Trivial Subgroup is Subgroup

Theorem

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

Then the trivial subgroup $\struct {\set e, \circ}$ is indeed a subgroup of $\struct {G, \circ}$.

Proof

Using the One-Step Subgroup Test:

$(1): \quad e \in \set e \leadsto \set e \ne \O$

$(2): \quad e \in \set e \leadsto e \circ e^{-1} = e \in \set e$

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.2$. Subgroups: Example $90$
• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Subgroups: Example $25$
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups
• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 36.1$: Subgroups
• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $4$: Subgroups: Definition $4.1$