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}

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.