Do These Functions Qualify as Group Homomorphisms?

[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?

 

[Dick]

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.

 

[umzung]

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?

 

[Dick]

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.