The group homomorphism $z mapsto (z/|z|)^2$ with an application of the homomorphism theorem.
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
asked Nov 28 '18 at 11:35
GMiiX
334
add a comment |
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
asked Nov 28 '18 at 11:35
GMiiX
334
add a comment |
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
asked Nov 28 '18 at 11:35
GMiiX
334
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
asked Nov 28 '18 at 11:35
GMiiX
334
asked Nov 28 '18 at 11:35
GMiiX
334
asked Nov 28 '18 at 11:35
GMiiX
334
asked Nov 28 '18 at 11:35
GMiiX
334
asked Nov 28 '18 at 11:35
GMiiX
334
334
add a comment |
add a comment |
1 Answer
1
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votes
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.
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
add a comment |
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1 Answer
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1 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.
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
add a comment |
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.
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
add a comment |
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.
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
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.
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
answered Nov 28 '18 at 11:52
астон вілла олоф мэллбэрг
37.4k33376
37.4k33376
add a comment |
add a comment |
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