Why in Quotient Group we need normal subgroup? [duplicate]

3

$begingroup$

This question already has an answer here:

• 
  Why do we define quotient groups for normal subgroups only?
  
  4 answers
  
  

For a group $G$ and a normal subgroup $N$ of $G$, the quotient group of $N$ in $G$, written $G/N$ and read "$G$ modulo $N$", is the set of cosets of $N$ in $G$.

Question : Why in the defintion of Quotient Group we need subgroup $N$ to be normal?

group-theory soft-question normal-subgroups quotient-group

share|cite|improve this question

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

asked Feb 17 '18 at 13:03

user275490

$endgroup$

marked as duplicate by rschwieb, Lord_Farin, Martin R, Martin Sleziak, Lord Shark the Unknown Jan 23 at 3:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is used to show that the product $(gN)(hN)=(gh)N$ is well defined.
  $endgroup$
  – John B
  Feb 17 '18 at 13:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try to define a similar quotient structure without assuming normality and see what the problems are.
  $endgroup$
  – the_fox
  Feb 17 '18 at 13:10
  

  
  
  
  

add a comment | 

3

$begingroup$

This question already has an answer here:

• 
  Why do we define quotient groups for normal subgroups only?
  
  4 answers
  
  

For a group $G$ and a normal subgroup $N$ of $G$, the quotient group of $N$ in $G$, written $G/N$ and read "$G$ modulo $N$", is the set of cosets of $N$ in $G$.

Question : Why in the defintion of Quotient Group we need subgroup $N$ to be normal?

group-theory soft-question normal-subgroups quotient-group

share|cite|improve this question

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

asked Feb 17 '18 at 13:03

user275490

$endgroup$

marked as duplicate by rschwieb, Lord_Farin, Martin R, Martin Sleziak, Lord Shark the Unknown Jan 23 at 3:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is used to show that the product $(gN)(hN)=(gh)N$ is well defined.
  $endgroup$
  – John B
  Feb 17 '18 at 13:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try to define a similar quotient structure without assuming normality and see what the problems are.
  $endgroup$
  – the_fox
  Feb 17 '18 at 13:10
  

  
  
  
  

add a comment | 

3

3

3

$begingroup$

This question already has an answer here:

• 
  Why do we define quotient groups for normal subgroups only?
  
  4 answers
  
  

For a group $G$ and a normal subgroup $N$ of $G$, the quotient group of $N$ in $G$, written $G/N$ and read "$G$ modulo $N$", is the set of cosets of $N$ in $G$.

Question : Why in the defintion of Quotient Group we need subgroup $N$ to be normal?

group-theory soft-question normal-subgroups quotient-group

share|cite|improve this question

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

asked Feb 17 '18 at 13:03

user275490

$endgroup$

This question already has an answer here:

• 
  Why do we define quotient groups for normal subgroups only?
  
  4 answers
  
  

For a group $G$ and a normal subgroup $N$ of $G$, the quotient group of $N$ in $G$, written $G/N$ and read "$G$ modulo $N$", is the set of cosets of $N$ in $G$.

Question : Why in the defintion of Quotient Group we need subgroup $N$ to be normal?

This question already has an answer here:

• 
  Why do we define quotient groups for normal subgroups only?
  
  4 answers
  
  

