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Question about the proof of Second Isomorphism Theorem

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Question about the proof of Second Isomorphism Theorem

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4

$begingroup$

The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$

There is the proof of Abstract Algebra Thomas by W. Judson:

Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi=hin H:hin N=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$

My question:

Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.

Thank you.

abstract-algebra group-theory group-isomorphism group-homomorphism

share|cite|improve this question

edited 4 hours ago

Andrews

1,2761421

asked 5 hours ago

NiaBieNiaBie

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$endgroup$

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4

$begingroup$

The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$

There is the proof of Abstract Algebra Thomas by W. Judson:

Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi=hin H:hin N=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$

My question:

Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.

Thank you.

abstract-algebra group-theory group-isomorphism group-homomorphism

share|cite|improve this question

edited 4 hours ago

Andrews

1,2761421

asked 5 hours ago

NiaBieNiaBie

232

New contributor

NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

The Second Isomorphism Theorem:
Let $H$ be a subgroup of a group $G$ and $N$ a normal subgroup of $G$. Then
$$H/(Hcap N)cong(HN)/N$$

There is the proof of Abstract Algebra Thomas by W. Judson:

Define a map $phi$ from $H$ to $HN/N$ by $Hmapsto hN$. The map $phi$ is onto, since any coset $hnN=hN$ is the image of $h$ in $H$. We also know that $phi$ is a homomorphism because
$$phi(hh')=hh'N=hNh'N=phi(h)phi(h')$$
By the First Isomorphism Theorem, the image of $phi$ is isomorphic to $H/kerphi$, that is
$$HN/N=phi(H)cong H/kerphi$$
Since
$$kerphi=hin H:hin N=Hcap N$$
$HN/N=phi(H)cong H/Hcap N$

My question:

Is it necessary to prove that the map $phi$ is onto? Can we only prove that $phi$ is well defined and the image of $phi$ is a subset of $HN/N$? And then we can use the First Isomorphism Theorem and continue the proof.

Thank you.

abstract-algebra group-theory group-isomorphism group-homomorphism

share|cite|improve this question

edited 4 hours ago

Andrews

1,2761421

asked 5 hours ago

NiaBieNiaBie

232

New contributor

NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 4 hours ago

Andrews

1,2761421

asked 5 hours ago

NiaBieNiaBie

232

New contributor

NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 4 hours ago

Andrews

1,2761421

asked 5 hours ago

NiaBieNiaBie

232

New contributor

NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 4 hours ago

Andrews

1,2761421

edited 4 hours ago

Andrews

1,2761421

asked 5 hours ago

NiaBieNiaBie

232

New contributor

NiaBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

NiaBieNiaBie

232

1 Answer
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3

$begingroup$

The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrmim(varphi) cong G/mathrmker(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrmim(phi) subseteq HN/N$, which does not finish the job.

share|cite|improve this answer

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028

$endgroup$

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3

$begingroup$

The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrmim(varphi) cong G/mathrmker(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrmim(phi) subseteq HN/N$, which does not finish the job.

share|cite|improve this answer

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028

$endgroup$

add a comment | 

3

$begingroup$

The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrmim(varphi) cong G/mathrmker(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrmim(phi) subseteq HN/N$, which does not finish the job.

share|cite|improve this answer

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028

$endgroup$

add a comment | 

$begingroup$

The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrmim(varphi) cong G/mathrmker(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrmim(phi) subseteq HN/N$, which does not finish the job.

share|cite|improve this answer

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028

answered 4 hours ago

Joshua MundingerJoshua Mundinger

2,7621028