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Find The Coset Of SubGroup [closed]

Question about Wikipedia Coset/Quotient Group Exampleleft and right coset verificationsWhat exactly is a coset?Coset of an infinite groupA subgroup such that at least one left coset is a right cosetNot a normal subgroup by left and right cosetProof varification: coset of the subgroup $Hcap K$Difference between coset and subgroupFind cosets and index of $GL_n$ and $SL_n$For the mapping x to ax+b, show that every left coset is a right coset

-2

$begingroup$

Denoting
$$
GL(n,mathbb R)=gin mathbb R^ntimes nmiddet(g)neq 0, quad SL(n,mathbb R):=gin GL(n,mathbb R)middet(g)=1,
$$
we get $SL(n,mathbb R) le GL(n,mathbb R)$. Let $A in GL(n,mathbb R)$

What are the right and left cosets
$$
A.SL(n,mathbb R),quad SL(n,mathbb R).A ?
$$

abstract-algebra group-theory

share|cite|improve this question

edited Mar 17 at 19:23

Jack

27.6k1782203

asked Oct 15 '18 at 11:23

129492129492

576

$endgroup$

closed as off-topic by Thomas Shelby, Leucippus, Lee David Chung Lin, Shailesh, Eevee Trainer Mar 18 at 3:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Thomas Shelby, Leucippus, Lee David Chung Lin, Shailesh, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $ to enclose your formulae.
  $endgroup$
  – DonAntonio
  Oct 15 '18 at 11:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give your definition of left and right coset? Do you mean the set $ Ag mid g in GL(nmathbbR)$, or do you mean $GL(n,mathbbR)/langle A rangle$, or something else entirely? Also, please show your work on that, this is not a place to get your homework done without own effort.
  $endgroup$
  – Dirk
  Oct 15 '18 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean the set $Ag∣g∈GL(nR)$
  $endgroup$
  – 129492
  Oct 15 '18 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you heard this result ? :$aH=H Longleftrightarrow a in H$
  $endgroup$
  – Chinnapparaj R
  Oct 15 '18 at 11:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have any idea on this exercise. I tried to find the set but i can't find any rule of this change .
  $endgroup$
  – 129492
  Oct 15 '18 at 11:31
  

  
  
  
  

 | 
show 4 more comments

-2

$begingroup$

Denoting
$$
GL(n,mathbb R)=gin mathbb R^ntimes nmiddet(g)neq 0, quad SL(n,mathbb R):=gin GL(n,mathbb R)middet(g)=1,
$$
we get $SL(n,mathbb R) le GL(n,mathbb R)$. Let $A in GL(n,mathbb R)$

What are the right and left cosets
$$
A.SL(n,mathbb R),quad SL(n,mathbb R).A ?
$$

abstract-algebra group-theory

share|cite|improve this question

edited Mar 17 at 19:23

Jack

27.6k1782203

asked Oct 15 '18 at 11:23

129492129492

576

$endgroup$

closed as off-topic by Thomas Shelby, Leucippus, Lee David Chung Lin, Shailesh, Eevee Trainer Mar 18 at 3:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Thomas Shelby, Leucippus, Lee David Chung Lin, Shailesh, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $ to enclose your formulae.
  $endgroup$
  – DonAntonio
  Oct 15 '18 at 11:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you give your definition of left and right coset? Do you mean the set $ Ag mid g in GL(nmathbbR)$, or do you mean $GL(n,mathbbR)/langle A rangle$, or something else entirely? Also, please show your work on that, this is not a place to get your homework done without own effort.
  $endgroup$
  – Dirk
  Oct 15 '18 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean the set $Ag∣g∈GL(nR)$
  $endgroup$
  – 129492
  Oct 15 '18 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you heard this result ? :$aH=H Longleftrightarrow a in H$
  $endgroup$
  – Chinnapparaj R
  Oct 15 '18 at 11:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have any idea on this exercise. I tried to find the set but i can't find any rule of this change .
  $endgroup$
  – 129492
  Oct 15 '18 at 11:31
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Denoting
$$
GL(n,mathbb R)=gin mathbb R^ntimes nmiddet(g)neq 0, quad SL(n,mathbb R):=gin GL(n,mathbb R)middet(g)=1,
$$
we get $SL(n,mathbb R) le GL(n,mathbb R)$. Let $A in GL(n,mathbb R)$

What are the right and left cosets
$$
A.SL(n,mathbb R),quad SL(n,mathbb R).A ?
$$

abstract-algebra group-theory

share|cite|improve this question

edited Mar 17 at 19:23

Jack

27.6k1782203

asked Oct 15 '18 at 11:23

129492129492

576

$endgroup$

share|cite|improve this question

edited Mar 17 at 19:23

Jack

27.6k1782203

asked Oct 15 '18 at 11:23

129492129492

576

share|cite|improve this question

edited Mar 17 at 19:23

Jack

27.6k1782203

asked Oct 15 '18 at 11:23

129492129492

576

edited Mar 17 at 19:23

Jack

27.6k1782203

edited Mar 17 at 19:23

Jack

27.6k1782203

asked Oct 15 '18 at 11:23

129492129492

576

asked Oct 15 '18 at 11:23

129492129492

576

2 Answers
2

active

oldest

votes

1

$begingroup$

Compute the determinant of any matrix in this coset. That should give you a rather good idea on how these cosets look like.

