Quick way to find the number of the group homomorphisms $\phi:{\bf Z}_3\to{\bf Z}_6$? Consider the following multiple choice problem:

Let $H$ be the set of all group homomorphsims $\phi:{\bf Z}_3\to{\bf Z}_6$. How many functions does $H$ contain?
A.1  B.2 C.3 D.4 E.6

Since $1$ generates ${\bf Z}_3$, one can analyze $\phi(1)$ case by case, which may be rather time consuming, for me, at least. Since this is a multiple choice problem, is there any quick way to solve it?
 A: Say $f$ be the homomorphism and say $f(1)=a\in \mathbb{Z}_6$
in $\mathbb{Z}_3$ we have  $1+1+1=3=0\Rightarrow f(1+1+1)=f(0)=0\Rightarrow 3f(1)=0\Rightarrow 3a=0$ in $\mathbb{Z}_6$ so $a=0,2,4$
so you have $3$ distinct homomorphism.
A: *

*A group homomorphism with cyclic domain is completely determined by the image of a generator.

*If $f\colon G\to H$ is a homomorphism, and $x\in G$, then the order of $f(x)$ must be a divisor of the order of $x$. 
Since the only divisors of $3$ are $1$ and $3$, the answer is one plus the number of elements of $\mathbb{Z}_6$ of order $3$ (the "one plus" comes from the trivial map). Are there any? Yes, so the answer is not A. How many? Two: 2 and 4. So the answer is C. 
A: In general the number of group homomorphisms $\varphi:\mathbb{Z}_{m} \to \mathbb{Z}_{n}$ is given by $\text{gcd}(m,n)$. So here you have $\text{gcd}(3,6)=3$.
The proof of this result can be found in Abstract Algebra Manual: Problems and Solutions
By Ayman Badawi.
A: I just learned that there are two general theorems about this question


*

*The number of group homomorphisms from $Z_m$ into $Z_n$ is $\text{gcd}(m,n)$.

*The number of ring homomorphisms from $Z_m$ into $Z_n$ is $2^{\omega(n)-\omega(n/\text{gcd}(m,n))}$ where $\omega(a)$ denotes the number of distinct prime divisors of the integer $a$.


from an article The Number of Homomorphisms from $Z_m$ into $Z_n$(American mathematical Monthly 91 (1984):196-197) by Gallian and Buskirk.
A: Order of 1 is 3 order of image of 1 divides 3 and #(f(1))/6 so so total choice is 1 and 3. Here there are there are two choice of order 3 element so total 3 homomorphism.
