Let's assume the strings have n bits and the zero string is the identity element. Then the number of different operations is (2^n-1)! divided by the number computed by the formula in this link with p=2. In the case of n=2 the answer is that XOR is unique as mentioned by Matchu, but in general there are many, many different operations that satisfy the conditions. For example, for n=5 there are 822,336,494,953,469,303,808,000,000 different operations.
Every commutative, associative, self-inverse binary operation on a finite set is essentially XOR on bit strings of a fixed length, by the classification of finite abelian groups. As egreg says, this turns the set into a vector space of dimension n over the field with two elements. Any two groups of this type with the same size are isomorphic so any function from the set of bit strings to itself that is one-to-one, onto, and fixes 0 gives another operation, which is possibly the same.
The operation is the same if and only if the function is an automorphism of the vector space, and the group of such automorphisms is called the general linear group, in this case of a vector space of dimension n over the finite field with two elements.
The group of all permutations of the nonzero elements, which has (2^n-1)! elements, acts on the set of satisfactory binary operations, and the stabilizer of an operation, which is a subgroup of the group of such permutations, has the same size as the general linear group. By the orbit-stabilizer theorem, the number of different operations is (2^n-1)!/(size of the general linear group).
Yes. But you don't need to add closure in these definitions For groups, for example, notice that an operation is, first of all, a function $\cdot :G \times G \to G$. And that its codomain is $G$ itself.
A vector space is a $4-$tuple $(\mathcal{V},{\Bbb K}, +, \cdot)$, where $$+: \mathcal{V}^2 \to \mathcal{V} \quad \text{and} \quad\cdot:\mathbb{K} \times \mathcal{V} \to \mathcal{V}$$ are the operations. The structure of a vector space is much richer than that of a group. A vector space has two operations and a underlying a field, while a group is only the set with one operation (satisfying conditions you well know). Given a vector space $(\mathcal{V},{\Bbb K}, +, \cdot)$, $(\mathcal{V},+)$ is an abelian group, always. Answering 4. along, given a field $\Bbb K$, $\Bbb K^n$ is both a vector field and an additive group, with respect to the operations of $\Bbb K$.
Vectors are elements of a vector space. It is just a name. Examples of vector spaces are:
- Polynomials with degree less or equal to $n$, with real coefficients: $\mathcal{P}_n(\Bbb R)$.
- All continuous functions from $[0,1]$ to $\Bbb R$: $\mathscr{C}^0([0,1],\Bbb R)$
- $\Bbb R^n$ itself.
- Matrices with real coefficients: $\mathbb{M}_{n \times m}(\Bbb R)$.
and a lot more stuff. I used $\Bbb R$ for concreteness, in general you can take an arbitrary field (for polynomials, matrices, etc). So a vector can be an arrow, a function, a polynomial, a matrix...
Best Answer
Yes. Because you know (a.b)'=b'.a', if its different than a'.b' then it's because the group isn't commutative. The group of invertible 2x2 matrices is an example of such a group