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We know that if $n = 2$ then the operation is called a binary operation.
$ \circ $ on set $X$ is a function $\circ : X \times X \rightarrow X$.
And the number of all associative binary operation on a finite set $X=\{1, 2\}$
with $|X|=2$ cardinality, is 8.
They are:

$$1) \begin{array}{c|ccccc} \circ & & \\ \hline & 1& 1& \\ & 1& 1& \end{array} 2) \begin{array}{c|ccccc} \circ & & \\ \hline & 1& 1& \\ & 1& 2& \end{array} 3) \begin{array}{c|ccccc} \circ & & \\ \hline & 1& 1& \\ & 2& 2& \end{array} 4) \begin{array}{c|ccccc} \circ & & \\ \hline & 1& 2& \\ & 1& 2& \end{array}$$ $$5) \begin{array}{c|ccccc} \circ & & \\ \hline & 2& 1& \\ & 1& 2& \end{array} 6) \begin{array}{c|ccccc} \circ & & \\ \hline & 1& 2& \\ & 2& 1& \end{array} 7) \begin{array}{c|ccccc} \circ & & \\ \hline & 1& 2& \\ & 2& 2& \end{array} 8) \begin{array}{c|ccccc} \circ & & \\ \hline & 2& 2& \\ & 2& 2& \end{array} $$ $$1,2 \in \mathbb X$$ Also I made a program which counts associative operations when $n=3$, I mean it counts ternary associative operations based on this condition: $a \circ b \circ (c \circ d \circ e) = a \circ (b \circ c \circ d) \circ e = (a \circ b \circ c) \circ d \circ e$
Where $ \circ $ on set $X$ is a function $\circ : X \times X \times X \rightarrow X$.
Also $a,b,c,d,e \in \mathbb X$ and when cardinality equals $|X|=2$ .I mean when $X=\{1, 2\}$ it showed me only 8 associative operations. They are:

$$1) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 1& & 1& \\ & 1& & 1& \\ & & 1& & 1& \\ & 1& & 1& \end{array} 2) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 1& & 1& \\ & 1& & 1& \\ & & 1& & 2& \\ & 1& & 1& \end{array} 3) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 2& & 2& \\ & 1& & 1& \\ & & 2& & 2& \\ & 1& & 1& \end{array} 4) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 1& & 2& \\ & 1& & 2& \\ & & 1& & 2& \\ & 1& & 2& \end{array} $$ $$5) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 2& & 1& \\ & 1& & 2& \\ & & 1& & 2& \\ & 2& & 1& \end{array} 6) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 2& & 2& \\ & 1& & 2& \\ & & 2& & 2& \\ & 2& & 2& \end{array} 7) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 1& & 2& \\ & 2& & 1& \\ & & 2& & 1& \\ & 1& & 2& \end{array} 8) \begin{array}{c|ccccc} \circ & & & & \\ \hline & & 2& & 2& \\ & 2& & 2& \\ & & 2& & 2& \\ & 2& & 2& \end{array} $$ So in this case I wrote result from arguments of a ternary operations in a cubic matrices as called in a tensors.

Things will get even more difficult when $n = 4$. I wrote such a program also based on this condition: $$a \circ b \circ c \circ (d \circ e \circ f \circ g) = a \circ b \circ (c \circ d \circ e \circ f) \circ g = a \circ (b \circ c \circ d \circ e) \circ f \circ g = (a \circ b \circ c \circ d) \circ e \circ f \circ g$$ Where $ \circ $ on set $X$ is a function $\circ : X \times X \times X \times X \rightarrow X$.
Also $a,b,c,d,e,f,g \in \mathbb X$ and when cardinality equals $|X|=2$ .I mean when $X=\{1, 2\}$ it showed me only 8 associative operations. They are:

$$1) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 1& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 1& \\ & & & & & & 1& & 1& \\ & & & & & 1& & 1& \\ & & & & & & 1& & 1& \\ & & & & & 1& & 1& \\ & & 1& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 1& \\ \end{array} 2) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 1& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 1& \\ & & & & & & 1& & 1& \\ & & & & & 1& & 1& \\ & & & & & & 1& & 2& \\ & & & & & 1& & 1& \\ & & 1& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 1& \\ \end{array} $$

