Exact sequence of groups: proof of injectivityIsomorphism 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
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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
$begingroup$
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
$f$ is homomorphism- $H'triangleleft G'$
- $H=f^-1(H')$
- $Htriangleleft G$
$f$ 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
edited Mar 22 at 9:54
Andrews
1,2812423
asked Mar 22 at 9:20
Aaron LenzAaron Lenz
7110
$endgroup$
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
$f$ is homomorphism- $H'triangleleft G'$
- $H=f^-1(H')$
- $Htriangleleft G$
$f$ 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
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
$f$ is homomorphism- $H'triangleleft G'$
- $H=f^-1(H')$
- $Htriangleleft G$
$f$ 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
edited Mar 22 at 9:54
Andrews
1,2812423
asked Mar 22 at 9:20
Aaron LenzAaron Lenz
7110
$endgroup$
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
$f$ is homomorphism- $H'triangleleft G'$
- $H=f^-1(H')$
- $Htriangleleft G$
$f$ 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
abstract-algebra group-theory group-homomorphism exact-sequence
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
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
edited Mar 22 at 9:54
Andrews
1,2812423
1,2812423
asked Mar 22 at 9:20
Aaron LenzAaron Lenz
7110
asked Mar 22 at 9:20
Aaron LenzAaron Lenz
7110
asked Mar 22 at 9:20
Aaron LenzAaron Lenz
7110
7110
$begingroup$
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 |
$begingroup$
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
$begingroup$
are the rows exact?
$endgroup$
– Alvin Lepik
Mar 22 at 9:32
$begingroup$
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 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
$begingroup$
@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
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
$begingroup$
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
$begingroup$
@FabioLucchini: perfect, grazie!
$endgroup$
– Aaron Lenz
Mar 22 at 10:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.
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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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.
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
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.
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
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.
answered Mar 22 at 10:11
Fabio LucchiniFabio Lucchini
9,50111426
$endgroup$
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
$Ker g=pi[pi^-1Ker g]=pi[H]=1$.
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
answered Mar 22 at 10:11
Fabio LucchiniFabio Lucchini
9,50111426
9,50111426
$begingroup$
cool!$phantomaaaaaaaaaaaaaa$
$endgroup$
– Aaron Lenz
Mar 22 at 10:17
add a comment |
$begingroup$
cool!$phantomaaaaaaaaaaaaaa$
$endgroup$
– Aaron Lenz
Mar 22 at 10:17
$begingroup$
cool!$phantomaaaaaaaaaaaaaa$
$endgroup$
– Aaron Lenz
Mar 22 at 10:17
$begingroup$
cool!$phantomaaaaaaaaaaaaaa$
$endgroup$
– Aaron Lenz
Mar 22 at 10:17
add a comment |
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$begingroup$
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