Quick way to find the number of the group homomorphisms $\phi:{\bf Z}_3\to{\bf Z}_6$?
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.
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.
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.