Trivial Subgroup is Normal

Theorem

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

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


Proof

First, by Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $G$.


To show $\struct {\set e, \circ}$ is normal in $G$:

$\forall a \in G: a \circ e \circ a^{-1} = a \circ a^{-1} = e$


Hence the result.

$\blacksquare$


Sources

  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.6$. Normal subgroups: Example $123$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
  • 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$: Example $35$
  • 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.7$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 49.1$ Normal subgroups
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Definition $7.3$: Remark
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