Number of homomorphisms from the cyclic additive group $Z_{15}$ to the cyclic additive group $Z_{10}$ 
Determine the number of homomorphisms from the additive group $Z_{15}$ to the additive group $Z_{10}$ where $Z_n$ is the cyclic group of order n.

As the order of an element is a divisor of the order of the group, the possible orders of the elements of $Z_{15}$ are $1,3,5,15$. The number of elements of order $1,3,5,15$ are $1,2,4,8$ respectively.
Similarly the possible orders of the elements of $Z_{10}$ are $1,2,5,10$. The number of elements of order $1,2,5,10$ are $1,1,4,4$ respectively.
In a homomorphism $f:A\rightarrow B$, $\;o[f(a)]$ must be a divisor of $o(a),\;\;\forall \;a \in A$. Here $o(a)$ refers to the order of $a$. And the identity element of $A$ will be mapped to that of $B$.
So we can map only the 12 elements of $Z_{15}$ to 4 elements of $Z_{10}$. Because the order of 4 elements of $Z_{10}$, which is 5 is a divisor of the orders 5, 15 of the 12 elements of $Z_{15}$. Thus there are $4^{12}$ possible mappings. Now I am not able to find how many of these preserve the additive composition to make homomorphisms?
Can some one please help me?
 A: Both groups are cyclic.  In particular, $\mathbb{Z}_{15}$ is.  Once you have decided where $1_{15}$ is mapped then the map is determined.  So, there are only $10$ to check.  Here's one for free $1_{15} \rightarrow 0_{10}$ is the trivial homomorphism - just $9$ to go.  You can use your observations on possible orders to eliminate some choices quickly.  
Extra information due to a comment.  
By $1_{15}$, I mean the element $1$ in $\mathbb{Z}_{15}$.  I added the suffix to distinguish it from the element $1$ in $\mathbb{Z}_{10}$.  $1_{15}$ generates $\mathbb{Z}_{15}$.  Similarly, $0_{10}$ means the element $0$ in $\mathbb{Z}_{10}$.  So, altogether I meant consider a map which sends the $1$ of $\mathbb{Z}_{15}$ to the $0$ of $\mathbb{Z}_{10}$. You can then quickly determine that everything in $\mathbb{Z}_{15}$ must map to $0$ and that it is a homomorphism, just a rather simple one.  
Next you can try $1_{15} \rightarrow 1_{10}$.  Again, the map is completely determined by this choice but you will find that it is not a homomorphism. You now have $8$ left to try.
