TITLE: General approach for finding how many group homomorphisms are there
QUESTION [8 upvotes]: So I've asked this type of questions for more than once, and still I don't get the method(s) I've been presented with. What's the general recommended method for finding how many homomorphisms are there, and finding them? 
I would probably understand better through an example: 

How many homomorphisms are there from $S_3$ to $\mathbb Z_2 \times \mathbb Z_2$. 

I would gladly appreciate full solutions.
You have my full gratitude for any sort of assistance, comment, insight or info you can provide.

REPLY [6 votes]: The other answers give correct and reasonably elegant proofs for the special example. 
There is a brute force, inelegant method for enumerating group homomorphisms $G \to H$ which will work whenever you have two things: 

a finite presentation of $G$; 
a method of calculation in $H$ that lets you determine when two elements are equal.

If you have these two things then here is what you do to enumerate homomorphisms $G \to H$. First, for each generator of $G$, choose its image in $H$; in other words, choose any function from the generating set of $G$ to $H$. Then check that the image in $H$ of each relator of $G$ is equal to the identity in $H$. Enumerating all choices that succeed, you obtain an enumeration of all homomorphisms $G \to H$.
For the special example, the domain $S_3$ has the presentation $\langle a,b \,\, | \,\, a^2=b^2=(ab)^3 = Id\rangle$, and in the range $\mathbb{Z}/2 \times \mathbb{Z}/2$ you can calculate using mod 2 arithmetic. Any map taking $a,b$ to the same element of the range satisfies all three of the relations, and any map taking $a,b$ to different elements of the range satisfies the two order 2 relations but not the order 3 relation. So there are four homomorphisms, each determined by choosing the common image of $a,b$.