Union of Subgroups

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Theorem

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

Let $H, K \le G$ be subgroups of $G$.


Let neither $H \subseteq K$ nor $K \subseteq H$.

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


Corollary 1

Let $H \cup K$ be a subgroup of $G$.


Then either $H \subseteq K$ or $K \subseteq H$.


Corollary 2

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


Then $H \vee K = H \cup K$ if and only if $H \subseteq K$ 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 $H \setminus K = \O$ 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 $G$.

$\blacksquare$


Examples

Subgroups of $S_3$

Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as:

$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$


Consider the subgroups $H, K \le G$:

$H = \set {e, \tuple {12} }$
$K = \set {e, \tuple {13} }$

We have that:

$H \cup K = \set {e, \tuple {12}, \tuple {13} }$

and:

$\tuple {12} \circ \tuple {13} = \tuple {123}$

But $\tuple {123} \notin H \cup K$.

Hence $H \cup K$ is not closed and so is not a group.

The result follows by definition of subgroup.

$\blacksquare$


Sources