# Group Homomorphism Question.

1. Nov 12, 2012

### Turnyface

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. Nov 12, 2012

### Number Nine

Note, also, that the integers aren't a group under multiplication.

3. Nov 12, 2012

### Turnyface

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)

Last edited: Nov 12, 2012
4. Nov 12, 2012

### Number Nine

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?

5. Nov 12, 2012

### Turnyface

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.

6. Nov 12, 2012

### micromass

As number 9 says, we don't have a group here, so we cannot talk about group morphisms. However, $\mathbb{Z}$ 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

$$\theta(n*m)=\theta(n)*\theta(m)$$

is this satisfied?

7. Nov 12, 2012

### Turnyface

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.

8. Nov 12, 2012

### Turnyface

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?

9. Nov 12, 2012

### Vargo

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).

10. Nov 12, 2012

### Turnyface

Figured. Hopefully partial credit is dished out.

I hate abstract algebra.

Thanks for the responses.