Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup
Jump to navigation
Jump to search
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 $12.15$