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Math Help - Homomorphism

  1. #1
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    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
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: Homomorphism

    Quote Originally Posted by Arron View Post
    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.
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  3. #3
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    Re: Homomorphism

    C* is an abelian group under multiplication, for any abelian group G: g\rightarrow g^2 is a homomorphism:

    (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
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    Post Re: 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.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Homomorphism

    Quote Originally Posted by Arron View Post
    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.
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  6. #6
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    Re: Homomorphism

    C is the set of complex numbers.
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  7. #7
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    Re: 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.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Re: Homomorphism

    Quote Originally Posted by Arron View Post
    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 \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, \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...)
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  9. #9
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    Re: 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.
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Re: Homomorphism

    Quote Originally Posted by Arron View Post
    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...
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  11. #11
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    Re: 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.
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  12. #12
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    Re: Homomorphism

    pick z_1 = z_2 = 0 = 0+0i
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  13. #13
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    Re: 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.
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  14. #14
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    Re: 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.
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  15. #15
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    Re: 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.

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