Union of Subgroups

Theorem
Let $$\left({G, \circ}\right)$$ be a group, and let $$H, K \le G$$ such that neither $$H \subseteq K$$ nor $$K \subseteq H$$.

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

Corollary
Let $$H \vee K$$ be the join of $$H$$ and $$K$$.

Then $$H \vee K = H \cup K$$ iff $$H \subseteq K$$ or $$K \subseteq H$$.

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

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

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

If $$\left({H \cup K, \circ}\right)$$ is a group, then it must be closed.

If $$\left({H \cup K, \circ}\right)$$ 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 $$K$$ and therefore $$\left({H \cup K, \circ}\right)$$ is not closed.

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

Proof of Corollary
From the definition of join, $$H \vee K$$ is the smallest subgroup of $$G$$ containing $$H \cup K$$.

The result follows from the main result.