If $H$ is normal in $G$ then $H$ is the kernel of a group homomorphism.

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Let $G$ a group and $H$ a subgroup of $G$. Show that if $H$ is normal, then it's the kernel of a group homomorphism.

Attempt

I proved that if $H$ is the kernel of a group homomorphism, then $H$ is normal. But how can I prove the converse ? I tried to construct something as $f:Gto G$ s.t. $f(gh)=g$. Then $ker f=H$, but I have difficulty to show that $f$ is indeed a homomorphism.

group-theory

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edited Jul 23 at 12:50

José Carlos Santos

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asked Jul 23 at 12:29

MathBeginner

695312

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up vote
0
down vote

favorite

Let $G$ a group and $H$ a subgroup of $G$. Show that if $H$ is normal, then it's the kernel of a group homomorphism.

Attempt

I proved that if $H$ is the kernel of a group homomorphism, then $H$ is normal. But how can I prove the converse ? I tried to construct something as $f:Gto G$ s.t. $f(gh)=g$. Then $ker f=H$, but I have difficulty to show that $f$ is indeed a homomorphism.

group-theory

share|cite|improve this question

edited Jul 23 at 12:50

José Carlos Santos

113k1698176

asked Jul 23 at 12:29

MathBeginner

695312

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $G$ a group and $H$ a subgroup of $G$. Show that if $H$ is normal, then it's the kernel of a group homomorphism.

Attempt

I proved that if $H$ is the kernel of a group homomorphism, then $H$ is normal. But how can I prove the converse ? I tried to construct something as $f:Gto G$ s.t. $f(gh)=g$. Then $ker f=H$, but I have difficulty to show that $f$ is indeed a homomorphism.

group-theory

share|cite|improve this question

edited Jul 23 at 12:50

José Carlos Santos

113k1698176

asked Jul 23 at 12:29

MathBeginner

695312

Let $G$ a group and $H$ a subgroup of $G$. Show that if $H$ is normal, then it's the kernel of a group homomorphism.

Attempt

I proved that if $H$ is the kernel of a group homomorphism, then $H$ is normal. But how can I prove the converse ? I tried to construct something as $f:Gto G$ s.t. $f(gh)=g$. Then $ker f=H$, but I have difficulty to show that $f$ is indeed a homomorphism.

group-theory

share|cite|improve this question

edited Jul 23 at 12:50

José Carlos Santos

113k1698176

asked Jul 23 at 12:29

MathBeginner

695312

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 12:50

José Carlos Santos

113k1698176

edited Jul 23 at 12:50

José Carlos Santos

113k1698176

edited Jul 23 at 12:50

José Carlos Santos

113k1698176

113k1698176

asked Jul 23 at 12:29

MathBeginner

695312

asked Jul 23 at 12:29

MathBeginner

695312

asked Jul 23 at 12:29

MathBeginner

695312

695312

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
4
down vote



accepted

$H$ is the kernel of the canonical surjection $G to G/H$ where $G/H$ is the quotient group.

share|cite|improve this answer

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

add a comment | 

up vote
2
down vote

Because $H$ is a normal subgroup of $G$, you know that the quotient $G/H$ comes with a canonical structure of group, such that the projection $pi: Grightarrow G/H$ is a group morphism, with kernel exactly $H$.

share|cite|improve this answer

answered Jul 23 at 12:31

Suzet

2,213427

add a comment | 

up vote
1
down vote

You send $gin G$ to the coset $gH=Hg$ (this equality is true by normality), and you can show that the obvious group operation on cosets is well-defined, and associative, with $H$ as identity and the obvious inverse. The kernel of this map, which is a homomorphism, is precisely $H$.

share|cite|improve this answer

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

$H$ is the kernel of the canonical surjection $G to G/H$ where $G/H$ is the quotient group.

share|cite|improve this answer

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

add a comment | 

up vote
4
down vote



accepted

$H$ is the kernel of the canonical surjection $G to G/H$ where $G/H$ is the quotient group.

share|cite|improve this answer

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

$H$ is the kernel of the canonical surjection $G to G/H$ where $G/H$ is the quotient group.

share|cite|improve this answer

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

$H$ is the kernel of the canonical surjection $G to G/H$ where $G/H$ is the quotient group.

share|cite|improve this answer

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

answered Jul 23 at 12:30

Kenny Lau

18.7k2157

18.7k2157

add a comment | 

add a comment | 

up vote
2
down vote

Because $H$ is a normal subgroup of $G$, you know that the quotient $G/H$ comes with a canonical structure of group, such that the projection $pi: Grightarrow G/H$ is a group morphism, with kernel exactly $H$.

share|cite|improve this answer

answered Jul 23 at 12:31

Suzet

2,213427

add a comment | 

up vote
2
down vote

Because $H$ is a normal subgroup of $G$, you know that the quotient $G/H$ comes with a canonical structure of group, such that the projection $pi: Grightarrow G/H$ is a group morphism, with kernel exactly $H$.

share|cite|improve this answer

answered Jul 23 at 12:31

Suzet

2,213427

add a comment | 

up vote
2
down vote

up vote
2
down vote

Because $H$ is a normal subgroup of $G$, you know that the quotient $G/H$ comes with a canonical structure of group, such that the projection $pi: Grightarrow G/H$ is a group morphism, with kernel exactly $H$.

share|cite|improve this answer

answered Jul 23 at 12:31

Suzet

2,213427

Because $H$ is a normal subgroup of $G$, you know that the quotient $G/H$ comes with a canonical structure of group, such that the projection $pi: Grightarrow G/H$ is a group morphism, with kernel exactly $H$.

share|cite|improve this answer

answered Jul 23 at 12:31

Suzet

2,213427

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 12:31

Suzet

2,213427

answered Jul 23 at 12:31

Suzet

2,213427

answered Jul 23 at 12:31

Suzet

2,213427

2,213427

add a comment | 

add a comment | 

up vote
1
down vote

You send $gin G$ to the coset $gH=Hg$ (this equality is true by normality), and you can show that the obvious group operation on cosets is well-defined, and associative, with $H$ as identity and the obvious inverse. The kernel of this map, which is a homomorphism, is precisely $H$.

share|cite|improve this answer

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

add a comment | 

up vote
1
down vote

You send $gin G$ to the coset $gH=Hg$ (this equality is true by normality), and you can show that the obvious group operation on cosets is well-defined, and associative, with $H$ as identity and the obvious inverse. The kernel of this map, which is a homomorphism, is precisely $H$.

share|cite|improve this answer

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

add a comment | 

up vote
1
down vote

up vote
1
down vote

You send $gin G$ to the coset $gH=Hg$ (this equality is true by normality), and you can show that the obvious group operation on cosets is well-defined, and associative, with $H$ as identity and the obvious inverse. The kernel of this map, which is a homomorphism, is precisely $H$.

share|cite|improve this answer

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

You send $gin G$ to the coset $gH=Hg$ (this equality is true by normality), and you can show that the obvious group operation on cosets is well-defined, and associative, with $H$ as identity and the obvious inverse. The kernel of this map, which is a homomorphism, is precisely $H$.

share|cite|improve this answer

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

answered Jul 23 at 12:42

Mark Bennet

76.4k773170

76.4k773170

add a comment | 

add a comment | 

 

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