$H trianglelefteq K$, $K/H$ abelian, $N trianglelefteq G$, then $KN/HN$ abelian
$begingroup$
Let $G$ be a group, and $H$, $K$, $N$, subgroups of $G$. $H$ is normal in $K$, $K/H$ is abelian, and $N$ is normal in $G$.
I can show that $HN$ is normal in $KN$, but how to see that $KN/HN$ is abelian if $K/H$ is abelian?
group-theory
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group, and $H$, $K$, $N$, subgroups of $G$. $H$ is normal in $K$, $K/H$ is abelian, and $N$ is normal in $G$.
I can show that $HN$ is normal in $KN$, but how to see that $KN/HN$ is abelian if $K/H$ is abelian?
group-theory
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
$endgroup$
1
$begingroup$
Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
$endgroup$
– user10354138
Dec 5 '18 at 4:14
add a comment |
$begingroup$
Let $G$ be a group, and $H$, $K$, $N$, subgroups of $G$. $H$ is normal in $K$, $K/H$ is abelian, and $N$ is normal in $G$.
I can show that $HN$ is normal in $KN$, but how to see that $KN/HN$ is abelian if $K/H$ is abelian?
group-theory
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
$endgroup$
Let $G$ be a group, and $H$, $K$, $N$, subgroups of $G$. $H$ is normal in $K$, $K/H$ is abelian, and $N$ is normal in $G$.
I can show that $HN$ is normal in $KN$, but how to see that $KN/HN$ is abelian if $K/H$ is abelian?
group-theory
group-theory
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
asked Dec 5 '18 at 4:11
RaekyeRaekye
24539
24539
1
$begingroup$
Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
$endgroup$
– user10354138
Dec 5 '18 at 4:14
add a comment |
1
$begingroup$
Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
$endgroup$
– user10354138
Dec 5 '18 at 4:14
1
1
$begingroup$
Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
$endgroup$
– user10354138
Dec 5 '18 at 4:14
$begingroup$
Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
$endgroup$
– user10354138
Dec 5 '18 at 4:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Although it may look like overkill, I believe it is advisable to split the problem in its elementary components.
First use the second isomorphism theorem:
$$
frac{K N}{H N}
=
frac{K H N}{H N}
cong
frac{K}{K cap H N},
$$
then Dedekind's identity (this step is unnecessary, see the comment beloew):
$$
frac{K}{K cap H N}
=
frac{K}{H (K cap N)},
$$
and finally the third isomorphism theorem:
$$
frac{K}{H (K cap N)}
cong
frac{dfrac{K}{H}}{dfrac{H (K cap N)}{H}}
$$
to show that $K N/H N$ is isomorphic to a quotient group of $K / H$. The argument in the other answer then applies.
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
add a comment |
$begingroup$
Use the hint in the comments to get a surjective homomorphism, and the fact that the homomorphic image of an abelian group is abelian: $ab=phi(g)phi(h)=phi(gh)=phi(hg)=phi(h)phi(g)=ba,,forall a,b$, where $phi(g)=a$ and $phi(h)=b$.
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
$endgroup$
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Although it may look like overkill, I believe it is advisable to split the problem in its elementary components.
First use the second isomorphism theorem:
$$
frac{K N}{H N}
=
frac{K H N}{H N}
cong
frac{K}{K cap H N},
$$
then Dedekind's identity (this step is unnecessary, see the comment beloew):
$$
frac{K}{K cap H N}
=
frac{K}{H (K cap N)},
$$
and finally the third isomorphism theorem:
$$
frac{K}{H (K cap N)}
cong
frac{dfrac{K}{H}}{dfrac{H (K cap N)}{H}}
$$
to show that $K N/H N$ is isomorphic to a quotient group of $K / H$. The argument in the other answer then applies.
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
add a comment |
$begingroup$
Although it may look like overkill, I believe it is advisable to split the problem in its elementary components.
First use the second isomorphism theorem:
$$
frac{K N}{H N}
=
frac{K H N}{H N}
cong
frac{K}{K cap H N},
$$
then Dedekind's identity (this step is unnecessary, see the comment beloew):
$$
frac{K}{K cap H N}
=
frac{K}{H (K cap N)},
$$
and finally the third isomorphism theorem:
$$
frac{K}{H (K cap N)}
cong
frac{dfrac{K}{H}}{dfrac{H (K cap N)}{H}}
$$
to show that $K N/H N$ is isomorphic to a quotient group of $K / H$. The argument in the other answer then applies.
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
add a comment |
$begingroup$
Although it may look like overkill, I believe it is advisable to split the problem in its elementary components.
