Basic Question about Operations on Sets

I'm just starting out in group theory, and I'm having a bit of trouble with this question.

Let S={a,b,c}.

1) How many elements does the set SxS contain?

I think the answer to this is 9, {a,a}{a,b}{a,c}{b,a}{b,b}{b,c}{c,a}{c,b}{c,c}. My only possible worry is that {a,b}={b,a}, but I'm basing my understanding of this as kind of like a graph, where the point {1,2} is not equal to {2,1}.

2) How many operations are there on S?

I don't really know how to go about this part. My understanding of an operation is that it takes two elements in the set and maps them to one element in the set. So maybe if I say that every element in SxS can be mapped to one of three elements, then there are 9 times 3 operations i.e. 27. Again though, I'm not totally clear on it.

3) Find the Caley table for an operation such that is a group with identity element a.

I think if I set up the table so that , this is a group. It has an identity, each element has an inverse, and I'm pretty sure it's associative.

4) Prove that the operation you have found is the only operation on S such that is a group with identity element a.

Well, a is the identity. a is also it's own inverse, while b and c are inverses of each other, so every element is invertible. Which leaves associativity. Is the only way to show this by computign every combination of elements? Also, I'm not sure how to show it it the "only" operation.

5) Finally, write down the Cayley table of each operation on S such that is a group, and determine which of these operations is commutative.

I know that I could just cycle the elements in the set and get analogues of the previous question with b and c as identity elements, instead of a. But I can't tell if there are any other operations that form a group. I know that when I get that part, if the Cayley table is symmetric, it is a commutative operation.

I'm just starting out on this topic, so any help/pointers would be much appreciated.

Thanks in advance.