# $n\bar k=\overline{nk}$ in $Z_m$?

I was reading Algebra of Hungerford, the (sketch of) proof of Theorem 3.6.

Definitions : Let $$m$$ be a positive integer. The equivalence relation '$$\equiv$$' modulo $$m$$ partitions $$\mathbb Z$$ into $$m$$ equivalence classes $$\bar a$$ for $$a=0,1,2,\cdots,m-1$$. The set $$\mathbb Z_m$$ of all such equivalence classes is a group under addition, defined by $$\bar a+\bar b=\overline{a+b}$$. Note that $$\bar a=\bar b$$ in $$\mathbb Z_m$$ iff $$a\equiv b$$ (mod $$m$$).

Then, is the following true for every integer $$n$$ and $$k$$? $$n\bar k=\overline{nk}\quad(\text{in }\mathbb Z_m)$$

examples : In $$\mathbb Z_6$$, $$3\bar5=\bar5+\bar5+\bar5=\overline{15}$$ $$0\bar5=\bar0$$ $$(-2)\bar5=-(\bar5+\bar5)=-\overline{10}=\overline{-10}.$$
$$n\bar k=\underbrace{\bar k+\cdots+\bar k}_{\text{n-times}}=\overline {\underbrace{k+\cdots+k}_{\text{n-times}}}=\overline{nk}$$, since $$\overline{a+b}=\bar a+\bar b$$ by definition.