# Are there examples of non-commutative inverses of non-commutative binary operations?

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:

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

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

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

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

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

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

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

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

• You require noncommutativity to work for all elements? You say division is not commutative but $2\div (2) = (2)\div 2$. – David Peterson Aug 16 at 5:28
• @DavidPeterson No, the operation simply not being commutative for all elements in enough. I've corrected the question. – Ryan1729 Aug 16 at 5:36
• 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 $a,b$. – bof Aug 16 at 5:54
• 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. – bof Aug 16 at 6:00
• 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 $s(a,m(b))\neq s(b,m(a))$. – Arturo Magidin Aug 16 at 6:58

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 $$x=y$$)