Fdtxjr

New order #4: World

Could a US political party gain complete control over the government by removing checks & balances?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

Prevent a directory in /tmp from being deleted

least quadratic residue under GRH: an EXPLICIT bound

Non-Jewish family in an Orthodox Jewish Wedding

Calculus Optimization - Point on graph closest to given point

Is it possible to make sharp wind that can cut stuff from afar?

What Brexit solution does the DUP want?

declaring a variable twice in IIFE

N.B. ligature in Latex

Can a German sentence have two subjects?

Can town administrative "code" overule state laws like those forbidding trespassing?

Why CLRS example on residual networks does not follows its formula?

Is there really no realistic way for a skeleton monster to move around without magic?

What is the command to reset a PC without deleting any files

whey we use polarized capacitor?

Can I make popcorn with any corn?

Why can't I see bouncing of a switch on an oscilloscope?

What is the meaning of "of trouble" in the following sentence?

How do I create uniquely male characters?

What defenses are there against being summoned by the Gate spell?

Is there a familial term for apples and pears?

Exact sequence of groups: proof of injectivity

Isomorphism on commutative diagrams of abelian groupsProof of first isomorphism theorem of groupSimple example of a short exact sequence of groupsDual of Schanuel lemmaSplitting short exact sequence of space groupsUnderstanding a proof about splitting of short exact sequences.Diagram with exact sequences of modulesfunctor $F$ satisfies $kerF(f)cong F(ker(f))$. $0to M'to Mto M''$ exact => $0to F(M')to F(M)to F(M'')$ exact?Miscellaneous Questions involving Tensor Products, Exact Sequences, and AlgebrasFlatness in a short exact sequence. If the

2

$begingroup$

There must be a duplicate being the question very introductory, but I was not able to find it.

We have the following diagram

$$beginarrayccccccccc
1&to& H &to & G &oversetpito & G/H &to & 1\
&& downarrow & & downarrow f && downarrow g & & \
1&to& H' &to & G'& oversetpi'to & G'/H' &to & 1\
endarray$$

Suppose that

1. 
   $f$ is homomorphism


2. $H'triangleleft G'$


3. $H=f^-1(H')$


4. $Htriangleleft G$


5. 
   $f$ is surjective

Prove that $g$ is an isomorphism.

Since $gcircpi=pi'circ f$ and $pi,pi',f$ are surjective, then $g$ is surjective.

The problem is with injectivity. (A) From $g(aH)=g(bH)$ I should derive $aH=bH$. Or (B) prove that $operatornameker g=H$.

(A) Working backwards I get
$$
aH=bH
iff
a^-1bin H
iff
f(a^-1b)in H'\
iff
(f(a))^-1f(b)in H'
iff
f(a)H'=f(b)H'
$$

Now, if $g(aH)=g(bH)$ how do I get $f(a)H'=f(b)H'$?

Method (B) would probably be nicer, but no idea how to start.

abstract-algebra group-theory group-homomorphism exact-sequence

share|cite|improve this question

edited Mar 22 at 9:54

Andrews

1,2812423

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are the rows exact?
  $endgroup$
  – Alvin Lepik
  Mar 22 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlvinLepik All the conditions are given (1 to 5), I suppose. Thus I think the answer is no.
  $endgroup$
  – Aaron Lenz
  Mar 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlvinLepik To be honest I don't know. The problem was in the chapter "exact sequences" of some book and I added it months ago to my notes. Today was something like the third failed attempt to solve it.
  $endgroup$
  – Aaron Lenz
  Mar 22 at 9:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $g(aH)=g(bH)$ implies $f(a)H'=f(b)H'$ by commutativity of right-handed square, since $f(a)H'=(pi'circ f)(a)=(gcircpi)(a)=g(aH)$ and, similarly, $f(b)H'=g(bH)$.
  $endgroup$
  – Fabio Lucchini
  Mar 22 at 10:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FabioLucchini: perfect, grazie!
  $endgroup$
  – Aaron Lenz
  Mar 22 at 10:03
  

  
  
  
  

add a comment | 

2

$begingroup$

There must be a duplicate being the question very introductory, but I was not able to find it.

We have the following diagram

$$beginarrayccccccccc
1&to& H &to & G &oversetpito & G/H &to & 1\
&& downarrow & & downarrow f && downarrow g & & \
1&to& H' &to & G'& oversetpi'to & G'/H' &to & 1\
endarray$$

Suppose that

1. 
   $f$ is homomorphism


2. $H'triangleleft G'$


3. $H=f^-1(H')$


4. $Htriangleleft G$


5. 
   $f$ is surjective

Prove that $g$ is an isomorphism.

Since $gcircpi=pi'circ f$ and $pi,pi',f$ are surjective, then $g$ is surjective.

The problem is with injectivity. (A) From $g(aH)=g(bH)$ I should derive $aH=bH$. Or (B) prove that $operatornameker g=H$.

(A) Working backwards I get
$$
aH=bH
iff
a^-1bin H
iff
f(a^-1b)in H'\
iff
(f(a))^-1f(b)in H'
iff
f(a)H'=f(b)H'
$$

Now, if $g(aH)=g(bH)$ how do I get $f(a)H'=f(b)H'$?

