Question

how to calculate the number of binary operations on any set A , say of 4 elements?

Answer 1

3down voteacceptedA binary operation on a set AA is a function from A×A→AA×A→A, which means that it assigns exactly one element of AA to every ordered pair in A×AA×A.In your example, A={a,b}A={a,b}, soA×A={(a,a),(a,b),(b,b),(b,a)}A×A={(a,a),(a,b),(b,b),(b,a)}You can describe an operation on AA by saying exactly where it sends the four pairs (see Cameron Williams's comment).Thus, for example, the second operation is the one that sends(a,a) to a, written (a∗a)=a(a,b) to b, written (a∗b)=b(b,a) to a, written (b∗a)=a(b,b) to b, written (b∗b)=b(a,a) to a, written (a∗a)=a(a,b) to b, written (a∗b)=b(b,a) to a, written (b∗a)=a(b,b) to b, written (b∗b)=bTo be commutative, an operation must send (x,y)(x,y) and (y,x)(y,x) to the same element for all x,y∈Ax,y∈A. In the operation notation, this means that∀x,y∈A,x∗y=y∗x∀x,y∈A,x∗y=y∗xIs this true for the second operation? There are only two distinct elements, so we just need to check that a∗b=b∗aa∗b=b∗a.But we have a∗b=a≠b=b∗aa∗b=a≠b=b∗a, which means the operation is not commutative.Now, existence of the identity: the identity is an element that fixes every element in AA. Is there an element ee in AA such thate∗x=x∗e=x∀x∈A?e∗x=x∗e=x∀x∈A?To check this, you need to look at each of the elements in AA:The equationa∗b=a≠b=b∗aa∗b=a≠b=b∗atells you that aa does not fix bb (because a∗b=aa∗b=a) and that bb does not fix aa (because b∗a=bb∗a=b). There are no more elements in AA, so we can conclude that this operation does not have an identity.Since it doesn't have an identity, it doesn't have inverses.To check associativity, you need to check if performing the operation in a different order affects the result, i.e., if different ways of parenthesizing affect the result. The operation is associative if(x∗y)∗z=x∗(y∗z)∀x,y,z∈A(x∗y)∗z=x∗(y∗z)∀x,y,z∈ANote that x,y,zx,y,z in the definition are not necessarily distinct.I'll let you check this one; the operation will be associative if all the following hold (just go through each using the definition of the operation, performing what's in parenthesis first):(a∗a)∗a=?a∗(a∗a)(a∗a)∗a=?a∗(a∗a)(b∗b)∗b=?b∗(b∗b)(b∗b)∗b=?b∗(b∗b)(a∗b)∗a=?a∗(b∗a)(a∗b)∗a=?a∗(b∗a)(b∗a)∗a=?a∗(b∗a)(b∗a)∗a=?a∗(b∗a)(a∗a)∗b=?a∗(a∗b)(a∗a)∗b=?a∗(a∗b)(a∗b)∗b=?a∗(b∗b)(a∗b)∗b=?a∗(b∗b)(b∗b)∗a=?b∗(b∗a)(b∗b)∗a=?b∗(b∗a)(b∗a)∗b=?b∗(a∗b)