Understanding proof about a group and homomorphism












0












$begingroup$


I have a book about group theory and there was the following question:




Let $G$ be a set of all the real matrices in the following form: $begin{pmatrix}a & b\
-b & a
end{pmatrix}$ when $a^2+b^2>0$.




  1. Prove that $G$ is a group.

  2. Prove that $Gcong (C^times,cdot )$.




I successfully proved that $G$ is a group. Now I'm trying to prove the second sub-question. In the book they suggested to declare the following function:



$$ f:begin{pmatrix}a & b\
-b & a
end{pmatrix} to a+ib$$



Also they wrote "obviously $f$ is bijection", and then they proved the Homomorphism equation. The only part that I didn't understand is why $f$ is a bijection, and why it is so obvious? How can I prove it formally?






group-theory complex-numbers group-homomorphism







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$endgroup$












  • $begingroup$
    Which book are you using?
    $endgroup$
    – Shaun
    Dec 21 '18 at 17:29
















0












$begingroup$


I have a book about group theory and there was the following question:




Let $G$ be a set of all the real matrices in the following form: $begin{pmatrix}a & b\
-b & a
end{pmatrix}$ when $a^2+b^2>0$.




  1. Prove that $G$ is a group.

  2. Prove that $Gcong (C^times,cdot )$.




I successfully proved that $G$ is a group. Now I'm trying to prove the second sub-question. In the book they suggested to declare the following function:



$$ f:begin{pmatrix}a & b\
-b & a
end{pmatrix} to a+ib$$



Also they wrote "obviously $f$ is bijection", and then they proved the Homomorphism equation. The only part that I didn't understand is why $f$ is a bijection, and why it is so obvious? How can I prove it formally?






group-theory complex-numbers group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which book are you using?
    $endgroup$
    – Shaun
    Dec 21 '18 at 17:29














0












0








0





$begingroup$


I have a book about group theory and there was the following question:




Let $G$ be a set of all the real matrices in the following form: $begin{pmatrix}a & b\
-b & a
end{pmatrix}$ when $a^2+b^2>0$.




  1. Prove that $G$ is a group.

  2. Prove that $Gcong (C^times,cdot )$.




I successfully proved that $G$ is a group. Now I'm trying to prove the second sub-question. In the book they suggested to declare the following function:



$$ f:begin{pmatrix}a & b\
-b & a
end{pmatrix} to a+ib$$



Also they wrote "obviously $f$ is bijection", and then they proved the Homomorphism equation. The only part that I didn't understand is why $f$ is a bijection, and why it is so obvious? How can I prove it formally?






group-theory complex-numbers group-homomorphism







share|cite|improve this question











$endgroup$




I have a book about group theory and there was the following question:




Let $G$ be a set of all the real matrices in the following form: $begin{pmatrix}a & b\
-b & a
end{pmatrix}$ when $a^2+b^2>0$.




  1. Prove that $G$ is a group.

  2. Prove that $Gcong (C^times,cdot )$.




I successfully proved that $G$ is a group. Now I'm trying to prove the second sub-question. In the book they suggested to declare the following function:



$$ f:begin{pmatrix}a & b\
-b & a
end{pmatrix} to a+ib$$



Also they wrote "obviously $f$ is bijection", and then they proved the Homomorphism equation. The only part that I didn't understand is why $f$ is a bijection, and why it is so obvious? How can I prove it formally?






group-theory complex-numbers group-homomorphism




group-theory complex-numbers group-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:32









Shaun

9,544113684




9,544113684










asked Dec 18 '18 at 16:12









vesiivesii

3828




3828












  • $begingroup$
    Which book are you using?
    $endgroup$
    – Shaun
    Dec 21 '18 at 17:29


















  • $begingroup$
    Which book are you using?
    $endgroup$
    – Shaun
    Dec 21 '18 at 17:29
















$begingroup$
Which book are you using?
$endgroup$
– Shaun
Dec 21 '18 at 17:29




$begingroup$
Which book are you using?
$endgroup$
– Shaun
Dec 21 '18 at 17:29










3 Answers
3






active

oldest

votes


















0












$begingroup$

Since $mathbb{C}^times=mathbb{C}setminus{0}$ and since each $zinmathbb{C}setminus{0}$ can be written in one and only one way as $a+bi$ with $a$ and $b$ not both equal to $0$ ($iff a^2+b^2>0$), it is clear that $f$ is a bijection.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    To see that the map is a bijection note that the kernel consists precisely of the identity matrix and hence the map is injective. Indeed if $f(A)=1=a+ib$ for some matrix $A=begin{bmatrix}
    a& b\
    -b& a
    end{bmatrix}
    $    ,
    then we have that $a=1$ and $b=0$. Conversely given any $zin mathbb{C}setminus {0}$, let $a=text{Re}(z)$, and $b=text{Im}(z)$, so that
    $$
    begin{bmatrix}
    a& b\
    -b& a
    end{bmatrix}to z
    $$
    whence the map is surjective.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      For every complex number $mathbb{C}-{0} ni x:=a+bi$,(hence $a neq 0$ or $b neq 0$). Consider the matrix
      $$y=begin{bmatrix}
      a & b \
      -b & a \
      end{bmatrix}$$      , The determinant of this matrix of $a^2+b^2 >0$ and it gets map to $a+bi$ by $f$(that is it! $f(y)=x$). So this shows surjectivity!