group-theory soft-question normal-subgroups quotient-group

group-theory soft-question normal-subgroups quotient-group

share|cite|improve this question

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

asked Feb 17 '18 at 13:03

user275490

share|cite|improve this question

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

asked Feb 17 '18 at 13:03

user275490

share|cite|improve this question

share|cite|improve this question

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

edited Jan 22 at 4:33

Martin Sleziak

44.8k10119273

44.8k10119273

asked Feb 17 '18 at 13:03

user275490

asked Feb 17 '18 at 13:03

user275490

asked Feb 17 '18 at 13:03

user275490

marked as duplicate by rschwieb, Lord_Farin, Martin R, Martin Sleziak, Lord Shark the Unknown Jan 23 at 3:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by rschwieb, Lord_Farin, Martin R, Martin Sleziak, Lord Shark the Unknown Jan 23 at 3:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is used to show that the product $(gN)(hN)=(gh)N$ is well defined.
  $endgroup$
  – John B
  Feb 17 '18 at 13:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try to define a similar quotient structure without assuming normality and see what the problems are.
  $endgroup$
  – the_fox
  Feb 17 '18 at 13:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is used to show that the product $(gN)(hN)=(gh)N$ is well defined.
  $endgroup$
  – John B
  Feb 17 '18 at 13:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try to define a similar quotient structure without assuming normality and see what the problems are.
  $endgroup$
  – the_fox
  Feb 17 '18 at 13:10
  

  
  
  
  

2

2

$begingroup$
This is used to show that the product $(gN)(hN)=(gh)N$ is well defined.
$endgroup$
– John B
Feb 17 '18 at 13:10

$begingroup$
This is used to show that the product $(gN)(hN)=(gh)N$ is well defined.
$endgroup$
– John B
Feb 17 '18 at 13:10

1

1

$begingroup$
Try to define a similar quotient structure without assuming normality and see what the problems are.
$endgroup$
– the_fox
Feb 17 '18 at 13:10

$begingroup$
Try to define a similar quotient structure without assuming normality and see what the problems are.
$endgroup$
– the_fox
Feb 17 '18 at 13:10

add a comment | 

4 Answers
4

active

oldest

votes

1

$begingroup$

Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $ain[a]$ and $bin[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'in[a]$ and $b'in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $ain[e]$ and $gin G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$ain[e]implies gag^{-1}in[e]$$
So $[e]$ is bound to be a normal subgroup.

share|cite|improve this answer

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

$endgroup$

add a comment | 

0

$begingroup$

We define the multiplication of two cosets $gN$ and $hN$ to be $ghN$. So, if $N$ is not a normal subgroup of $G$, then this multiplication will not be well defined, i.e. there may be two different and non-equal values for $(gN)(hN)$.

share|cite|improve this answer

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
  $endgroup$
  – ancientmathematician
  Feb 17 '18 at 13:41
  

  
  
  
  

add a comment | 

0

$begingroup$

That is to have a well-defined group law on the quotient: for $a,bin G$, we need to check that , if $a'equiv amod H$, $b'equiv bmod H$*, we want that $a'b'equiv abmod H$.

Explicitly, this means that, if $a'=ah$, $;b'=bk;$ for some $h,kin H$, we must have
$$a'b'(=ahmkern1.5mubk)=ab mkern1.5muellquadtext{for some }enspace ellin H.$$
In particular, we must have
$$a'a^{-1}=aha^{-1}in Henspacetext{for any }hin H,~ain G, $$
which exactly means $H$ is normal in $G$.

share|cite|improve this answer

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

$endgroup$

add a comment | 

0

$begingroup$

Just as an example: take the (not normal) subgroup ${i,(23)}=Hsubset S_3$, where $i$ is the identity. Indeed, $$H_{(132)}={(132),(12)}=H_{(12)}$$ and $$H_{(123)}={ (123),(13)}=H_{(13)}.$$ However, $H_{(132)circ (123)}=H_i$, which is different from $H_{(12)circ (13)}=H_{(132)}$, so that our operation isn't well-defined.

share|cite|improve this answer

edited Feb 17 '18 at 14:33

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

$endgroup$

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $ain[a]$ and $bin[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'in[a]$ and $b'in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $ain[e]$ and $gin G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$ain[e]implies gag^{-1}in[e]$$
So $[e]$ is bound to be a normal subgroup.

share|cite|improve this answer

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

$endgroup$

add a comment | 

1

$begingroup$

Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $ain[a]$ and $bin[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'in[a]$ and $b'in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $ain[e]$ and $gin G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$ain[e]implies gag^{-1}in[e]$$
So $[e]$ is bound to be a normal subgroup.

