TITLE: Union of subgroups is a subgroup if and only if one subgroup is a subset of the other
QUESTION [7 upvotes]: Let $H$ and $K$ denote two subgroups of a group $G$.  Prove that the union $H \cup K$ is a subgroup of $G$ if and only if $H \subset K$ or $K \subset H$.

REPLY [14 votes]: I shall use multiplicative notation for the group law.
If $H$ or $K$ is the trivial subgroup $\{1\}$ consisting only of the multiplicative identity, the result is trivial.  So, take $H, K$ to be subgroups of $G$ such that $H,K \neq \{1\}$.
$\Leftarrow$
If WLOG $H \subset K$, then $H \cup K=K$.  Because $K$ is a subgroup of $G$, $H \cup K$ is a subgroup of $G$.
$\Rightarrow$
Suppose for sake of contradiction that we have neither $H \subset K$ nor $K \subset H$ but that $H \cup K$ is a subgroup of $G$.  This implies that there exists $h \in H$ such that $h \notin K$; likewise, $\exists k \in K$ such that $k \notin H$.  Because $H \cup K$ is a subgroup of $G$, $hk \in H \cup K$, which implies $hk \in H \vee hk \in K$.  Say, WLOG, $hk \in H$.  Well, since $h \in H$, $h^{-1} \in H$.  Thus, $h^{-1}hk=k \in H$, contradiction.