We know what is called an anti-isomorphic operation on a set S.
it is just a one two one $ g $ function mapping from $S$ to $S$.
$ g: S \rightarrow S$. and it satisfy this condition
$ g(xy)= g(y)g(x) $.

let`s say $\circ$ and $*$ is an anti-isomorphic binary operation on a set S.
if $S=\{1,2,3\}$ or if cardinality of $S=3$, and if it is a finite set, and
$ \begin{array}{c|ccccc} \circ & & & \\ \hline & 1& 1& 1& \\ & 1& 2& 3& \\ & 1& 1& 1& \end{array} $
Then it is an associative operation and it has only one anti-isomorphic operation!
$ \begin{array}{c|ccccc} * & & & \\ \hline & 1& 1& 1& \\ & 1& 2& 1& \\ & 1& 3& 1& \end{array} $
Which is again associative!
We know that basically an anti-isomorphic operations Cayley tables are transpose matrices to each other.

I need to know if a binary operation has a lot of anti-isomorphism operation, then will all of them associative?
Also what about if S is an infinite set or uncountable set?
If they will be associative as well then I need to know How to prove it?


Let $g\colon (S,\circ)\to (S,*)$ be the antiisomorphism and $h\colon S\to S$ be its inverse. By assumption $$ g(x\circ y)=g(y)*g(x) $$ and it's easy to prove that $h$ is an antiisomorphism as well: $$ h(x*y)=h(y)\circ h(x) $$ (prove it).

Now, suppose $*$ is associative. Then \begin{align} x\circ (y\circ z)&=h(g(x))\circ(h(g(y)\circ h(g(z)))\\ &=h(g(x))\circ(h(g(z)*g(y)))\\ &=h(g(x))\circ(h(g(y\circ z)))\\ &=h(g(y\circ z)*g(x)))\\ &=h((g(z)*g(y))*g(x))\\ &=h(g(z)*(g(y)*g(x))) &&\text{by associativity of $*$}\\ &=\dots \end{align}


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