More specifically, if I'm looking at a map: $f : (\mathbb{Q}, +) \to (\mathbb{Q}, +)$, and $f$ maps $x$ to $bx$, where $b$ is some rational number, why would this be homomorphic?
Here's what I've tried: $$f(pq) = bpq$$ but $$f(p) \cdot f(q) = ap \cdot cq $$ (I guess $ac$ can equal to $b$ but how do I know for sure?)
$$f(p+q) = b(p+q)$$ $$f(p) + f(q) = ap + cq$$ How do I prove that $a=c=b$?
Note that the groups are written with addition. Therefore to get a homomorphism it has to be $f(x+y)=f(x)+f(y)$. Then $f(x+y)=b(x+y)=bx+by=f(x)+f(y)$.