Exercise 2.1.8 ($H \cup K$ is a subgroup of $G$ iff $H \subseteq K$ or $K \subseteq H$)

Question

Let $H$ and $K$ be subgroup of $G$. Prove that $H \cup K$ is a subgroup if and only if either $H \subseteq K$ or $K \subseteq H$.

Richard Ganaye · Accepted Answer

\begin{proof} 
If $H \subseteq K$ , then $H \cup K = K$ is a subgroup of $G$, and similarly if $K \subseteq H$, $H \cup K = H$ is a subgroup of $G$.

\bigskip
Conversely, suppose that $H \cup K$ is a subgroup of $G$. Assume for the sake of contradiction that $H 
ot \subseteq K$ and $K 
ot \subseteq H$. Then there exist elements $h$ and $k$ such that
\begin{align}
h \in H,\ h 
ot \in K,\qquad k \in K, \ k 
ot \in H.
\end{align}
Then $h \in H \cup K$ and $k \in H \cup K$. Since $H \cup K$ is a subgroup by hypothesis, $h k \in H \cup K$, so $hk \in H$ or $hk \in K$.
\begin{itemize}
\item If $hk \in H$, then $k = h^{-1} (hk) \in H$: this contradicts (1).
\item If $hk \in K$, then $h =(hk)k^{-1} \in K$: this is also in contradiction  with (1).
\end{itemize}
This contradiction shows that either $H \subseteq K$ or $K \subseteq H$.

\bigskip
If $H$ and $K$ are subgroups of $G$, then $H \cup K$ is a subgroup if and only if either $H \subseteq K$ or $K \subseteq H$.
\end{proof}
(Think at the example where $H$ and $K$ are two vectorial lines in $\mathbb{R}^2$, in additive notations).

Exercise 2.1.8 ($H \cup K$ is a subgroup of $G$ iff $H \subseteq K$ or $K \subseteq H$)

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Exercise 2.1.8 ($H \cup K$ is a subgroup of $G$ iff $H \subseteq K$ or $K \subseteq H$)

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