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Number of right residue classes equal to left ones

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0

1

$begingroup$

Let $G$ is a group and $H leq G$ and $x in G$

Show that number of $G$'s right residue classes are equal to left ones wrt $H$

1) There is a hint in my notes " use $f: Hx to x^-1H$ but using $f : Hx to xH$ looks better for me. Is it wrong?

If it is not, should I put elements in function like below?

2) For instant, should I write $" f(h_1x)=f(h_2x) Rightarrow h_1=h_2$ when showing being (1-1)?

3) Last, how can I show surjectivity in the best way? We have never explain surjectivity in detail in our examples.

In addition there is a little exercise.

Take $G=S_4$ and $H=I,(12)(34),(13)(24),(14)(23)$ where $I$ is identity function. Find all right residue classes wrt $H$

4) We know by Lagrange's Theorem that the number of right residue classes is $6$. Do I have to multiply all elements one by one or is there anything that I can do quickly?

Any help'd be very beneficial for me. Thanks in advance

abstract-algebra group-theory permutations division-algebras

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asked Mar 29 at 21:57

user519955user519955

328111

$endgroup$

add a comment | 

0

1

$begingroup$

Let $G$ is a group and $H leq G$ and $x in G$

Show that number of $G$'s right residue classes are equal to left ones wrt $H$

1) There is a hint in my notes " use $f: Hx to x^-1H$ but using $f : Hx to xH$ looks better for me. Is it wrong?

If it is not, should I put elements in function like below?

2) For instant, should I write $" f(h_1x)=f(h_2x) Rightarrow h_1=h_2$ when showing being (1-1)?

3) Last, how can I show surjectivity in the best way? We have never explain surjectivity in detail in our examples.

In addition there is a little exercise.

Take $G=S_4$ and $H=I,(12)(34),(13)(24),(14)(23)$ where $I$ is identity function. Find all right residue classes wrt $H$

4) We know by Lagrange's Theorem that the number of right residue classes is $6$. Do I have to multiply all elements one by one or is there anything that I can do quickly?

Any help'd be very beneficial for me. Thanks in advance

abstract-algebra group-theory permutations division-algebras

share|cite|improve this question

asked Mar 29 at 21:57

user519955user519955

328111

$endgroup$

add a comment | 

$begingroup$

Let $G$ is a group and $H leq G$ and $x in G$

Show that number of $G$'s right residue classes are equal to left ones wrt $H$

1) There is a hint in my notes " use $f: Hx to x^-1H$ but using $f : Hx to xH$ looks better for me. Is it wrong?

If it is not, should I put elements in function like below?

2) For instant, should I write $" f(h_1x)=f(h_2x) Rightarrow h_1=h_2$ when showing being (1-1)?

3) Last, how can I show surjectivity in the best way? We have never explain surjectivity in detail in our examples.

In addition there is a little exercise.

Take $G=S_4$ and $H=I,(12)(34),(13)(24),(14)(23)$ where $I$ is identity function. Find all right residue classes wrt $H$

4) We know by Lagrange's Theorem that the number of right residue classes is $6$. Do I have to multiply all elements one by one or is there anything that I can do quickly?

Any help'd be very beneficial for me. Thanks in advance

abstract-algebra group-theory permutations division-algebras

share|cite|improve this question

asked Mar 29 at 21:57

user519955user519955

328111

$endgroup$

share|cite|improve this question

asked Mar 29 at 21:57

user519955user519955

328111

share|cite|improve this question

asked Mar 29 at 21:57

user519955user519955

328111

asked Mar 29 at 21:57

user519955user519955

328111

asked Mar 29 at 21:57

user519955user519955

328111

1 Answer
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0

$begingroup$

The reason why you would want to use $f: Hx rightarrow x^-1H$ is because you can show that it is simply a bijection where:

$$f(Hx) = (hx)^-1 : h in H = x^-1h^-1 : h in H = x^-1H $$

So it's similar to yours, except in your approach there is no clear way to identify the left and right actions on $H$. Instead, I simply use the fact that $x$ has an inverse that we can immediately use.

share|cite|improve this answer

answered Mar 29 at 22:13

gdepaulgdepaul

1537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for answer. Did you say there is no clear way for 4th question? I’m confused a little bit. Should I fix h’s or x’s or none of them while taking 2 different elements in showing bijectivity?
  $endgroup$
  – user519955
  Mar 30 at 8:29
  

  
  
  
  

add a comment | 

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0

$begingroup$

The reason why you would want to use $f: Hx rightarrow x^-1H$ is because you can show that it is simply a bijection where:

$$f(Hx) = (hx)^-1 : h in H = x^-1h^-1 : h in H = x^-1H $$

So it's similar to yours, except in your approach there is no clear way to identify the left and right actions on $H$. Instead, I simply use the fact that $x$ has an inverse that we can immediately use.

share|cite|improve this answer

answered Mar 29 at 22:13

gdepaulgdepaul

1537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for answer. Did you say there is no clear way for 4th question? I’m confused a little bit. Should I fix h’s or x’s or none of them while taking 2 different elements in showing bijectivity?
  $endgroup$
  – user519955
  Mar 30 at 8:29
  

  
  
  
  

add a comment | 

0

$begingroup$

The reason why you would want to use $f: Hx rightarrow x^-1H$ is because you can show that it is simply a bijection where:

$$f(Hx) = (hx)^-1 : h in H = x^-1h^-1 : h in H = x^-1H $$

So it's similar to yours, except in your approach there is no clear way to identify the left and right actions on $H$. Instead, I simply use the fact that $x$ has an inverse that we can immediately use.

share|cite|improve this answer

answered Mar 29 at 22:13

gdepaulgdepaul

1537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for answer. Did you say there is no clear way for 4th question? I’m confused a little bit. Should I fix h’s or x’s or none of them while taking 2 different elements in showing bijectivity?
  $endgroup$
  – user519955
  Mar 30 at 8:29
  

  
  
  
  

add a comment | 

$begingroup$

The reason why you would want to use $f: Hx rightarrow x^-1H$ is because you can show that it is simply a bijection where:

$$f(Hx) = (hx)^-1 : h in H = x^-1h^-1 : h in H = x^-1H $$

So it's similar to yours, except in your approach there is no clear way to identify the left and right actions on $H$. Instead, I simply use the fact that $x$ has an inverse that we can immediately use.

share|cite|improve this answer

answered Mar 29 at 22:13

gdepaulgdepaul

1537

$endgroup$

share|cite|improve this answer

answered Mar 29 at 22:13

gdepaulgdepaul

1537

answered Mar 29 at 22:13

gdepaulgdepaul

1537

answered Mar 29 at 22:13

gdepaulgdepaul

1537