lpetrich
Contributor
I'd started Abstract Algebra long ago, but I want to post on something more specific: how many abstract-algebra entities there are. I will consider only finite entities, those with finite sets of possible operation arguments. One can easily show that the number of each kind of entity must be finite, because the number of possible operation tables without constraints is an upper limit on it.
For a unary operation and order (set size) n, that upper limit is n^n: 1, 4, 27, 256, 3125, 46656, 823543, 16777216, 387420489, 10000000000, ...
For a binary operation and order n, that upper limit is n^(n^2): 1, 16, 19683, 4294967296, 298023223876953125, ...
But there is a problem. If one relabels the set members, one will get an equivalent operation, so one has to consider all the possible relabelings.
I will start with a unary function f to make it easy. The only constraint that I will impose is closure. For a set S, f(S) must be a subset of S. For the empty set, one gets the empty algebra. For a one-element set, one gets the trivial algebra: (f,{a}) has f(a) = a. For two elements, one has 4 possibilities:
f(a} = a, f(b) = a / f(a} = a, f(b) = b / f(a} = b, f(b) = a / f(a} = b, f(b) = b
Turning a and b into 1 and 2, I will use this shorthand: 11, 12, 21, 22
If we impose the constraint that f(S) = S, then the possible tables are 12, 21.
For finite sets, f(S) = S implies that f is a bijection, meaning that it is invertible. One can define an inverse function g such that f(g(x)) = g(f(x)) = x.
For a unary operation and order (set size) n, that upper limit is n^n: 1, 4, 27, 256, 3125, 46656, 823543, 16777216, 387420489, 10000000000, ...
For a binary operation and order n, that upper limit is n^(n^2): 1, 16, 19683, 4294967296, 298023223876953125, ...
But there is a problem. If one relabels the set members, one will get an equivalent operation, so one has to consider all the possible relabelings.
I will start with a unary function f to make it easy. The only constraint that I will impose is closure. For a set S, f(S) must be a subset of S. For the empty set, one gets the empty algebra. For a one-element set, one gets the trivial algebra: (f,{a}) has f(a) = a. For two elements, one has 4 possibilities:
f(a} = a, f(b) = a / f(a} = a, f(b) = b / f(a} = b, f(b) = a / f(a} = b, f(b) = b
Turning a and b into 1 and 2, I will use this shorthand: 11, 12, 21, 22
If we impose the constraint that f(S) = S, then the possible tables are 12, 21.
For finite sets, f(S) = S implies that f is a bijection, meaning that it is invertible. One can define an inverse function g such that f(g(x)) = g(f(x)) = x.