Binary operations in an algebra

Question

Is there a binary operation ° for the algebra <{1,...,n},°> such that for each $k \in \{1,...,n\}$ there are exactly $k-1$ left inverse elements?

Answer 1

Yes, it is possible to define such an operation as long as $n\geq 2$. It is only possible to define an associative one if $n=2$.

We can define one by letting $xy $ be given as
$x $ if $y=2$
$y $ if $x=2$
$1$ if $x=y=1$
$2$ if $x\leq y $, $x\neq 2$ and $y\geq 3$
$1$ if $x > y $

It is easy to check that the above works. To see that it can only be associative when $n=2$, note that $1$ is the unique element with no left inverse. Assume that some $x$ has more than one left inverse (which will be the case if $n\geq 3$). Then at least one of these, let's call it $y$, will itself have a left inverse, let's call it $z$. I claim that $z = x$ and thus that $y$ is in fact a twosided inverse of $x$.

This is the case because $x = (zy)x = z(yx) = z$. But if $x$ has a twosided inverse $y$ and $w$ is some left inverse of $x$ then $w = y$ since $w = w(xy) = (wx)y = y$. This contradicts the assumption that $x$ had more than one left inverse.