Proving Map with Order Condition is Group Morphism Let $(G, \circ)$ and $(H,\ast)$ be groups. Suppose $G$ is cyclic with a order $n \in \Bbb N $ and $g$ is a generator.  Let $ h \in H$ be an element with $\text{order}(h)\ |\ n$.  
Show that the map $\phi: G \rightarrow H $ determined by $\phi(g)= h $ is a group homomorphism.  
I tried this way: Take $x,y \in G$ then there are $r,s \in \mathbb{Z}$ such that $x = g^r $ and $ y =g^s$. Next $\phi( g^r\circ g^s)= \phi(g^{r+s})= \phi(g)^{r+s}= h^{r+s}=h^r\ast h^s = \phi(g)^r \ast \phi(g)^s= \phi(g^r) \ast \phi(g^s)$. 
So I did not use the fact that $\text{order}(h) \ |\ n$. So I must be wrong somewhere. Someone who can point it out for me? Thanks in advance!   
 A: The potential problem with $\phi$ is not the homomorphism property which you correctly verified but rather the well-defined-ness of $\phi$.
See, when you determine $\phi$ with the single relation $\phi(g) = h$, you are implicitly defining $\phi(g^N) = h^N$ for any $N \in \Bbb N$.
But in doing that you run the risk of assigning multiple values to the same element in $G$. After all since $G$ has order $n = \text{order}(g) \in \Bbb N$, the element $g^3$, say, is the same thing as $g^{3+n}$. How are you sure that their images under $\phi$: $h^3$ and $h^{3+n}$ are the same? If they are not the same, then $\phi$ would fail to be a function as functions must assign exactly one image per domain element.
So to verify that this does not happen suppose $g^s = g^t$ for potentially different $s, t \in \Bbb N$.
You must show $\phi(g^s) = h^s = h^t = \phi(h^t)$.
Well since $g^s = g^t$, by a well known fact of cyclic groups, we must have $n = \text{order}(g) \ \big|\ s - t$. But you assumed $\text{order}(h) \ \big|\ n$. So $\text{order}(h) \ \big|\ s - t$ i.e $\text{order}(h) \cdot k = s - t$ for some $k \in \Bbb N$. Hence $$h^{s - t} = h^{\text{order}(h) \cdot k} = \big(h^{\text{order}(h)}\big)^k = e_H^k = e_H$$ and multiplying both sides above by $h^t$ we get $h^s = h^t$ as desired.
Edit:
Also here is an example where $\phi$ fails to be well-defined if $\text{order}(h) \ \big|\ n = \text{order}(g)$ is not satisfied.
Consider the "map" $\phi$ generated by $\phi(g) = h$ where $\text{order}(g) = 2$ and $\text{order}(h) = 3$. If you like you can think of $g$ being the generator of the cyclic group $C_2$ and $h$ being the generator of cyclic group $C_3$. So what's the problem with "map" defined above?
Well $\phi(g^3) = \phi(g)^3 = h^3 = e_H$. But at the same time $\phi(g^3) = \phi(g^2 \circ g) = \phi(e_G \circ g) = \phi(g) = h$. In other words $h = e_H$ which is contradiction because the order of $h$ is $3$ while that of $e_H$ is $1$. So you can see that $g^3 = g$ is being assigned multiple values here and $\phi$ cannot really be a map.
A: Here is a more precise statement:

Let $G$ be a cyclic group of finite order $n$ generated by $g$. Let $H$ be a group and let $h \in H$. Then, 
  there is a homomorphism $\phi : G \to H$ with $\phi(g)=h$ iff $h$ has finite order $m$ and $m$ divides $n$.

Indeed, if there is such a homomorphism, then $1=\phi(1)=\phi(g^n)=\phi(g)^n=h^n$. So $h$ has finite order dividing $n$.
Conversely, $\phi(g)=h$ restricts the choice of $\phi$ to $\phi(g^i)=h^i$. This map $\phi$ is clearly a homomorphism, provided it is well-defined, that is, $g^i=g^j$ implies $\phi(g^i)=\phi(g^j)$. This happens if $m$ divides $n$.
The point of the argument above is that $g^i$ does not determine $i$, only the class of $i$ mod $n$.
