Inner Automorphism Maps Subgroup to Itself iff Normal/Sufficient Condition

Theorem

Let $G$ be a group.

For $x \in G$, let $\kappa_x$ denote the inner automorphism of $x$ in $G$.

Let $H$ be a normal subgroup of $G$.

Then:

$\forall x \in G: \kappa_x \sqbrk H = H$

Proof

Let $H$ be a normal subgroup of $G$.

Let $x \in G$ be arbitrary.

By definition, $\kappa_x: G \to G$ is a mapping defined as:

$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$

Let $n \in N$.

Then:

\(\ds \map {\kappa_x} n\)

\(=\)

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

\(\ds \)

\(\in\)

\(\ds N\)

Definition of Normal Subgroup

$\blacksquare$

Sources

• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$: Theorem $27$