Index of subgroup of Real number in Complex Number

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Complex Number is group under addition. Real Number is its subgroup. It is the trivially normal subgroup. I wanted to find its index.

I thought one 2 dimension plane with X axis and Y axis. Its coset can be given in form like $(0,a)+(R,0)$ Where $a$ is any real Number.
This then says that is index is same as the cardinality of R.

Is I am right?

abstract-algebra group-theory normal-subgroups

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asked Jul 28 at 12:19

SRJ

1,041317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Exactly right. . .
  – Lubin
  Jul 28 at 12:28
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

1

Complex Number is group under addition. Real Number is its subgroup. It is the trivially normal subgroup. I wanted to find its index.

I thought one 2 dimension plane with X axis and Y axis. Its coset can be given in form like $(0,a)+(R,0)$ Where $a$ is any real Number.
This then says that is index is same as the cardinality of R.

Is I am right?

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked Jul 28 at 12:19

SRJ

1,041317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Exactly right. . .
  – Lubin
  Jul 28 at 12:28
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

Complex Number is group under addition. Real Number is its subgroup. It is the trivially normal subgroup. I wanted to find its index.

I thought one 2 dimension plane with X axis and Y axis. Its coset can be given in form like $(0,a)+(R,0)$ Where $a$ is any real Number.
This then says that is index is same as the cardinality of R.

Is I am right?

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked Jul 28 at 12:19

SRJ

1,041317

Complex Number is group under addition. Real Number is its subgroup. It is the trivially normal subgroup. I wanted to find its index.

I thought one 2 dimension plane with X axis and Y axis. Its coset can be given in form like $(0,a)+(R,0)$ Where $a$ is any real Number.
This then says that is index is same as the cardinality of R.

Is I am right?

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked Jul 28 at 12:19

SRJ

1,041317

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 12:19

SRJ

1,041317

asked Jul 28 at 12:19

SRJ

1,041317

asked Jul 28 at 12:19

SRJ

1,041317

1,041317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Exactly right. . .
  – Lubin
  Jul 28 at 12:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Exactly right. . .
  – Lubin
  Jul 28 at 12:28
  

  
  
  
  

Exactly right. . .
– Lubin
Jul 28 at 12:28

Exactly right. . .
– Lubin
Jul 28 at 12:28

add a comment | 

2 Answers
2

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3
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accepted

You are right, the index is $|Bbb R|$ because that's how many cosets $Bbb R$ has in $Bbb C$. Geometrically, each coset is a horizontal line in the complex plane, and each horizontal line is a coset. Your argument is an exact algebraic analogue of that.

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answered Jul 28 at 13:02

Arthur

98.4k793174

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Another way to see this: as abelian additive groups $mathbbC=mathbbR oplus mathbbR$. Here $mathbbR cong (r,0): r in mathbbR$. Now define a map $mathbbC rightarrow mathbbR$, by $z mapsto Im(z)$. This is a surjective homomorphism, with kernel $mathbbR$. Hence $mathbbC/mathbbR cong mathbbR$ as additive groups.

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answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

You are right, the index is $|Bbb R|$ because that's how many cosets $Bbb R$ has in $Bbb C$. Geometrically, each coset is a horizontal line in the complex plane, and each horizontal line is a coset. Your argument is an exact algebraic analogue of that.

share|cite|improve this answer

answered Jul 28 at 13:02

Arthur

98.4k793174

add a comment | 

up vote
3
down vote



accepted

You are right, the index is $|Bbb R|$ because that's how many cosets $Bbb R$ has in $Bbb C$. Geometrically, each coset is a horizontal line in the complex plane, and each horizontal line is a coset. Your argument is an exact algebraic analogue of that.

share|cite|improve this answer

answered Jul 28 at 13:02

Arthur

98.4k793174

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

You are right, the index is $|Bbb R|$ because that's how many cosets $Bbb R$ has in $Bbb C$. Geometrically, each coset is a horizontal line in the complex plane, and each horizontal line is a coset. Your argument is an exact algebraic analogue of that.

share|cite|improve this answer

answered Jul 28 at 13:02

Arthur

98.4k793174

You are right, the index is $|Bbb R|$ because that's how many cosets $Bbb R$ has in $Bbb C$. Geometrically, each coset is a horizontal line in the complex plane, and each horizontal line is a coset. Your argument is an exact algebraic analogue of that.

share|cite|improve this answer

answered Jul 28 at 13:02

Arthur

98.4k793174

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 13:02

Arthur

98.4k793174

answered Jul 28 at 13:02

Arthur

98.4k793174

answered Jul 28 at 13:02

Arthur

98.4k793174

98.4k793174

add a comment | 

add a comment | 

up vote
3
down vote

Another way to see this: as abelian additive groups $mathbbC=mathbbR oplus mathbbR$. Here $mathbbR cong (r,0): r in mathbbR$. Now define a map $mathbbC rightarrow mathbbR$, by $z mapsto Im(z)$. This is a surjective homomorphism, with kernel $mathbbR$. Hence $mathbbC/mathbbR cong mathbbR$ as additive groups.

share|cite|improve this answer

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

add a comment | 

up vote
3
down vote

Another way to see this: as abelian additive groups $mathbbC=mathbbR oplus mathbbR$. Here $mathbbR cong (r,0): r in mathbbR$. Now define a map $mathbbC rightarrow mathbbR$, by $z mapsto Im(z)$. This is a surjective homomorphism, with kernel $mathbbR$. Hence $mathbbC/mathbbR cong mathbbR$ as additive groups.

share|cite|improve this answer

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

add a comment | 

up vote
3
down vote

up vote
3
down vote

Another way to see this: as abelian additive groups $mathbbC=mathbbR oplus mathbbR$. Here $mathbbR cong (r,0): r in mathbbR$. Now define a map $mathbbC rightarrow mathbbR$, by $z mapsto Im(z)$. This is a surjective homomorphism, with kernel $mathbbR$. Hence $mathbbC/mathbbR cong mathbbR$ as additive groups.

share|cite|improve this answer

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

Another way to see this: as abelian additive groups $mathbbC=mathbbR oplus mathbbR$. Here $mathbbR cong (r,0): r in mathbbR$. Now define a map $mathbbC rightarrow mathbbR$, by $z mapsto Im(z)$. This is a surjective homomorphism, with kernel $mathbbR$. Hence $mathbbC/mathbbR cong mathbbR$ as additive groups.

share|cite|improve this answer

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

answered Jul 29 at 12:51

Nicky Hekster

26.9k53052

26.9k53052

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add a comment | 

 

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