Identity of Group is Unique/Proof 2

Theorem

Let $\struct {G, \circ}$ be a group which has an identity element $e \in G$.

Then $e$ is unique.


Proof

Let $e$ and $f$ both be identity elements of a group $\struct {G, \circ}$.

Then:

\(\ds e\) \(=\) \(\ds e \circ f\) $f$ is an identity
\(\ds \) \(=\) \(\ds f\) $e$ is an identity

So $e = f$ and there is only one identity after all.

$\blacksquare$


Sources

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory
  • 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$: Lemma $2$
  • 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: The Group Property
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Proposition $3.1$
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