Subgroup lattices are union-free

Theorem. let $G$ be a group, then $Sub(G)$ is union-free. In particular, if $A,B \subseteq G$ such that $A \not\subseteq B$ and $B \not\subseteq B$ then $A \cup B$ is not a group.

Proof. since $A$ and $B$ are incomparable there must be some $a$ in $A-B$ and some $b$ in $B-A$. Then we can consider their product $ab$, since it is in $A \cup B$ it must be in either $A$ or $B$. If it is in $A$ then $a^{-1}ab = b$ is in $A$ which contradicts the supposition that $b$ is not in $A$, likewise if it is in $B$ then $abb^{-1}=a$ is in $B$ which contradicts the supposition that $a$ is not in $B$. So by proof by contradiction, the union of subgroups is not a subgroup.

Corollaries:
Modules, ideals, rings, or any structure more algebraically complicated then a group is also union-free. The union of any of them cannot even be an additive subgroup, much less satisfy any other conditions. This theorem lets us show that the union of two ideals is not an ideal using only group theory.