Union of Subgroups/Corollary 1

Theorem

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

Let $H, K \le G$.

Let $H \cup K$ be a subgroup of $G$.

Then either $H \subseteq K$ or $K \subseteq H$.

Proof

Aiming for a contradiction, suppose neither $H \subseteq K$ nor $K \subseteq H$.

Then from Union of Subgroups it follows that $H \cup K$ is not a subgroup of $G$.

The result follows by Proof by Contradiction.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $5$: Subgroups: Exercise $7$