Join of Subgroups is Group Generated by Union
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Theorem
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $S$ be the set of words of $H \cup K$.
Then $S$ is a subgroup of $K$ such that:
- $S = \gen {H \cup K} = H \vee K$
where $H \vee K$ denotes the join of $H$ and $K$.
Proof
By definition, the set of words in $H \cup K$ is:
- $S = \map W {H \cup K} := \set {s_1 \circ s_2 \circ \cdots \circ s_n: n \in \N_{>0}: s_i \in H \cup K 1 \le i \le n}$
Let $h \in H$.
Then setting $n = 1$ in the above definition and letting $s_1 = h$ it follows that $H \subseteq S$.
Similarly it is seen that $K \subseteq S$
So from Union is Smallest Superset it follows that $H \cup K \subseteq S$.
From Set of Words Generates Group it follows that $S$ is the subgroup of $G$ generated by $H \cup K$.
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $19$