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 Mar 24 at 2:22
Andrews
1,2812422
asked Mar 24 at 1:47
NiaBieNiaBie
233
$endgroup$
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 Mar 24 at 2:22
Andrews
1,2812422
asked Mar 24 at 1:47
NiaBieNiaBie
233
$endgroup$
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 Mar 24 at 2:22
Andrews
1,2812422
asked Mar 24 at 1:47
NiaBieNiaBie
233
$endgroup$
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 Mar 24 at 2:22
Andrews
1,2812422
asked Mar 24 at 1:47
NiaBieNiaBie
233
edited Mar 24 at 2:22
Andrews
1,2812422
asked Mar 24 at 1:47
NiaBieNiaBie
233
edited Mar 24 at 2:22
Andrews
1,2812422
edited Mar 24 at 2:22
Andrews
1,2812422
edited Mar 24 at 2:22
Andrews
1,2812422
1,2812422
asked Mar 24 at 1:47
NiaBieNiaBie
233
asked Mar 24 at 1:47
NiaBieNiaBie
233
asked Mar 24 at 1:47
NiaBieNiaBie
233
233
add a comment |
add a comment |
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 Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
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1 Answer
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1 Answer
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active
oldest
votes
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 Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
$endgroup$
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 Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
$endgroup$
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 Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
$endgroup$
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 Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
answered Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
answered Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
answered Mar 24 at 1:59
Joshua MundingerJoshua Mundinger
2,9071028
2,9071028
add a comment |
add a comment |
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