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 $G$.

$\blacksquare$

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.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.2$: Example $94$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $8.13$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$ (passim), Exercise $5.8$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$: Exercise $\text{J}$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $6.7$