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Subtraction and division are non-commutative binary operations. But their inverses are both commutative, (addition and multiplication respectively). Are there any examples of non-commutative binary operations that have inverse operations which are themselves non-commutative?

After a little searching it appears that my notion of inverse operations is not widely used under those terms. So to be more precise, are there any functions $f$ and $g$ such that there exist at least one $x$ and $y$ such that $f(x, y) \neq f(y, x)$ and there exist at least one $z$ and $w$ such that $g(z, w) \neq g(w, z)$, (non-commutativity) and for all $a$ and $b$ at least one of the following is true:

$g(f(a, b), a) = a$

$g(f(a, b), a) = b$

$g(a, f(a, b)) = a$

$g(a, f(a, b)) = b$

$g(f(a, b), b) = a$

$g(f(a, b), b) = b$

$g(b, f(a, b)) = a$

$g(b, f(a, b)) = b$

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  • $\begingroup$ You require noncommutativity to work for all elements? You say division is not commutative but $2\div (2) = (2)\div 2$. $\endgroup$
    – David P
    Commented Aug 16, 2019 at 5:28
  • $\begingroup$ @DavidPeterson No, the operation simply not being commutative for all elements in enough. I've corrected the question. $\endgroup$
    – Ryan1729
    Commented Aug 16, 2019 at 5:36
  • $\begingroup$ Take any noncommutative group and define $f(x,y)=xy$ and $g(x,y)=xy^{-1}$. Then $g(f(a,b),b)=a$ for all $a,b$. $\endgroup$
    – bof
    Commented Aug 16, 2019 at 5:54
  • $\begingroup$ There was another nice example in the comments but it went away: $f(x,y)=g(x,y)=y$ on any set eith more than one element. $\endgroup$
    – bof
    Commented Aug 16, 2019 at 6:00
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    $\begingroup$ Perhaps a better way to think of subtraction is that it is not a binary operation, but rather the composition of a binary operation (addition) and a unary operation (that takes every element to its additive inverse). Letting $s(a,b) = a+b$, and $m(a) = -a$, $a-b = s(a,m(b))$. $a-b\neq b-a$ is actually $s(m(a),b) \neq s(a,m(b))$ (because $s$ is commutative) or $s(a,m(b))\neq s(b,m(a))$. $\endgroup$
    – Arturo Magidin
    Commented Aug 16, 2019 at 6:58

1 Answer 1

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If your domain has at least two elements, a simple example of two functions meeting your conditions are $f(x,y) = x$ and $g(x,y) = y$. Picking $a\neq b$ gives you $f(a,b) = a \neq b = f(b,a)$, and $g(a,b) = b \neq a = g(b,a)$.

Moreover, for any $a,b$ we have: $$\begin{align*} g(f(a,b),a) &= a &&\text{(your first equation)}\\ g(a,f(a,b)) &= a &&\text{(your third equation)}\\ g(f(a,b),b) &= b &&\text{(your sixth equation)}\\ g(b,f(a,b)) &= a &&\text{(your seventh equation)} \end{align*}$$

(If your domain has exactly one element, then you cannot find “noncommuting” functions, since $f(x,y) = f(y,x)$ as we must have $x=y$)

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