Associative Binary Operation(composition) is anticommutative iff idempotent...

Question

if Binary Operation, $\Delta$, defined on $\mathbb{E}$ is associative, then $\Delta$ is anticommutative iff $\Delta$ is idempotent and $x \Delta y \Delta x=x$, ∀$x,y \in \mathbb{E}$.

Answer 1

$\Longrightarrow$

If $\Delta$ is associative and anticommutative, then $\Delta$ is idempotent and $x \Delta y \Delta x=x$ ∀$x,y,z \in \mathbb{E}$.

Proving $\Delta$ is idempotent:

We want to show $x\Delta x =x$, $x\in \mathbb{E}$.

let $x\Delta x=y$ for some $y \in \mathbb{E}$.

$x\Delta x=y$ $\Longrightarrow$ (right-multiply by x) $x\Delta x\Delta x=y\Delta x$ $\Longrightarrow$ (associativity of $\Delta$) $x\Delta (x\Delta x)=y\Delta x$ $\Longrightarrow$ $x\Delta y=y \Delta x$

The last statement violates the anticommutativity of $\Delta$ forcing $x=y$. ∴ we have $x\Delta x =x$.

Proving $x \Delta y \Delta x =x$ ∀$x,y,z \in \mathbb{E}$ :

Let $x \Delta y \Delta x =z$ for some $x,y,z \in \mathbb{E}$

Left multiplying by x we get:

(1) $x\Delta x \Delta y \Delta x =z$ $\Longrightarrow$($\Delta$ was just shown to be idempotent) $x \Delta y \Delta x =x\Delta z$ $\Longrightarrow$ $z=x\Delta z$

Right multiplying by x we get:

$(2)$ $x \Delta y \Delta x\Delta x =z\Delta x$ $\Longrightarrow$ $x \Delta y \Delta x =z\Delta x$ $\Longrightarrow$ $z=z\Delta x$.

Putting (1) and (2) together we get: $x\Delta z=z\Delta x$. By anticommutativity of $\Delta$ we have $x=z$. ∴ $x \Delta y \Delta x =x$.

$\Longleftarrow$

If $\Delta$ is associative on $\mathbb{E}$ and is idempotent and $x \Delta y \Delta x=x$, ∀$x,y \in \mathbb{E}$, then $\Delta$ is anticommutative.

We want to show ∀$x,y\in \mathbb{E}$ $x\Delta y=y\Delta x$ implies $y=x$

Right multiplying $x\Delta y=y\Delta x$ by $x$ yields $x=y\Delta x$. Right multiplying the same expression by $y$ yields $x\Delta y=y$.

∴ We have $x=y\Delta x$ $\Longrightarrow$ $x=x\Delta y\Delta x$ $\Longrightarrow$ $x=y$.

∴ $\Delta$ is anticommutative.

QED