$H trianglelefteq K$, $K/H$ abelian, $N trianglelefteq G$, then $KN/HN$ abelian












0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
    $endgroup$
    – user10354138
    Dec 5 '18 at 4:14
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: Write down the obvious homomorphism $K/Hto KN/HN$.
    $endgroup$
    – user10354138
    Dec 5 '18 at 4:14














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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 $phi(h)=b$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 10:03

























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


















  • $begingroup$
    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











1












$begingroup$

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 $phi(h)=b$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $phi(h)=b$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $phi(h)=b$.






share|cite|improve this answer









$endgroup$



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 $phi(h)=b$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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How to change which sound is reproduced for terminal bell?

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3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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