Thread: Homomorphism

I am not sure if the following is correct, could someone confirm

O : C* ---> C*

z ----> zz

O (z1Z2) = z1z2 x z1z2 = z1z1 x z2z2 = O (z1) O (z2)

Thus O is a homomorphism

Originally Posted by Arron

I am not sure if the following is correct, could someone confirm

O : C* ---> C*

z ----> zz

O (z1Z2) = z1z2 x z1z2 = z1z1 x z2z2 = O (z1) O (z2)

Thus O is a homomorphism

What is C*? Are you wanting to prove O is a homomorphism of groups? Or of Rings?

However, assuming you mean C* to be the group of complex numbers under multiplication, and you are wanting to prove it is a homomorphism of groups, what you have done is correct.

C* is an abelian group under multiplication, for any abelian group is a homomorphism:

.

the fact that C* is indeed abelian under multiplication is worth verifying once

Sorry I do mean C as in complex numbers. Thanks.

Can anyone tell me how to get the advanced features.

One other thing, is the following correct?

O : C ----> C
z ---> 3/z

O(z1 x z2) = 3/z1 x z2 = 3/z1 x 3/z2 = O(z1) O(z2)

thus O is a homomorphism.

Originally Posted by Arron

One other thing, is the following correct?

O : C ----> C
z ---> 3/z

O(z1 x z2) = 3/z1 x z2 = 3/z1 x 3/z2 = O(z1) O(z2)

thus O is a homomorphism.

Define C.

C is the set of complex numbers.

think about this: if h: C*--->C* is defined by h(z) = 3/z,

then h(zw) = 3/(zw), but h(z)h(w) = (3/z)(3/w) = 9/(zw).

last time i checked 3 ≠ 9.

Originally Posted by Arron

Can anyone tell me how to get the advanced features.

The advanced features are LaTeX. There is a LaTeX tutorial somewhere on this site...alternatively, search for `LaTeX commands' on google (but forget all the stuff about dollar signs and \[s 'cause they aren't used here).

To put it in LaTeX, put [tex] at the start of what you are wanting to make into LaTeX and [/ tex] at the end (without the space). For example,

[tex]\phi: \mathbb{C}\rightarrow \mathbb{C}, z\mapsto 3/z[/ tex] will give you , and if you want to be ultra fancy, do a \frac{3}{z} to make the 3/z into a fraction, . (the \mathbb{...} puts the thing in the brackets into blackboard bold, and everything is case-sensitive, so \rightarrow is different from \Rightarrow...)

Thanks for that.

How about the following, C is the group of complex numbers under addition.

O : C ---> C
z ---> z + i

For all z1,z2 which is an element C

O (z1+z2) = ((z1+i) + (z2+i)) = O (z1) + O (z2)

Thus O is a homomorphism.

Originally Posted by Arron

Thanks for that.

How about the following, C is the group of complex numbers under addition.

O : C ---> C
z ---> z + i

For all z1,z2 which is an element C

O (z1+z2) = ((z1+i) + (z2+i)) = O (z1) + O (z2)

Thus O is a homomorphism.

Remember, a homomorphism sends the identity to the identity...

So it is not a homomorphism as it does not preserve the identity. How do i show this.

I.e I need to find 2 elemnt s z1 and z2 which do not hold.

pick

Just want make sure I get this right. Are you saying

C is the group of all complex numbers under addition.

H : C ---> C
z -----> z + i

The function H is not a homomorphism.
For example
H(z1+z2) = 0

but H (z1) H(z2) = 0+0i

Hence H does not have the homomorphism property.

I have one question dont these both add up to 0.

H(0+0) is NOT 0.

H(0+0) = H(0) = 0 + i = i

H(0) + H(0) = 0+i + 0 + i = i+i =2i.

these are not equal, so H cannot be a homomorphism of the additive group of C.

Not only does H not map 0 to 0, but it also doesn't have the additive property required of being a group homomorphism.

.