# How to prove that: If two binary operations are anti-isomorphic and one of them is associative then the second one also will be associative?

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}