share|cite|improve this answer

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $ain[a]$ and $bin[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'in[a]$ and $b'in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $ain[e]$ and $gin G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$ain[e]implies gag^{-1}in[e]$$
So $[e]$ is bound to be a normal subgroup.

share|cite|improve this answer

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

$endgroup$

Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $ain[a]$ and $bin[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'in[a]$ and $b'in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $ain[e]$ and $gin G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$ain[e]implies gag^{-1}in[e]$$
So $[e]$ is bound to be a normal subgroup.

share|cite|improve this answer

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

share|cite|improve this answer

share|cite|improve this answer

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

answered Feb 17 '18 at 14:03

drhabdrhab

103k545136

103k545136

add a comment | 

add a comment | 

0

$begingroup$

We define the multiplication of two cosets $gN$ and $hN$ to be $ghN$. So, if $N$ is not a normal subgroup of $G$, then this multiplication will not be well defined, i.e. there may be two different and non-equal values for $(gN)(hN)$.

share|cite|improve this answer

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
  $endgroup$
  – ancientmathematician
  Feb 17 '18 at 13:41
  

  
  
  
  

add a comment | 

0

$begingroup$

We define the multiplication of two cosets $gN$ and $hN$ to be $ghN$. So, if $N$ is not a normal subgroup of $G$, then this multiplication will not be well defined, i.e. there may be two different and non-equal values for $(gN)(hN)$.

share|cite|improve this answer

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
  $endgroup$
  – ancientmathematician
  Feb 17 '18 at 13:41
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

We define the multiplication of two cosets $gN$ and $hN$ to be $ghN$. So, if $N$ is not a normal subgroup of $G$, then this multiplication will not be well defined, i.e. there may be two different and non-equal values for $(gN)(hN)$.

share|cite|improve this answer

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

$endgroup$

We define the multiplication of two cosets $gN$ and $hN$ to be $ghN$. So, if $N$ is not a normal subgroup of $G$, then this multiplication will not be well defined, i.e. there may be two different and non-equal values for $(gN)(hN)$.

share|cite|improve this answer

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

share|cite|improve this answer

share|cite|improve this answer

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

answered Feb 17 '18 at 13:12

QurultayQurultay

611313

611313

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
  $endgroup$
  – ancientmathematician
  Feb 17 '18 at 13:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
  $endgroup$
  – ancientmathematician
  Feb 17 '18 at 13:41
  

  
  
  
  

$begingroup$
@shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
$endgroup$
– ancientmathematician
Feb 17 '18 at 13:41

$begingroup$
@shi, take the smallest possible example: $G=S_3$ and $H={e, (12)}$ with $g=h=(123)$.
$endgroup$
– ancientmathematician
Feb 17 '18 at 13:41

add a comment | 

0

$begingroup$

That is to have a well-defined group law on the quotient: for $a,bin G$, we need to check that , if $a'equiv amod H$, $b'equiv bmod H$*, we want that $a'b'equiv abmod H$.

Explicitly, this means that, if $a'=ah$, $;b'=bk;$ for some $h,kin H$, we must have
$$a'b'(=ahmkern1.5mubk)=ab mkern1.5muellquadtext{for some }enspace ellin H.$$
In particular, we must have
$$a'a^{-1}=aha^{-1}in Henspacetext{for any }hin H,~ain G, $$
which exactly means $H$ is normal in $G$.

share|cite|improve this answer

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

$endgroup$

add a comment | 

0

$begingroup$

That is to have a well-defined group law on the quotient: for $a,bin G$, we need to check that , if $a'equiv amod H$, $b'equiv bmod H$*, we want that $a'b'equiv abmod H$.

Explicitly, this means that, if $a'=ah$, $;b'=bk;$ for some $h,kin H$, we must have
$$a'b'(=ahmkern1.5mubk)=ab mkern1.5muellquadtext{for some }enspace ellin H.$$
In particular, we must have
$$a'a^{-1}=aha^{-1}in Henspacetext{for any }hin H,~ain G, $$
which exactly means $H$ is normal in $G$.

share|cite|improve this answer

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

$endgroup$

add a comment | 

0

0

0

$begingroup$

That is to have a well-defined group law on the quotient: for $a,bin G$, we need to check that , if $a'equiv amod H$, $b'equiv bmod H$*, we want that $a'b'equiv abmod H$.

