More than One Right Zero then No Left Zero

Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

If $\struct {S, \circ}$ has more than one left zero, then it has no right zero.

Likewise, if $\struct {S, \circ}$ has more than one right zero, then it has no left zero.

Proof

Let $\struct {S, \circ}$ be an algebraic structure with more than one left zero.

Take any two of these, and call them $z_{L_1}$ and $z_{L_2}$, where $z_{L_1} \ne z_{L_2}$.

Suppose $\struct {S, \circ}$ has a right zero.

Call it $z_R$.

Then, by the behaviour of $z_R$, $z_{L_1}$ and $z_{L_2}$:

$z_{L_1} = z_{L_1} \circ z_R = z_R$

$z_{L_2} = z_{L_2} \circ z_R = z_R$

So $z_{L_1} = z_R = z_{L_2}$, which contradicts the supposition that $z_{L_1}$ and $z_{L_2}$ are different.

Therefore, in an algebraic structure with more than one left zero, there can be no right zero.

$\blacksquare$

The same argument can be applied to an algebraic structure with more than one right zero.