Checking for homomorphism ($\mathbb{Z}_6 \rightarrow \mathbb{Z}_6$)?

Question

I'm trying to define a homomorphism $\mathbb{Z}_6 \rightarrow \mathbb{Z}_6$.

I've got $f(x)=ax$, $a \in \mathbb{Z}_6$ as a candidate, but how am I supposed to show this as a homomorphism?

Merely (laborously) show that $$f(ab)=f(a)f(b)$$ for all $a,b \in \mathbb{Z}_6$?

Answer 1

Yes -- but it won't be that laborious; you can do it symbolically once and for all.

Take care not to use the letter $a$ both for the parameter of the homomorphism and one of the elements you're checking the homomorphism condition for. And remember that the group operation on both ends of the homomorphism is (modular) addition rather than multiplication.

Answer 2

Just pick the generator of $Z_6$ say $a$ and try to map it to the elements whose order divides the order of$ a$. i.e you can map a to elements of order $1,2,3,6$. So in total you get all of $Z_6.$ $$f(1)=a$$ for all $a$ in $Z_6$ is a homomorphism.