More than one Left Identity then no Right Identity

Theorem
Let $$\left({S, \circ}\right)$$ be an algebraic structure.

If $$\left({S, \circ}\right)$$ has more than one left identity, then it has no right identity.

Likewise, if $$\left({S, \circ}\right)$$ has more than one right identity, then it has no left identity.

Proof

 * Let $$\left({S, \circ}\right)$$ be an algebraic structure with more than one left identity.

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

Suppose $$\left({S, \circ}\right)$$ has a right identity. Call it $$e_R$$.

Then, by the behaviour of $$e_R$$, $$e_{L_1}$$ and $$e_{L_2}$$:


 * $$e_{L_1} = e_{L_1} \circ e_R = e_R$$
 * $$e_{L_2} = e_{L_2} \circ e_R = e_R$$

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

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

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