      For injectivity, we have if $0neq a+bi=f(A)=f(B)=c+di neq 0$, then $a=c, b=d$. so
      $$A=begin{bmatrix}
      a & b \
      -b & a \
      end{bmatrix}=begin{bmatrix}
      c & d \
      -d & c \
      end{bmatrix}=B$$
      And so $f$ is bijective!






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
        $endgroup$
        – vesii
        Dec 18 '18 at 17:21










      • $begingroup$
        I edited a bit my answer I don't know if that helps?
        $endgroup$
        – nafhgood
        Dec 18 '18 at 17:35











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      Since $mathbb{C}^times=mathbb{C}setminus{0}$ and since each $zinmathbb{C}setminus{0}$ can be written in one and only one way as $a+bi$ with $a$ and $b$ not both equal to $0$ ($iff a^2+b^2>0$), it is clear that $f$ is a bijection.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Since $mathbb{C}^times=mathbb{C}setminus{0}$ and since each $zinmathbb{C}setminus{0}$ can be written in one and only one way as $a+bi$ with $a$ and $b$ not both equal to $0$ ($iff a^2+b^2>0$), it is clear that $f$ is a bijection.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Since $mathbb{C}^times=mathbb{C}setminus{0}$ and since each $zinmathbb{C}setminus{0}$ can be written in one and only one way as $a+bi$ with $a$ and $b$ not both equal to $0$ ($iff a^2+b^2>0$), it is clear that $f$ is a bijection.






          share|cite|improve this answer









          $endgroup$



          Since $mathbb{C}^times=mathbb{C}setminus{0}$ and since each $zinmathbb{C}setminus{0}$ can be written in one and only one way as $a+bi$ with $a$ and $b$ not both equal to $0$ ($iff a^2+b^2>0$), it is clear that $f$ is a bijection.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 16:16









          José Carlos SantosJosé Carlos Santos

          168k22132236




          168k22132236























              0












              $begingroup$

              To see that the map is a bijection note that the kernel consists precisely of the identity matrix and hence the map is injective. Indeed if $f(A)=1=a+ib$ for some matrix $A=begin{bmatrix}
              a& b\
              -b& a
              end{bmatrix}
              $              ,
              then we have that $a=1$ and $b=0$. Conversely given any $zin mathbb{C}setminus {0}$, let $a=text{Re}(z)$, and $b=text{Im}(z)$, so that
              $$
              begin{bmatrix}
              a& b\
              -b& a
              end{bmatrix}to z
              $$
              whence the map is surjective.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                To see that the map is a bijection note that the kernel consists precisely of the identity matrix and hence the map is injective. Indeed if $f(A)=1=a+ib$ for some matrix $A=begin{bmatrix}
                a& b\
                -b& a
                end{bmatrix}
                $                ,
                then we have that $a=1$ and $b=0$. Conversely given any $zin mathbb{C}setminus {0}$, let $a=text{Re}(z)$, and $b=text{Im}(z)$, so that
                $$
                begin{bmatrix}
                a& b\
                -b& a
                end{bmatrix}to z
                $$
                whence the map is surjective.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  To see that the map is a bijection note that the kernel consists precisely of the identity matrix and hence the map is injective. Indeed if $f(A)=1=a+ib$ for some matrix $A=begin{bmatrix}
                  a& b\
                  -b& a
                  end{bmatrix}
                  $                  ,
                  then we have that $a=1$ and $b=0$. Conversely given any $zin mathbb{C}setminus {0}$, let $a=text{Re}(z)$, and $b=text{Im}(z)$, so that
                  $$
                  begin{bmatrix}
                  a& b\
                  -b& a
                  end{bmatrix}to z
                  $$
                  whence the map is surjective.






                  share|cite|improve this answer









                  $endgroup$



                  To see that the map is a bijection note that the kernel consists precisely of the identity matrix and hence the map is injective. Indeed if $f(A)=1=a+ib$ for some matrix $A=begin{bmatrix}
                  a& b\
                  -b& a
                  end{bmatrix}
                  $                  ,
                  then we have that $a=1$ and $b=0$. Conversely given any $zin mathbb{C}setminus {0}$, let $a=text{Re}(z)$, and $b=text{Im}(z)$, so that
                  $$
                  begin{bmatrix}
                  a& b\
                  -b& a
                  end{bmatrix}to z
                  $$
                  whence the map is surjective.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 16:18









                  Foobaz JohnFoobaz John

                  22.7k41452




                  22.7k41452























                      0












                      $begingroup$

                      For every complex number $mathbb{C}-{0} ni x:=a+bi$,(hence $a neq 0$ or $b neq 0$). Consider the matrix
                      $$y=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}$$                      , The determinant of this matrix of $a^2+b^2 >0$ and it gets map to $a+bi$ by $f$(that is it! $f(y)=x$). So this shows surjectivity!