Explicitly, this means that, if $a'=ah$, $;b'=bk;$ for some $h,kin H$, we must have
$$a'b'(=ahmkern1.5mubk)=ab mkern1.5muellquadtext{for some }enspace ellin H.$$
In particular, we must have
$$a'a^{-1}=aha^{-1}in Henspacetext{for any }hin H,~ain G, $$
which exactly means $H$ is normal in $G$.

share|cite|improve this answer

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

$endgroup$

That is to have a well-defined group law on the quotient: for $a,bin G$, we need to check that , if $a'equiv amod H$, $b'equiv bmod H$*, we want that $a'b'equiv abmod H$.

Explicitly, this means that, if $a'=ah$, $;b'=bk;$ for some $h,kin H$, we must have
$$a'b'(=ahmkern1.5mubk)=ab mkern1.5muellquadtext{for some }enspace ellin H.$$
In particular, we must have
$$a'a^{-1}=aha^{-1}in Henspacetext{for any }hin H,~ain G, $$
which exactly means $H$ is normal in $G$.

share|cite|improve this answer

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

share|cite|improve this answer

share|cite|improve this answer

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

answered Feb 17 '18 at 13:36

BernardBernard

122k741116

122k741116

add a comment | 

add a comment | 

0

$begingroup$

Just as an example: take the (not normal) subgroup ${i,(23)}=Hsubset S_3$, where $i$ is the identity. Indeed, $$H_{(132)}={(132),(12)}=H_{(12)}$$ and $$H_{(123)}={ (123),(13)}=H_{(13)}.$$ However, $H_{(132)circ (123)}=H_i$, which is different from $H_{(12)circ (13)}=H_{(132)}$, so that our operation isn't well-defined.

share|cite|improve this answer

edited Feb 17 '18 at 14:33

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

$endgroup$

add a comment | 

0

$begingroup$

Just as an example: take the (not normal) subgroup ${i,(23)}=Hsubset S_3$, where $i$ is the identity. Indeed, $$H_{(132)}={(132),(12)}=H_{(12)}$$ and $$H_{(123)}={ (123),(13)}=H_{(13)}.$$ However, $H_{(132)circ (123)}=H_i$, which is different from $H_{(12)circ (13)}=H_{(132)}$, so that our operation isn't well-defined.

share|cite|improve this answer

edited Feb 17 '18 at 14:33

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

$endgroup$

add a comment | 

0

0

0

$begingroup$

Just as an example: take the (not normal) subgroup ${i,(23)}=Hsubset S_3$, where $i$ is the identity. Indeed, $$H_{(132)}={(132),(12)}=H_{(12)}$$ and $$H_{(123)}={ (123),(13)}=H_{(13)}.$$ However, $H_{(132)circ (123)}=H_i$, which is different from $H_{(12)circ (13)}=H_{(132)}$, so that our operation isn't well-defined.

share|cite|improve this answer

edited Feb 17 '18 at 14:33

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

$endgroup$

Just as an example: take the (not normal) subgroup ${i,(23)}=Hsubset S_3$, where $i$ is the identity. Indeed, $$H_{(132)}={(132),(12)}=H_{(12)}$$ and $$H_{(123)}={ (123),(13)}=H_{(13)}.$$ However, $H_{(132)circ (123)}=H_i$, which is different from $H_{(12)circ (13)}=H_{(132)}$, so that our operation isn't well-defined.

share|cite|improve this answer

edited Feb 17 '18 at 14:33

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

share|cite|improve this answer

share|cite|improve this answer

edited Feb 17 '18 at 14:33

edited Feb 17 '18 at 14:33

edited Feb 17 '18 at 14:33

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

answered Feb 17 '18 at 13:40

TPaceTPace

5111318

5111318

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here