nov 19 at 23:30
Diglett
879520
asked Nov 19 at 23:30
Diglett
879520
asked Nov 19 at 23:30
Diglett
879520
879520
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
Nov 20 at 0:43
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
Nov 20 at 2:37
add a comment |
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
Nov 20 at 0:43
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
Nov 20 at 2:37
2
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
Nov 20 at 0:43
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
Nov 20 at 0:43
1
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
Nov 20 at 2:37
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
Nov 20 at 2:37
add a comment |
1 Answer
1
active
oldest
votes
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered Nov 20 at 6:12
R.C.Cowsik
31514
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%2f3005685%2fif-we-divide-a-group-and-a-proper-subgroup-of-this-group-by-the-same-normal-subg%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
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered Nov 20 at 6:12
R.C.Cowsik
31514
add a comment |
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered Nov 20 at 6:12
R.C.Cowsik
31514
add a comment |
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered Nov 20 at 6:12
R.C.Cowsik
31514
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered Nov 20 at 6:12
R.C.Cowsik
31514
answered Nov 20 at 6:12
R.C.Cowsik
31514
answered Nov 20 at 6:12
R.C.Cowsik
31514
answered Nov 20 at 6:12
R.C.Cowsik
31514
31514
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3005685%2fif-we-divide-a-group-and-a-proper-subgroup-of-this-group-by-the-same-normal-subg%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
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
Nov 20 at 0:43
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
Nov 20 at 2:37