Union of Subgroups/Corollary 1
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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 $7$