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.