The group homomorphism $z mapsto (z/|z|)^2$ with an application of the homomorphism theorem.












1














Question

Let $G = (mathbb{C}setminus{0}, cdot)$.

1) Show that $N = (mathbb{R}setminus {0},cdot)$ is a normal subgroup.

2) Show that $f:G rightarrow G$ with $z mapsto (z/|z|)^2$ is a group homomorphism.

3) Determine $H$ s.t. $G/N cong H$ (using $f$ from 2.)

My solution

I think 1) is obvious bc. $N$ is a subgroup of $G$ and $G$ is commutative. Therefore $N$ is a normal subgroup. For 2) i take $w,z in G$ and calculate $f(wz)$ and $f(w)f(z)$ and look if they are identical.
$f(wz)=(wz/|wz|)^2$
$f(w)f(z)=(w/|w|)^2 cdot (z/|z|)^2 = (wz/|wz|)^2$,

so $f$ is a homomorphism.

in 3) i would say that $H = {z : |z| = 1}$, so i would say $H$ is the unit circle bc i only get complex numbers with absolute value $1$ and i get every point on the unit circle.
Is my solution correct?






abstract-algebra group-theory normal-subgroups group-homomorphism







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    Question

    Let $G = (mathbb{C}setminus{0}, cdot)$.

    1) Show that $N = (mathbb{R}setminus {0},cdot)$ is a normal subgroup.

    2) Show that $f:G rightarrow G$ with $z mapsto (z/|z|)^2$ is a group homomorphism.

    3) Determine $H$ s.t. $G/N cong H$ (using $f$ from 2.)

    My solution

    I think 1) is obvious bc. $N$ is a subgroup of $G$ and $G$ is commutative. Therefore $N$ is a normal subgroup. For 2) i take $w,z in G$ and calculate $f(wz)$ and $f(w)f(z)$ and look if they are identical.
    $f(wz)=(wz/|wz|)^2$
    $f(w)f(z)=(w/|w|)^2 cdot (z/|z|)^2 = (wz/|wz|)^2$,

    so $f$ is a homomorphism.

    in 3) i would say that $H = {z : |z| = 1}$, so i would say $H$ is the unit circle bc i only get complex numbers with absolute value $1$ and i get every point on the unit circle.
    Is my solution correct?






    abstract-algebra group-theory normal-subgroups group-homomorphism







    share|cite|improve this question

























      1












      1








      1


      0





      Question

      Let $G = (mathbb{C}setminus{0}, cdot)$.

      1) Show that $N = (mathbb{R}setminus {0},cdot)$ is a normal subgroup.

      2) Show that $f:G rightarrow G$ with $z mapsto (z/|z|)^2$ is a group homomorphism.

      3) Determine $H$ s.t. $G/N cong H$ (using $f$ from 2.)

      My solution

      I think 1) is obvious bc. $N$ is a subgroup of $G$ and $G$ is commutative. Therefore $N$ is a normal subgroup. For 2) i take $w,z in G$ and calculate $f(wz)$ and $f(w)f(z)$ and look if they are identical.
      $f(wz)=(wz/|wz|)^2$
      $f(w)f(z)=(w/|w|)^2 cdot (z/|z|)^2 = (wz/|wz|)^2$,

      so $f$ is a homomorphism.

      in 3) i would say that $H = {z : |z| = 1}$, so i would say $H$ is the unit circle bc i only get complex numbers with absolute value $1$ and i get every point on the unit circle.
      Is my solution correct?






      abstract-algebra group-theory normal-subgroups group-homomorphism







      share|cite|improve this question













      Question

      Let $G = (mathbb{C}setminus{0}, cdot)$.

      1) Show that $N = (mathbb{R}setminus {0},cdot)$ is a normal subgroup.

      2) Show that $f:G rightarrow G$ with $z mapsto (z/|z|)^2$ is a group homomorphism.

      3) Determine $H$ s.t. $G/N cong H$ (using $f$ from 2.)

      My solution

      I think 1) is obvious bc. $N$ is a subgroup of $G$ and $G$ is commutative. Therefore $N$ is a normal subgroup. For 2) i take $w,z in G$ and calculate $f(wz)$ and $f(w)f(z)$ and look if they are identical.
      $f(wz)=(wz/|wz|)^2$
      $f(w)f(z)=(w/|w|)^2 cdot (z/|z|)^2 = (wz/|wz|)^2$,

      so $f$ is a homomorphism.

      in 3) i would say that $H = {z : |z| = 1}$, so i would say $H$ is the unit circle bc i only get complex numbers with absolute value $1$ and i get every point on the unit circle.
      Is my solution correct?






      abstract-algebra group-theory normal-subgroups group-homomorphism




      abstract-algebra group-theory normal-subgroups group-homomorphism






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      asked Nov 28 '18 at 11:35









      GMiiX

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          The first isomorphism theorem tells you that if you have a homomorphism from $G$ to $H$ with exactly $N$ as kernel, then $G/N$ is isomorphic to the image of this homomorphism.



          Therefore, your task should be to locate a homomorphism which has kernel exactly $N$, and then find the image of this homomorphism.



