Left and Right Identity are the Same

Theorem

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

Let $e_L \in S$ be a left identity, and $e_R \in S$ be a right identity.


Then:

$e_L = e_R$

that is, both the left identity and right identity are the same, and are therefore an identity $e$.

Furthermore, $e$ is the only left identity and right identity for $\circ$.


Proof

Let $\struct {S, \circ}$ be an algebraic structure such that:

$\exists e_L \in S: \forall x \in S: e_L \circ x = x$
$\exists e_R \in S: \forall x \in S: x \circ e_R = x$


Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result.

The uniqueness of the left and right identity is a direct result of Identity is Unique.

$\blacksquare$


Sources

  • 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.4: \ 13$
  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $5$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.3 \ \text{(a)}$
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