Group Homomorphism Question.

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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?
 
801
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What about the identity mapping?
Note, also, that the integers aren't a group under multiplication.
 
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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:
801
23
So if we take n=0 under addition, then it is a homomorphism, no?
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.

Does the operation have to be preserved for all n in order to be a homomorphism?
Can you recite the definition of homomorphism?
 
6
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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.
 
21,992
3,269
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?
 
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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.
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.
 
6
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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?
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?
 
349
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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).
 
6
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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).
Figured. Hopefully partial credit is dished out.

I hate abstract algebra.

Thanks for the responses.
 

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