# Group Homomorphism Question.

#### 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?

Related Linear and Abstract Algebra News on Phys.org

#### Number Nine

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

#### 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:

#### Number Nine

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?

#### 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.

#### 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?

#### Turnyface

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.

#### Turnyface

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?
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?

#### 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).

#### Turnyface

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.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving