Identity of Group is Unique/Proof 2
Jump to navigation
Jump to search
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 $3.1$