Identity is Unique/Proof 1

Theorem

Let $\struct {S, \circ}$ be an algebraic structure that has an identity element $e \in S$.


Then $e$ is unique.


Proof

Suppose $e_1$ and $e_2$ are both identity elements of $\struct {S, \circ}$.

Then by the definition of identity element:

$\forall s \in S: s \circ e_1 = s = e_2 \circ s$

Then:

$e_1 = e_2 \circ e_1 = e_2$

So:

$e_1 = e_2$

and there is only one identity element after all.

$\blacksquare$


Sources

  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.3$. Units and zeros
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.1$
  • 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2$
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