Group is Normal in Itself

Theorem

Let $\struct {G, \circ}$ be a group.

Then $\struct {G, \circ}$ is a normal subgroup of itself.

Proof

First, by Group is Subgroup of Itself, $\struct {G, \circ}$ is a subgroup of itself.

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

$\forall a, g \in G: a \circ g \circ a^{-1} \in G$

as $G$ is closed by definition.

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