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Group Homomorphism Question.

  1. Nov 12, 2012 #1
    Consider θ:Z -> Z is a mapping where θ(n) = n^3 and it's homomorphism under multiplication. In this case, it's not a homomorphism under addition.

    So my question is this. In general, if we show that a group is homomorphic under multiplication, does this imply that it is not under addition and vice versa?
     
  2. jcsd
  3. Nov 12, 2012 #2
    What about the identity mapping?
    Note, also, that the integers aren't a group under multiplication.
     
  4. Nov 12, 2012 #3
    So if we take n=0 under addition, then it is a homomorphism, no? But for θ(n) = n^3 for elements not equal to 0, this is false. Does the operation have to be preserved for all n in order to be a homomorphism?

    Edit: It has to be for all elements in Z. So it's not under addition, even considering the identity.

    And to answer my own question:

    Z->Z, θ(n)= n so θ(n+m) = θ(n)+θ(n)
    Z->Z, θ(n)= n so θ(n*m) = θ(n)*θ(m)

    Thus, the answer is no.
     
    Last edited: Nov 12, 2012
  5. Nov 12, 2012 #4
    I'm not sure what you mean by this.

    Regarding the OP, again, the integers are not a group under multiplication, so it makes no sense to talk about group homomorphisms here.

    Can you recite the definition of homomorphism?
     
  6. Nov 12, 2012 #5
    Let H be a subgroup of G. Let * be an operation under G and # be an operation under H. Then for all a,b in G:

    θ(a*b)=θ(a)#θ(b)

    is a homomorphism.
     
  7. Nov 12, 2012 #6

    micromass

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    As number 9 says, we don't have a group here, so we cannot talk about group morphisms. However, [itex]\mathbb{Z}[/itex] does form a semigroup under multiplication so we can talk about semigroup morphisms.

    For that, we just need to check the same thing, namely that

    [tex]\theta(n*m)=\theta(n)*\theta(m)[/tex]

    is this satisfied?
     
  8. Nov 12, 2012 #7
    So back to the original problem where θ(n) = n^3.

    So if n=0 and the homomorphism theorem above.

    θ(m+n) = θ(0+0) = (0+0)^3 = 0^3 + 0^3, since 0=0. This is not true for any other elements in Z.

    For the same mapping, θ(n) = n^3, θ(mn) = (mn)^3 = (m)^3 * (n)^3. So this is a homomorphism under multiplication.
     
  9. Nov 12, 2012 #8
    The original question is this:

    "Let θ: Z->Z. Let θ(n) = n^3. Is this a homomorphism under addition?"

    For the answer to the question, I showed that with θ: Z->Z and θ(n) = n^3, this is a homomorphism is under multiplication.

    Is this not correct?
     
  10. Nov 12, 2012 #9
    No, it is not correct. The question is asking about addition. So its behavior under multiplication is not relevant. You just need to show that for some value of n,m
    theta(n+m) is not equal to theta(n)+theta(m).
     
  11. Nov 12, 2012 #10
    Figured. Hopefully partial credit is dished out.

    I hate abstract algebra.

    Thanks for the responses.
     
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