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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

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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