Identity is only Idempotent Element in Group/Proof 2

Theorem

Every group has exactly one idempotent element: the identity.

Proof

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $x \in G$ such that $x \circ x = x$.

\(\ds e\)

\(=\)

\(\ds x \circ x^{-1}\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds \paren {x \circ x} \circ x^{-1}\)

by hypothesis: $x \circ x = x$

\(\ds \)

\(=\)

\(\ds x \circ \paren {x \circ x^{-1} }\)

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds x \circ e\)

Definition of Inverse Element

\(\ds \)

\(=\)

\(\ds x\)

Definition of Identity Element

$\blacksquare$

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: $(1.14)$
• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$: Lemma $1$
• 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(i)}$