          Now, what is the kernel of $f$? Note that $|z|^2 = z bar z$, so $frac{z^2}{|z|^2} = frac z{bar z}$ ,which equals $1$ if and only if $z = bar z$, if and only if $z in mathbb R$. In other words, the kernel of $f$ is indeed $N$.



          As for the image, as you have pointed out, since $|z|= |bar z|$, we see that the image is certainly contained within $H = {|x|= 1}$. However, the converse needs to be shown i.e. that $H$ is contained in the image. This is not very difficult : let $y in H$, and let $z$ be a square root of $y$, which always exists by the fundamental theorem of algebra (or using the polar form). Then $|z| = 1$, and $f(z) = y$.



          Therefore, your statements are correct, but here is the formal verification anyway.






          share|cite|improve this answer





















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            The first isomorphism theorem tells you that if you have a homomorphism from $G$ to $H$ with exactly $N$ as kernel, then $G/N$ is isomorphic to the image of this homomorphism.



            Therefore, your task should be to locate a homomorphism which has kernel exactly $N$, and then find the image of this homomorphism.



            Now, what is the kernel of $f$? Note that $|z|^2 = z bar z$, so $frac{z^2}{|z|^2} = frac z{bar z}$ ,which equals $1$ if and only if $z = bar z$, if and only if $z in mathbb R$. In other words, the kernel of $f$ is indeed $N$.



            As for the image, as you have pointed out, since $|z|= |bar z|$, we see that the image is certainly contained within $H = {|x|= 1}$. However, the converse needs to be shown i.e. that $H$ is contained in the image. This is not very difficult : let $y in H$, and let $z$ be a square root of $y$, which always exists by the fundamental theorem of algebra (or using the polar form). Then $|z| = 1$, and $f(z) = y$.



            Therefore, your statements are correct, but here is the formal verification anyway.






            share|cite|improve this answer


























              3














              The first isomorphism theorem tells you that if you have a homomorphism from $G$ to $H$ with exactly $N$ as kernel, then $G/N$ is isomorphic to the image of this homomorphism.



              Therefore, your task should be to locate a homomorphism which has kernel exactly $N$, and then find the image of this homomorphism.



              Now, what is the kernel of $f$? Note that $|z|^2 = z bar z$, so $frac{z^2}{|z|^2} = frac z{bar z}$ ,which equals $1$ if and only if $z = bar z$, if and only if $z in mathbb R$. In other words, the kernel of $f$ is indeed $N$.



              As for the image, as you have pointed out, since $|z|= |bar z|$, we see that the image is certainly contained within $H = {|x|= 1}$. However, the converse needs to be shown i.e. that $H$ is contained in the image. This is not very difficult : let $y in H$, and let $z$ be a square root of $y$, which always exists by the fundamental theorem of algebra (or using the polar form). Then $|z| = 1$, and $f(z) = y$.



              Therefore, your statements are correct, but here is the formal verification anyway.






              share|cite|improve this answer
























                3












                3








                3






                The first isomorphism theorem tells you that if you have a homomorphism from $G$ to $H$ with exactly $N$ as kernel, then $G/N$ is isomorphic to the image of this homomorphism.



                Therefore, your task should be to locate a homomorphism which has kernel exactly $N$, and then find the image of this homomorphism.



                Now, what is the kernel of $f$? Note that $|z|^2 = z bar z$, so $frac{z^2}{|z|^2} = frac z{bar z}$ ,which equals $1$ if and only if $z = bar z$, if and only if $z in mathbb R$. In other words, the kernel of $f$ is indeed $N$.



                As for the image, as you have pointed out, since $|z|= |bar z|$, we see that the image is certainly contained within $H = {|x|= 1}$. However, the converse needs to be shown i.e. that $H$ is contained in the image. This is not very difficult : let $y in H$, and let $z$ be a square root of $y$, which always exists by the fundamental theorem of algebra (or using the polar form). Then $|z| = 1$, and $f(z) = y$.



                Therefore, your statements are correct, but here is the formal verification anyway.






                share|cite|improve this answer












                The first isomorphism theorem tells you that if you have a homomorphism from $G$ to $H$ with exactly $N$ as kernel, then $G/N$ is isomorphic to the image of this homomorphism.



                Therefore, your task should be to locate a homomorphism which has kernel exactly $N$, and then find the image of this homomorphism.



                Now, what is the kernel of $f$? Note that $|z|^2 = z bar z$, so $frac{z^2}{|z|^2} = frac z{bar z}$ ,which equals $1$ if and only if $z = bar z$, if and only if $z in mathbb R$. In other words, the kernel of $f$ is indeed $N$.



                As for the image, as you have pointed out, since $|z|= |bar z|$, we see that the image is certainly contained within $H = {|x|= 1}$. However, the converse needs to be shown i.e. that $H$ is contained in the image. This is not very difficult : let $y in H$, and let $z$ be a square root of $y$, which always exists by the fundamental theorem of algebra (or using the polar form). Then $|z| = 1$, and $f(z) = y$.



                Therefore, your statements are correct, but here is the formal verification anyway.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 11:52









                астон вілла олоф мэллбэрг

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                37.4k33376






























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