Let $1< m,n \in \Bbb N$. The map $g: \Bbb Z_{m} \to \Bbb Z_{mk}$ given by $x \mapsto kx$ is a monomorphism

Question

Let $1< m,n \in \Bbb N$. The map $g: \Bbb Z_{m} \to \Bbb Z_{mk}$ given by $x \mapsto kx$ is a monomorphism

Example from Hungerford's Algebra.

Clearly this is injective, but why is it a homomorphism? I don't see how to get a factor of $k$ for each equivalence class.

$g(\bar{x}\bar{x_{1}}) = g(\overline{xx_{1}}) = \overline{kxx_{1}}$ = ?

Answer 1

$g(x+y)=k(x+y)=kx+ky=g(x)+g(y)$. I have switched to additive notation (you had actually sort of mixed the two, since in multiplicative it would be $g(xy)=(xy)^k=x^ky^k=g(x)g(y)$).

Answer 2

$g(\bar{x}+\bar{x_{1}}) = g(\overline{x+x_{1}}) = \overline{k(x+x_{1}}) = \overline{kx} + \overline {kx_{1}} = g(\bar{x}) + g(\bar{x_{1}})$.