A Group Homomorphism: Verifying ø(gh) = ø(g) + ø(h) for ø: Z → Z

[PsychonautQQ]

Homework Statement

For any integer K, the map ø_k: Z → Z given by ø_k(n) = kn is a homomorpism. Verify this

Homework Equations

if ø(gh) = ø(g)ø(h) for all g,h in G then the map ø: G → H is a group homomorpism

The Attempt at a Solution

So I have barely any linear algebra so many taking this summer course wasn't the best idea, but I have PF so I'm good.
So...
if ø(xy) = ø(x)ø(y) for all x,y in Z then the map ø: Z → Z is a group homomorphism. ø_k(x) = kx and ø_k(y) = ky

but then
ø(xy) = kxy and ø(x)ø(y) = (xy)k^2

i'm new to this type of thinking, can anyone help me out here?

 

[pasmith]

Homework Statement

For any integer K, the map ø_k: Z → Z given by ø_k(n) = kn is a homomorpism. Verify this

Homework Equations

if ø(gh) = ø(g)ø(h) for all g,h in G then the map ø: G → H is a group homomorpism

The Attempt at a Solution

So I have barely any linear algebra so many taking this summer course wasn't the best idea, but I have PF so I'm good.
So...
if ø(xy) = ø(x)ø(y) for all x,y in Z then the map ø: Z → Z is a group homomorphism. ø_k(x) = kx and ø_k(y) = ky

but then
ø(xy) = kxy and ø(x)ø(y) = (xy)k^2

i'm new to this type of thinking, can anyone help me out here?

The group operation for the integers is addition, not multiplication.

 

[PsychonautQQ]

The group operation for the integers is addition, not multiplication.

So you are saying it should read ø(gh) = ø(g) + ø(h)
?

 

[CAF123]

So you are saying it should read ø(gh) = ø(g) + ø(h)
?

No, he meant $\phi(g + h) = \phi(g) + \phi(h)$. The + in $\phi(g+h)$ is the group operation of the domain (here Z) and the + in $\phi(g) + \phi(h)$ is the group operation of the codomain (here also Z).