# More than one Left Identity then no Right Identity

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## Theorem

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

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

## Proof

Let $\struct {S, \circ}$ 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 $\struct {S, \circ}$ 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.

$\blacksquare$

## Also see

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Semigroups: Exercise $3 \ \text{(ii)}$