Method (B) would probably be nicer, but no idea how to start.

abstract-algebra group-theory group-homomorphism exact-sequence

share|cite|improve this question

edited Mar 22 at 9:54

Andrews

1,2812423

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are the rows exact?
  $endgroup$
  – Alvin Lepik
  Mar 22 at 9:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlvinLepik All the conditions are given (1 to 5), I suppose. Thus I think the answer is no.
  $endgroup$
  – Aaron Lenz
  Mar 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlvinLepik To be honest I don't know. The problem was in the chapter "exact sequences" of some book and I added it months ago to my notes. Today was something like the third failed attempt to solve it.
  $endgroup$
  – Aaron Lenz
  Mar 22 at 9:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $g(aH)=g(bH)$ implies $f(a)H'=f(b)H'$ by commutativity of right-handed square, since $f(a)H'=(pi'circ f)(a)=(gcircpi)(a)=g(aH)$ and, similarly, $f(b)H'=g(bH)$.
  $endgroup$
  – Fabio Lucchini
  Mar 22 at 10:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FabioLucchini: perfect, grazie!
  $endgroup$
  – Aaron Lenz
  Mar 22 at 10:03
  

  
  
  
  

add a comment | 

$begingroup$

There must be a duplicate being the question very introductory, but I was not able to find it.

We have the following diagram

$$beginarrayccccccccc
1&to& H &to & G &oversetpito & G/H &to & 1\
&& downarrow & & downarrow f && downarrow g & & \
1&to& H' &to & G'& oversetpi'to & G'/H' &to & 1\
endarray$$

Suppose that

1. 
   $f$ is homomorphism


2. $H'triangleleft G'$


3. $H=f^-1(H')$


4. $Htriangleleft G$


5. 
   $f$ is surjective

Prove that $g$ is an isomorphism.

Since $gcircpi=pi'circ f$ and $pi,pi',f$ are surjective, then $g$ is surjective.

The problem is with injectivity. (A) From $g(aH)=g(bH)$ I should derive $aH=bH$. Or (B) prove that $operatornameker g=H$.

(A) Working backwards I get
$$
aH=bH
iff
a^-1bin H
iff
f(a^-1b)in H'\
iff
(f(a))^-1f(b)in H'
iff
f(a)H'=f(b)H'
$$

Now, if $g(aH)=g(bH)$ how do I get $f(a)H'=f(b)H'$?

Method (B) would probably be nicer, but no idea how to start.

abstract-algebra group-theory group-homomorphism exact-sequence

share|cite|improve this question

edited Mar 22 at 9:54

Andrews

1,2812423

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

$endgroup$

share|cite|improve this question

edited Mar 22 at 9:54

Andrews

1,2812423

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

share|cite|improve this question

edited Mar 22 at 9:54

Andrews

1,2812423

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

edited Mar 22 at 9:54

Andrews

1,2812423

edited Mar 22 at 9:54

Andrews

1,2812423

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

asked Mar 22 at 9:20

Aaron LenzAaron Lenz

7110

1 Answer
1

active

oldest

votes

2

$begingroup$

To prove directly that $DeclareMathOperatorKerKerKer g$ is trivial, note that
beginalign
pi^-1Ker g
&=Ker(gcircpi)\
&=Ker(pi'circ f)\
&=f^-1Kerpi'\
&=f^-1H'\
&=H
endalign
Since $pi$ is surjective, we get
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.

share|cite|improve this answer

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  cool!$phantomaaaaaaaaaaaaaa$
  $endgroup$
  – Aaron Lenz
  Mar 22 at 10:17
  

  
  
  
  

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3157940%2fexact-sequence-of-groups-proof-of-injectivity%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

To prove directly that $DeclareMathOperatorKerKerKer g$ is trivial, note that
beginalign
pi^-1Ker g
&=Ker(gcircpi)\
&=Ker(pi'circ f)\
&=f^-1Kerpi'\
&=f^-1H'\
&=H
endalign
Since $pi$ is surjective, we get
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.

share|cite|improve this answer

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  cool!$phantomaaaaaaaaaaaaaa$
  $endgroup$
  – Aaron Lenz
  Mar 22 at 10:17
  

  
  
  
  

add a comment | 

2

$begingroup$

To prove directly that $DeclareMathOperatorKerKerKer g$ is trivial, note that
beginalign
pi^-1Ker g
&=Ker(gcircpi)\
&=Ker(pi'circ f)\
&=f^-1Kerpi'\
&=f^-1H'\
&=H
endalign
Since $pi$ is surjective, we get
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.

share|cite|improve this answer

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  cool!$phantomaaaaaaaaaaaaaa$
  $endgroup$
  – Aaron Lenz
  Mar 22 at 10:17
  

  
  
  
  

add a comment | 

$begingroup$

To prove directly that $DeclareMathOperatorKerKerKer g$ is trivial, note that
beginalign
pi^-1Ker g
&=Ker(gcircpi)\
&=Ker(pi'circ f)\
&=f^-1Kerpi'\
&=f^-1H'\
&=H
endalign
Since $pi$ is surjective, we get
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.

share|cite|improve this answer

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

share|cite|improve this answer

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426

answered Mar 22 at 10:11

Fabio LucchiniFabio Lucchini

9,50111426