Question about the proof of Second Isomorphism Theorem
$begingroup$
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
edited yesterday
Andrews
1,2812422
asked yesterday
NiaBieNiaBie
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$endgroup$
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
edited yesterday
Andrews
1,2812422
asked yesterday
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
edited yesterday
Andrews
1,2812422
asked yesterday
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.
$endgroup$
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
abstract-algebra group-theory group-isomorphism group-homomorphism
edited yesterday
Andrews
1,2812422
asked yesterday
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 yesterday
Andrews
1,2812422
asked yesterday
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 yesterday
Andrews
1,2812422
edited yesterday
Andrews
1,2812422
edited yesterday
Andrews
1,2812422
1,2812422
asked yesterday
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 yesterday
NiaBieNiaBie
232
asked yesterday
NiaBieNiaBie
232
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.
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.
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Check out our Code of Conduct.
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1 Answer
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$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
$endgroup$
add a comment |
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
$endgroup$
add a comment |
$begingroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
$endgroup$
The First Isomorphism Theorem states that if $varphi: G to G'$, then $mathrm{im}(varphi) cong G/mathrm{ker}(varphi)$. If we do not know that your $phi$ is surjective, then the First Isomorphism Theorem only shows us that $H/H cap N cong mathrm{im}(phi) subseteq HN/N$, which does not finish the job.
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
answered yesterday
Joshua MundingerJoshua Mundinger
2,8621028
2,8621028
add a comment |
add a comment |
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