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

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- Jun 20th 2011, 04:06 AMArronHomomorphism
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 - Jun 20th 2011, 05:21 AMSwlabrRe: Homomorphism
- Jun 20th 2011, 06:24 AMDevenoRe: Homomorphism
C* is an abelian group under multiplication, for any abelian group $\displaystyle G: g\rightarrow g^2$ is a homomorphism:

$\displaystyle (gh)^2 = (gh)(gh) = g(hg)h = g(gh)h = (gg)(hh) = g^2h^2$.

the fact that C* is indeed abelian under multiplication is worth verifying once :) - Jun 20th 2011, 10:35 AMArronRe: Homomorphism
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. - Jun 20th 2011, 11:07 AMFernandoRevillaRe: Homomorphism
- Jun 20th 2011, 12:06 PMArronRe: Homomorphism
C is the set of complex numbers.

- Jun 20th 2011, 01:17 PMDevenoRe: Homomorphism
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. - Jun 21st 2011, 01:51 AMSwlabrRe: Homomorphism
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 $\displaystyle \phi: \mathbb{C}\rightarrow \mathbb{C}, z\mapsto 3/z$, and if you want to be ultra fancy, do a \frac{3}{z} to make the 3/z into a fraction, $\displaystyle \frac{3}{z}$. (the \mathbb{...} puts the thing in the brackets into blackboard bold, and everything is case-sensitive, so \rightarrow is different from \Rightarrow...) - Jun 21st 2011, 06:01 AMArronRe: Homomorphism
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. - Jun 21st 2011, 06:07 AMSwlabrRe: Homomorphism
- Jun 21st 2011, 01:16 PMArronRe: Homomorphism
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. - Jun 21st 2011, 03:30 PMDevenoRe: Homomorphism
pick $\displaystyle z_1 = z_2 = 0 = 0+0i$

- Jun 24th 2011, 12:07 AMArronRe: Homomorphism
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. - Jun 24th 2011, 10:07 PMDevenoRe: Homomorphism
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. - Jun 24th 2011, 11:09 PMtopspin1617Re: Homomorphism
Not only does H not map 0 to 0, but it also doesn't have the additive property required of being a group homomorphism.

$\displaystyle H(z_1+z_2)=z_1+z_2+i\neq z_1+i+z_2+i=H(z_1)+H(z_2)$.