4

Suppose $*$ is a binary operator on a set $A$ such that $\forall x,y\in A,$ we have $$x*(x*y)=y$$ and $$(y*x)*x=y.$$

Is $*$ associative?

I can show that $*$ is commutative, because each of the following implies the next $$\begin{eqnarray} x & = & x \\ y*(y*x) & = & [x*(y*x)]*(y*x)\\ [y*(y*x)]*(y*x) & = & \{[x*(y*x)]*(y*x)\}*(y*x)\\ y & = & x*(y*x)\\ x*y & = & x*(x*(y*x))\\ x*y &= & y*x \end{eqnarray}$$

Could you show me whether $*$ is associative?

Thanks.

Community
  • 1
sai ten
  • 43
  • 3
  • 1
    See this post for further comments concerning commutativity. – Dietrich Burde May 26 '20 at 10:41
  • I'm not sure, but if it is, then $\langle A,\rangle$ is a commutative group. To see it, fix any $x\in A.$ If $$ is associative, then for any $y\in A,$ we would have that $$(x * x) * y = x * (x * y) = y,$$ and by commutativity, we would likewise have $$y * (x * x) = y.$$ In other words, $x * x$ is an identity element, say $e.$ Now, take any $z\in A.$ We can see that $$(z * z) = (z * z) * e = z * (z * e) = e$$ by our temporary assumption of associativity. Thus, each element of $A$ would be its own inverse, too. – Cameron Buie May 26 '20 at 12:06
  • One key thing to note is that $$ has not actually been explicitly defined, though you originally stated your question as if it had. The question then becomes whether we can define an operation $$ with the two given properties, but without giving a group structure to $A$. I'd try to explicitly define an operation $*$ on some small sets $A$ with the given properties. It might be that once $A$ is "big enough," there's a way to do it without having an identity and inverses. – Cameron Buie May 26 '20 at 12:13

1 Answers1

8

See the addendum for a generalization of the below construction for any finite size of $A$.

No, the operation need not be associative.

A counterexample is the 3 element set $A=\{a,b,c\}$ with the operation defined as

$$ \begin{array}{c|ccc} * & a & b & c \\ \hline a & a & c & b\\ b & c & b & a\\ c & b & a & c \end{array} $$

This $*$ operation yields the input value if both input values are the same, otherwise it yields the remaining element of $A$ not among the input values.

It's easy to see from the table that this $*$ is commutative, so only the first of conditions needs to be shown, the other follows from commutativity.

If $x=y$, we have

$$x*(x*y)=y*(y*y)=y*y=y,$$

as required. If $x \neq y$, then let $z$ be the only element of $A-\{x,y\}$ and we get:

$$x*(x*y)=x*z=y,$$ again as required, as $x\neq z$.

So this operation $*$ fullfills all the required conditons, but is not associative as

$$a=a*a=a*(b*c), \text { but } c=c*c=(a*b)*c.$$

Addendum:

Looking at the above operation I found how to interpret that as operation mod $3$ and how to generalize that to any $n$ element set $A$.

So for $n\ge 1$ let $A:=\mathbb Z_n$ and define $x*y:=-(x+y)$. Then we have

$$x*(x*y)=x*(-(x+y))=-(x - (x+y))=-(-y)=y$$

and $$(y*x)*x = (-(x+y))*x)= - (-(x+y) + x)=-(-y)=y,$$

so the required condtions are fullfilled.

But we have

$$0*(1*(-1))=0*(-(1+(-1)))=0*0=0$$

and

$$(0*1)*(-1)=(-(0+1))*(-1)=(-1)*(-1)=-((-1)+(-1))=-(-2)=2.$$

Only for $n=1,2$ do we have $0=2$, so this is an example that such a non-associative operation exists for any finite size $n$ of $A$ for $n\ge3$.

It's easy to see that for $n=1,2$ the operation is always associative. For $n=1$ it's trivial. For $n=2$, note that $f_x(y)=x*y$ must be injective (that's true for any $n$), so each row in the operator table is a permutation.

There are only two permutations of a two-element set, and it's easy to see that the only possible operators are $x*y:=x+y$ and $x*y:=x+y+1$, both assuming $x,y \in \mathbb Z_2$, which are associative.

Ingix
  • 14,494