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.

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 this right identity $e_R$.

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.

Also see

 * More than one Right Identity then no Left Identity