Is there a semigroup which admits no involution?

Question

A semigroup with involution is a semigroup $(S;*)$ equipped with a unary function $f$ such that $f(f(x))=x$ and $f(x*y)=f(y)*f(x)$. I want to know, does there exist a semigroup for which there exists no corresponding involution? I would be especially interested if there is a finite semigroup which is a counterexample.

Answer 1

Let $$S = \{x, y\},$$ and let the multiplication $*$ for $S$ be given by

1. $x*x = x$
2. $x*y = x$
3. $y*x = y$
4. $y*y = y.$

It is not difficult to check that this is indeed a semigroup.

Suppose, for a contradiction, that there exists an involution $f \colon S \to S.$ From the condition that $f(f(z)) = z$ for all $z \in S,$ it follows that $f$ is a bijection. So, we must either have $f(x) = x$ and $f(y) = y,$ or $f(x) = y$ and $f(y) = x.$

First, we show that $f(x) = y$ and $f(y) = x$ is impossible. Suppose not. Since $f$ is an involution, we have $$f(x*y) = f(y)*f(x) = x*y.$$ So, $f$ fixes $x*y$. However, we have $x*y = x,$ which is clearly not fixed by $f,$ contradiction.

So, we only have to consider the case $f(x) =x$ and $f(y) = y.$ Once again, since $f$ is an involution, we have $$f(x*y) = f(y)*f(x) = y*x.$$ However, since $$f(x*y) = f(x) = x$$ and $$y*x = y,$$ we also have a contradiction.

Since both cases lead to contradiction, we conclude that the given $(S, *)$ does not admit any involution.