# Do These Functions Qualify as Group Homomorphisms?

• umzung
In summary: Is it just all complex numbers of the form ##z=a+ib## with ##a=-b##?In summary, we have four functions φ1, φ2, φ3, and φ4, where φ1 and φ4 are not homomorphisms, and φ2 and φ3 are homomorphisms. The identity element under addition is zero for φ2 and the identity element under multiplication is 1 for φ3. The kernel of φ2 is the set of complex numbers of the form z = a + ib where a = -b, and the image is not all complex numbers. The kernel of φ3 is the set of non-zero complex numbers, and the image is all non-zero complex numbers.
umzung

## Homework Statement

Are these functions homomorphisms, determine the kernel and image, and identify the quotient group up to isomorphism?
C^∗ is the group of non-zero complex numbers under multiplication, and C is the group of all complex numbers under addition.

## Homework Equations

φ1 : C−→C
z −→ (Re(z))^2;
φ2 : C−→C
z −→ z^macron (conjugate of z) + iz;
φ3 : C^∗ −→ C^∗
z −→ (z^macron (conjugate of z))^2;
φ4 : C∗ −→ C∗
z −→ i/z

## The Attempt at a Solution

I found elements in each of φ1 and φ4 to show they are not homomorphisms.
In φ2, I find that φ(z1 + z2) = (z1^macron + iz1) + (z2^macron + iz2) = φ(z1) + φ(z2), hence a homomorphism.
Identity element under addition is zero, hence Ker(φ) is z^macron + iz = 0, so z^macron = -iz. Not sure if this is correct. So φ is onto and the image is C, and not sure of quotient group?
In φ3, I find that φ(z1z2) = φ(z1)φ(z2), hence a homomorphism.
Identity element under multiplication is 1, Ker(φ) is Z^macron = 1. So φ is onto and the image is C^∗ , and not sure of quotient group?

umzung said:

## Homework Statement

Are these functions homomorphisms, determine the kernel and image, and identify the quotient group up to isomorphism?
C^∗ is the group of non-zero complex numbers under multiplication, and C is the group of all complex numbers under addition.

## Homework Equations

φ1 : C−→C
z −→ (Re(z))^2;
φ2 : C−→C
z −→ z^macron (conjugate of z) + iz;
φ3 : C^∗ −→ C^∗
z −→ (z^macron (conjugate of z))^2;
φ4 : C∗ −→ C∗
z −→ i/z

## The Attempt at a Solution

I found elements in each of φ1 and φ4 to show they are not homomorphisms.
In φ2, I find that φ(z1 + z2) = (z1^macron + iz1) + (z2^macron + iz2) = φ(z1) + φ(z2), hence a homomorphism.
Identity element under addition is zero, hence Ker(φ) is z^macron + iz = 0, so z^macron = -iz. Not sure if this is correct. So φ is onto and the image is C, and not sure of quotient group?
In φ3, I find that φ(z1z2) = φ(z1)φ(z2), hence a homomorphism.
Identity element under multiplication is 1, Ker(φ) is Z^macron = 1. So φ is onto and the image is C^∗ , and not sure of quotient group?

You aren't being very clear about exactly what your specific questions are. But I'd start with the second one, which looks like it supposed to be ##\phi_2(z)=z^*+iz## (where the '*' is complex conjugate). Try and come up with a more concrete description of the kernel, say in terms of conditions on ##x## and ##y##, where ##x+iy=z##. ##\phi_2## is certainly NOT onto.

Ker(ϕ2) = {z is in C: ϕ(z) = 0}
= {z is in C: z* (complex conjugate) + iz = 0}
= {z is in C: z* = -iz.
Im(ϕ2) = the set of complex numbers.

Not sure if that makes sense.

Ker(ϕ3) = {z is in C*: ϕ(z) = 1}
= {z is in C*: (z* (complex conjugate))^2 = 1}
= {z is in C*: z* = 1}
Im(ϕ3) = the set of all non-zero complex numbers.

Any closer?

umzung said:
Ker(ϕ2) = {z is in C: ϕ(z) = 0}
= {z is in C: z* (complex conjugate) + iz = 0}
= {z is in C: z* = -iz.
Im(ϕ2) = the set of complex numbers.

Not sure if that makes sense.

Ker(ϕ3) = {z is in C*: ϕ(z) = 1}
= {z is in C*: (z* (complex conjugate))^2 = 1}
= {z is in C*: z* = 1}
Im(ϕ3) = the set of all non-zero complex numbers.

Any closer?

You are just writing ##z^*= -iz## and then jumping to the conclusion that the image of ##\phi_2## is all complex numbers. It's not. Just try and give me a solution to the equation ##\phi_2(z)=1##. Work it out. Then give me a better description of the kernel.

## 1. What is a homomorphism?

A homomorphism is a function that preserves the structure of algebraic objects. It maps elements from one algebraic structure to another in a way that respects the operations and relationships between elements.

## 2. How do you determine if two functions are homomorphisms?

To determine if two functions are homomorphisms, you must check if they preserve the operations and relationships between elements. This means that if f(a) and f(b) are the images of elements a and b under the function, then f(a + b) must equal f(a) + f(b) and f(ab) must equal f(a)f(b).

## 3. What are some examples of homomorphisms?

Some examples of homomorphisms include addition and multiplication in algebra, differentiation and integration in calculus, and rotation and translation in geometry.

## 4. Can a homomorphism be an isomorphism?

Yes, a homomorphism can be an isomorphism if it is both one-to-one (injective) and onto (surjective). This means that the function preserves both the structure and the elements of the algebraic objects, making it a bijective mapping.

## 5. How are homomorphisms used in science?

Homomorphisms are used in various branches of science, including mathematics, physics, and computer science. They are particularly useful in studying the relationships between different objects or systems and can help simplify complex structures by preserving their essential features.

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