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$$3) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 1& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 1& \\ & & & & & & 2& & 2& \\ & & & & & 2& & 2& \\ & & & & & & 2& & 2& \\ & & & & & 2& & 2& \\ & & 1& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 1& \\ \end{array} 4) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 1& & & & & & & & & & 2& \\ & 1& & & & & & & & & & 2& \\ & & & & & & 1& & 2& \\ & & & & & 1& & 2& \\ & & & & & & 1& & 2& \\ & & & & & 1& & 2& \\ & & 1& & & & & & & & & & 2& \\ & 1& & & & & & & & & & 2& \\ \end{array}$$

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$$5) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 2& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 2& \\ & & & & & & 1& & 2& \\ & & & & & 2& & 1& \\ & & & & & & 2& & 1& \\ & & & & & 1& & 2& \\ & & 1& & & & & & & & & & 2& \\ & 2& & & & & & & & & & 1& \\ \end{array} 6) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 2& & & & & & & & & & 2& \\ & 1& & & & & & & & & & 2& \\ & & & & & & 2& & 2& \\ & & & & & 2& & 2& \\ & & & & & & 2& & 2& \\ & & & & & 2& & 2& \\ & & 2& & & & & & & & & & 2& \\ & 2& & & & & & & & & & 2& \\ \end{array} $$

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$$7) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 1& & & & & & & & & & 2& \\ & 2& & & & & & & & & & 1& \\ & & & & & & 2& & 1& \\ & & & & & 1& & 2& \\ & & & & & & 1& & 2& \\ & & & & & 2& & 1& \\ & & 2& & & & & & & & & & 1& \\ & 1& & & & & & & & & & 2& \\ \end{array} 8) \begin{array}{c|ccccc} \circ & & & & & & & & \\ \hline & & 2& & & & & & & & & & 2& \\ & 2& & & & & & & & & & 2& \\ & & & & & & 2& & 2& \\ & & & & & 2& & 2& \\ & & & & & & 2& & 2& \\ & & & & & 2& & 2& \\ & & 2& & & & & & & & & & 2& \\ & 2& & & & & & & & & & 2& \\ \end{array} $$

So in this case I wrote result from arguments of operations in a Hypercube as called in a "Tesseract" in order to see it more easily.

I did not tried it when $n = 5$ because we will have $2^{(2^5)} = 4294967296 $ algebraic operations and to check it`s associative property it takes a long time!

Will it be again 8 if $n = 5$?
So my question is: If this my program tells me true, then how to prove it using only math, I mean without programming, that the number of all associative $n$-ary algebraic operations on a finite set with 2 cardinality will be always 8?

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  • $\begingroup$ Excellent question! $\endgroup$ – Nicky Hekster Aug 6 '14 at 14:42
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The paper "Classification of all associative mono-$n$-ary algebras with $2$ elements" by S.D.Andres (Int. J. Math. Math. Sci. (2009)) http://www.emis.de/journals/HOA/IJMMS/Volume2009/678987.pdf classifies all associative $n$-ary operations on a $2$-element set up to isomorphism, and you can easily deduce from the classification in Section 3.1 that if you count isomorphic ones separately then there are $8$ of them for any $n>1$.

The eight operations are:

  • $f(x_1,\dots,x_n)=1$ for all $(x_1,\dots,x_n)$.
  • $f(x_1,\dots,x_n)=2$ for all $(x_1,\dots,x_n)$.
  • $f(1,\dots,1)=1$, and otherwise $f(x_1,\dots,x_n)=2$.
  • $f(2,\dots,2)=2$, and otherwise $f(x_1,\dots,x_n)=1$.
  • $f(x_1,\dots,x_n)=x_1$.
  • $f(x_1,\dots,x_n)=x_n$.
  • $f(x_1,\dots,x_n)=1$ iff an odd number of $x_i$ are equal to 1.
  • $f(x_1,\dots,x_n)=1$ iff an even number of $x_i$ are equal to 1.

The first and second are isomorphic, as are the third and fourth, and the seventh and eighth are isomorphic if $n$ is even, so up to isomorphism there are five for even $n$ and six for odd $n$.

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  • $\begingroup$ Surprising result! $\endgroup$ – Nicky Hekster Aug 6 '14 at 14:42

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