Inner Automorphism Maps Subgroup to Itself iff Normal

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 subgroup of $G$.

Then:

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

if and only if:

$H$ is a normal subgroup of $G$.

Proof

Sufficient Condition

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

$\Box$

Necessary Condition

Suppose that:

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

Let $x \in G$ be arbitrary.

By definition of inner automorphism of $x$ in $G$:

$\forall h \in H: x h x^{-1} \in H$

So, by definition, $H$ is a normal subgroup of $G$

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.10$