                      For injectivity, we have if $0neq a+bi=f(A)=f(B)=c+di neq 0$, then $a=c, b=d$. so
                      $$A=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}=begin{bmatrix}
                      c & d \
                      -d & c \
                      end{bmatrix}=B$$
                      And so $f$ is bijective!






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
                        $endgroup$
                        – vesii
                        Dec 18 '18 at 17:21










                      • $begingroup$
                        I edited a bit my answer I don't know if that helps?
                        $endgroup$
                        – nafhgood
                        Dec 18 '18 at 17:35
















                      0












                      $begingroup$

                      For every complex number $mathbb{C}-{0} ni x:=a+bi$,(hence $a neq 0$ or $b neq 0$). Consider the matrix
                      $$y=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}$$                      , The determinant of this matrix of $a^2+b^2 >0$ and it gets map to $a+bi$ by $f$(that is it! $f(y)=x$). So this shows surjectivity!



                      For injectivity, we have if $0neq a+bi=f(A)=f(B)=c+di neq 0$, then $a=c, b=d$. so
                      $$A=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}=begin{bmatrix}
                      c & d \
                      -d & c \
                      end{bmatrix}=B$$
                      And so $f$ is bijective!






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
                        $endgroup$
                        – vesii
                        Dec 18 '18 at 17:21










                      • $begingroup$
                        I edited a bit my answer I don't know if that helps?
                        $endgroup$
                        – nafhgood
                        Dec 18 '18 at 17:35














                      0












                      0








                      0





                      $begingroup$

                      For every complex number $mathbb{C}-{0} ni x:=a+bi$,(hence $a neq 0$ or $b neq 0$). Consider the matrix
                      $$y=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}$$                      , The determinant of this matrix of $a^2+b^2 >0$ and it gets map to $a+bi$ by $f$(that is it! $f(y)=x$). So this shows surjectivity!



                      For injectivity, we have if $0neq a+bi=f(A)=f(B)=c+di neq 0$, then $a=c, b=d$. so
                      $$A=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}=begin{bmatrix}
                      c & d \
                      -d & c \
                      end{bmatrix}=B$$
                      And so $f$ is bijective!






                      share|cite|improve this answer











                      $endgroup$



                      For every complex number $mathbb{C}-{0} ni x:=a+bi$,(hence $a neq 0$ or $b neq 0$). Consider the matrix
                      $$y=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}$$                      , The determinant of this matrix of $a^2+b^2 >0$ and it gets map to $a+bi$ by $f$(that is it! $f(y)=x$). So this shows surjectivity!



                      For injectivity, we have if $0neq a+bi=f(A)=f(B)=c+di neq 0$, then $a=c, b=d$. so
                      $$A=begin{bmatrix}
                      a & b \
                      -b & a \
                      end{bmatrix}=begin{bmatrix}
                      c & d \
                      -d & c \
                      end{bmatrix}=B$$
                      And so $f$ is bijective!







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 18 '18 at 17:35

























                      answered Dec 18 '18 at 16:21









                      nafhgoodnafhgood

                      1,803422




                      1,803422












                      • $begingroup$
                        I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
                        $endgroup$
                        – vesii
                        Dec 18 '18 at 17:21










                      • $begingroup$
                        I edited a bit my answer I don't know if that helps?
                        $endgroup$
                        – nafhgood
                        Dec 18 '18 at 17:35


















                      • $begingroup$
                        I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
                        $endgroup$
                        – vesii
                        Dec 18 '18 at 17:21










                      • $begingroup$
                        I edited a bit my answer I don't know if that helps?
                        $endgroup$
                        – nafhgood
                        Dec 18 '18 at 17:35
















                      $begingroup$
                      I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
                      $endgroup$
                      – vesii
                      Dec 18 '18 at 17:21




                      $begingroup$
                      I understand why $f$ is injective, but I don't understand why $f$ is surjective. I know that I need to show that for every $yin Y$ there is $xin X$ so $f(x)=y$. From similar question I say that I need to point for that exact $x$ to prove the example. Which $x$ should I use?
                      $endgroup$
                      – vesii
                      Dec 18 '18 at 17:21












                      $begingroup$
                      I edited a bit my answer I don't know if that helps?
                      $endgroup$
                      – nafhgood
                      Dec 18 '18 at 17:35




                      $begingroup$
                      I edited a bit my answer I don't know if that helps?
                      $endgroup$
                      – nafhgood
                      Dec 18 '18 at 17:35


















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