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














up vote
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down vote

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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




share|cite|improve this question



















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












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











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




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
















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















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




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










2 Answers
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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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    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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      2 Answers
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      2 Answers
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      up vote
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      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

























        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























          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













          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




          98.4k793174




















              up vote
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              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

























                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























                  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













                  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

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