More than one Right Identity then no Left Identity

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

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 right identity.

Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$.

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

Call this left identity $e_L$.

Then, by the behaviour of $e_L$, $e_{R_1}$ and $e_{R_2}$:


 * $e_{R_1} = e_{R_1} \circ e_L = e_L$
 * $e_{R_2} = e_{R_2} \circ e_L = e_L$

So $e_{R_1} = e_L = e_{R_2}$, which contradicts the supposition that $e_{R_1}$ and $e_{R_2}$ are different.

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

Also see

 * More than one Left Identity then no Right Identity