Suppose $G=H\cup K$, where $H$ and $K$ are subgroups. Show that either $H=G$ or $K=G$.
What I did: For finite $G$, if $H\neq G$ and $K\neq G$, then $|H|,|K|\le |G|/2$. But they clearly share the identity element, so $H\cup K\neq G$.
How can I extend it to $G$ infinite?
We show that if $H\cup K$ is a group, then either $H\subseteq K$ or $K\subseteq H$.
If neither $H\subseteq K$ nor $K\subseteq H$ is true, then there is an $h\in H$ which is not in $K$, and a $k\in K$ which is not in $H$.
If $H\cup K$ is a group, then the product $hk$ is in $H\cup K$. Thus the product is in $H$ or in $K$. If $hk\in H$, then $k\in H$, contradicting the choice of $k$. We get a similar contradiction if $hk\in K$.
