Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup

Theorem

Let $G$ be a group.

Let $H$ and $K$ be normal subgroups of $G$.

Let $H \subseteq K$.

Then $H$ is a normal subgroup of $K$.

Proof

\(\ds \forall g \in G: \, \)

\(\ds g K\)

\(=\)

\(\ds K g\)

Definition of Normal Subgroup

\(\ds \forall g \in G: \, \)

\(\ds g H\)

\(=\)

\(\ds H g\)

Definition of Normal Subgroup

\(\ds \leadsto \ \ \)

\(\ds \forall g \in K: \, \)

\(\ds g H\)

\(=\)

\(\ds H g\)

as $K \subseteq G$

Hence the result by definition of normal subgroup.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.15$