Recent content by Turnyface

Figured. Hopefully partial credit is dished out. I hate abstract algebra. Thanks for the responses.

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?

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...

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.

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...

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...