share|cite|improve this answer

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

$endgroup$

add a comment | 

1

$begingroup$

Here $$A.SL(n,BbbR)=AB: B in SL(n,BbbR)$$

Claim: $A.SL(n,BbbR)=BigB in GL(n, BbbR): textdet(B)=k Big$ if $textdet(A)=k neq 0$

Proof.
Let $B in GL(n,BbbR)$ be an arbitrary matrix with $textdet(B)=k neq 0$. Then $$textdet(A^-1B) =1 $$ so $A^-1B in SL(n,BbbR)$ and so $B in A cdot SL(n,BbbR)$. Thus $$BigB in GL(n, BbbR): textdet(B)=k Big subseteq A.SL(n,BbbR)$$

On the other hand, for any $AB in A.SL(n,BbbR)$, we have $$textdet(AB)=textdet(A) cdot textdet(B)=k cdot 1=k$$
so $$ A.SL(n,BbbR) subseteq BigB in GL(n, BbbR): textdet(B)=k Big;;;blacksquare$$

Summary: The set $A.SL(n,BbbR)$ is precisely of all matrices having determinant $k$ if $A$ has determinat $k neq 0$

share|cite|improve this answer

edited Oct 15 '18 at 12:21

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928

$endgroup$

add a comment | 

1

$begingroup$

Compute the determinant of any matrix in this coset. That should give you a rather good idea on how these cosets look like.

share|cite|improve this answer

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

$endgroup$

add a comment | 

1

$begingroup$

Compute the determinant of any matrix in this coset. That should give you a rather good idea on how these cosets look like.

share|cite|improve this answer

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

$endgroup$

add a comment | 

$begingroup$

Compute the determinant of any matrix in this coset. That should give you a rather good idea on how these cosets look like.

share|cite|improve this answer

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

$endgroup$

share|cite|improve this answer

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

answered Oct 15 '18 at 11:45

DirkDirk

4,438218

1

$begingroup$

Here $$A.SL(n,BbbR)=AB: B in SL(n,BbbR)$$

Claim: $A.SL(n,BbbR)=BigB in GL(n, BbbR): textdet(B)=k Big$ if $textdet(A)=k neq 0$

Proof.
Let $B in GL(n,BbbR)$ be an arbitrary matrix with $textdet(B)=k neq 0$. Then $$textdet(A^-1B) =1 $$ so $A^-1B in SL(n,BbbR)$ and so $B in A cdot SL(n,BbbR)$. Thus $$BigB in GL(n, BbbR): textdet(B)=k Big subseteq A.SL(n,BbbR)$$

On the other hand, for any $AB in A.SL(n,BbbR)$, we have $$textdet(AB)=textdet(A) cdot textdet(B)=k cdot 1=k$$
so $$ A.SL(n,BbbR) subseteq BigB in GL(n, BbbR): textdet(B)=k Big;;;blacksquare$$

Summary: The set $A.SL(n,BbbR)$ is precisely of all matrices having determinant $k$ if $A$ has determinat $k neq 0$

share|cite|improve this answer

edited Oct 15 '18 at 12:21

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928

$endgroup$

add a comment | 

1

$begingroup$

Here $$A.SL(n,BbbR)=AB: B in SL(n,BbbR)$$

Claim: $A.SL(n,BbbR)=BigB in GL(n, BbbR): textdet(B)=k Big$ if $textdet(A)=k neq 0$

Proof.
Let $B in GL(n,BbbR)$ be an arbitrary matrix with $textdet(B)=k neq 0$. Then $$textdet(A^-1B) =1 $$ so $A^-1B in SL(n,BbbR)$ and so $B in A cdot SL(n,BbbR)$. Thus $$BigB in GL(n, BbbR): textdet(B)=k Big subseteq A.SL(n,BbbR)$$

On the other hand, for any $AB in A.SL(n,BbbR)$, we have $$textdet(AB)=textdet(A) cdot textdet(B)=k cdot 1=k$$
so $$ A.SL(n,BbbR) subseteq BigB in GL(n, BbbR): textdet(B)=k Big;;;blacksquare$$

Summary: The set $A.SL(n,BbbR)$ is precisely of all matrices having determinant $k$ if $A$ has determinat $k neq 0$

share|cite|improve this answer

edited Oct 15 '18 at 12:21

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928

$endgroup$

add a comment | 

$begingroup$

Here $$A.SL(n,BbbR)=AB: B in SL(n,BbbR)$$

Claim: $A.SL(n,BbbR)=BigB in GL(n, BbbR): textdet(B)=k Big$ if $textdet(A)=k neq 0$

Proof.
Let $B in GL(n,BbbR)$ be an arbitrary matrix with $textdet(B)=k neq 0$. Then $$textdet(A^-1B) =1 $$ so $A^-1B in SL(n,BbbR)$ and so $B in A cdot SL(n,BbbR)$. Thus $$BigB in GL(n, BbbR): textdet(B)=k Big subseteq A.SL(n,BbbR)$$

On the other hand, for any $AB in A.SL(n,BbbR)$, we have $$textdet(AB)=textdet(A) cdot textdet(B)=k cdot 1=k$$
so $$ A.SL(n,BbbR) subseteq BigB in GL(n, BbbR): textdet(B)=k Big;;;blacksquare$$

Summary: The set $A.SL(n,BbbR)$ is precisely of all matrices having determinant $k$ if $A$ has determinat $k neq 0$

share|cite|improve this answer

edited Oct 15 '18 at 12:21

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928

$endgroup$

share|cite|improve this answer

edited Oct 15 '18 at 12:21

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928

answered Oct 15 '18 at 11:52

Chinnapparaj RChinnapparaj R

5,8082928