Union of Subgroups

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

Let $H, K \le G$ be subgroups of $G$ such that neither $H \subseteq K$ nor $K \subseteq H$.

Then $H \cup K$ is not a subgroup of $G$.

Proof
As neither $H \subseteq K$ nor $K \subseteq H$, it follows from Set Difference with Superset is Empty Set‎ that neither $H \setminus K = \O$ nor $K \setminus H = \O$.

So, let $h \in H \setminus K, k \in K \setminus H$.

Thus, $h \notin K, k \notin H$.

If $\struct {H \cup K, \circ}$ is a group, then it must be closed.

If $\struct {H \cup K, \circ}$ is closed, then $h \circ k \in H \cup K \implies h \circ k \in H \lor h \circ k \in K$.

If $h \circ k \in H$ then $h^{-1} \circ h \circ k \in H \implies k \in H$.

If $h \circ k \in K$ then $h \circ k \circ k^{-1} \in K \implies h \in K$.

So $h \circ k$ can be in neither $H$ nor $K$.

Therefore $\struct {H \cup K, \circ}$ is not closed.

Therefore $H \cup K$ is not a subgroup of $G$.