Show that a map is well-defined and homomorphism Define a function $f_n:\mathbb{Z}_m \to \mathbb{Z}_m$ as a map $\bar{a}\mapsto n\cdot \bar{a}$.  Show that it is both well-defined and a group homomorphism.
For the well-defined part, I know that I need to somehow show that when $a_1=b_1 \in \mathbb{Z}_m$, then $f(a_1)=f(b_1)\in \mathbb{Z}_m$.  I just don't know where to start.  For the homomorphism, $f(a_1b_1)=f(a_1)f(b_1)$, I'm also unclear as how to proceed.
 A: Who on earth has created this "exercise"?
The used notation $n  \cdot \overline{a}$ means that well-definedness is already assumed! It would be different for $\overline{n \cdot a}$, but this expression doesn't appear in the statement. Besides: This has nothing to do with $\mathbb{Z}_m$ (which seems to be the usual wrong/ambigious notation for $\mathbb{Z}/m\mathbb{Z}$).
If $A$ is an arbitrary abelian group, and $n \in \mathbb{N}$, then $A \to A, a \mapsto n \cdot a$ is a homomorphism. Hint for the proof: Recall the inductive definition of $n \cdot a$. Show $n \cdot (a+b)=n \cdot a + n \cdot b$ by induction on $n$.
One gains nothing and it doesn't get easier by restricting to this artificial special case $A=\mathbb{Z}/m\mathbb{Z}$.
A: Here is a rather long argument for what you are trying to prove, but I thought it could perhaps be interesting to see that your homomorphism arises in a very natural manner.
First note that 
$$
a\longmapsto na
$$
is a homomorphism from $\mathbb{Z}$ to itself.
Then compose with the canonical surjection of $\mathbb{Z}$ onto $\mathbb{Z}/m\mathbb{Z}$ to get a homomorphism
$$
a\longmapsto \overline{na}=n\overline{a}
$$
from $\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$.
Finally observe that $m\mathbb{Z}$ is contained in the kernel of the latter, so that you homomorphism factors through the quotient and leads to a homomorphism
$$
\overline{a}\longmapsto n\overline{a}
$$
from $\mathbb{Z}/m\mathbb{Z}$ into itself.
A: Assume $a_1=b_1\,\in\Bbb Z_m$. What does it mean? That $a_1\equiv b_1 \pmod m$, i.e. $m|\,b_1-a_1$. Then $f(b_1)=[n\cdot b_1]$ where $[x]$ denotes the equivalence class of $x$ w.r.t. mod $m$... You will have to prove that $[n\cdot a_1]=[n\cdot b_1]$.
Similarly, playing with the equivalence classes by arbitrary representants will help you in the other part.
Note that a homomorphic relation $R$ between (additively written) groups $A$ and $B$, such that, for all $a$ there is $b$ such that $a$ and $b$ are in the relation $R$ --written as $aRb$--, is actually a function (homomorphism) iff $0Rb\Rightarrow b=0$, i.e. it is enough to verify that $0$ can go only to $0$.