First use the second isomorphism theorem:
$$
frac{K N}{H N}
=
frac{K H N}{H N}
cong
frac{K}{K cap H N},
$$
then Dedekind's identity (this step is unnecessary, see the comment beloew):
$$
frac{K}{K cap H N}
=
frac{K}{H (K cap N)},
$$
and finally the third isomorphism theorem:
$$
frac{K}{H (K cap N)}
cong
frac{dfrac{K}{H}}{dfrac{H (K cap N)}{H}}
$$
to show that $K N/H N$ is isomorphic to a quotient group of $K / H$. The argument in the other answer then applies.
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
Although it may look like overkill, I believe it is advisable to split the problem in its elementary components.
First use the second isomorphism theorem:
$$
frac{K N}{H N}
=
frac{K H N}{H N}
cong
frac{K}{K cap H N},
$$
then Dedekind's identity (this step is unnecessary, see the comment beloew):
$$
frac{K}{K cap H N}
=
frac{K}{H (K cap N)},
$$
and finally the third isomorphism theorem:
$$
frac{K}{H (K cap N)}
cong
frac{dfrac{K}{H}}{dfrac{H (K cap N)}{H}}
$$
to show that $K N/H N$ is isomorphic to a quotient group of $K / H$. The argument in the other answer then applies.
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
edited Dec 6 '18 at 10:03
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
answered Dec 5 '18 at 8:16
Andreas CarantiAndreas Caranti
56.6k34395
56.6k34395
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
add a comment |
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
Is it necessary to use dedekind's identity (it's my first time seeing it)? Since $H$ is normal in $K$, $H subseteq HN$, then $H leq K cap HN leq K$, and $H trianglelefteq K cap HN$, so we can apply the third isomorphism theorem from there to get $K/(K cap HN) cong (K/H)/((K cap HN)/H)$. Am I missing something?
$endgroup$
– Raekye
Dec 5 '18 at 17:25
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
$begingroup$
@Raekye, you are perfectly right, it is unnecessary. Thanks for pointing this out.
$endgroup$
– Andreas Caranti
Dec 6 '18 at 10:02
add a comment |
$begingroup$
Use the hint in the comments to get a surjective homomorphism, and the fact that the homomorphic image of an abelian group is abelian: $ab=phi(g)phi(h)=phi(gh)=phi(hg)=phi(h)phi(g)=ba,,forall a,b$, where $phi(g)=a$ and $phi(h)=b$.
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
$endgroup$
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
add a comment |
$begingroup$
Use the hint in the comments to get a surjective homomorphism, and the fact that the homomorphic image of an abelian group is abelian: $ab=phi(g)phi(h)=phi(gh)=phi(hg)=phi(h)phi(g)=ba,,forall a,b$, where $phi(g)=a$ and $phi(h)=b$.
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
$endgroup$
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
add a comment |
$begingroup$
Use the hint in the comments to get a surjective homomorphism, and the fact that the homomorphic image of an abelian group is abelian: $ab=phi(g)phi(h)=phi(gh)=phi(hg)=phi(h)phi(g)=ba,,forall a,b$, where $phi(g)=a$ and $phi(h)=b$.
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
$endgroup$
Use the hint in the comments to get a surjective homomorphism, and the fact that the homomorphic image of an abelian group is abelian: $ab=phi(g)phi(h)=phi(gh)=phi(hg)=phi(h)phi(g)=ba,,forall a,b$, where $phi(g)=a$ and $phi(h)=b$.
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 6:17
Chris CusterChris Custer
13.9k3827
13.9k3827
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
add a comment |
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
The "obvious" homomorphism is the projection given by $k + H in K/H mapsto k + HN in KN/HN$. My understanding is this is well defined because $H leq HN$? Because different elements of the coset $k + H$ all map to the same $k + HN$. But it's not clear to me why this is surjective; given $kn + HN in KN/HN$, what maps to it?
$endgroup$
– Raekye
Dec 5 '18 at 17:16
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
$kn_1$ and $kn_2$ differ by an element of $N$ by normality of $N$ in $G$. Hence they differ by an element of $HN$. Hence they give the same coset. So we send $k+H$ to that coset.
$endgroup$
– Chris Custer
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Is it true that $kn + HN = k + HN$ because $n in HN$? (Sorry for the inconsistent coset notation, have recently been working with rings so I was thinking of cosets with the addition notation). Edit: posted comment at the same time... thinking about yours
$endgroup$
– Raekye
Dec 5 '18 at 17:59
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
$begingroup$
Even better, I think. Normality seems to be unnecessary here, on second thought.
$endgroup$
– Chris Custer
Dec 5 '18 at 19:17
add a comment |
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1
$begingroup$
Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
$endgroup$
– user10354138
Dec 5 '18 at 4:14