I am trying to prove the following result.
No group $G$ can be the union of two proper subgroups.
The first point of confusion is that I have two different definitions of proper subgroup. My professor defined it as a subgroup $H \subsetneqq G$, while the textbook (Artin) defines it as a subgroup $H \subset G$ that is neither $\{e\}$ or $G$. The former seems more standard to me. Is that correct? I'm going to stick with that in the below attempt.
Suppose for the sake of contradiction that $G = H \cup K$ for subgroups $H,K \subsetneqq G$. Then $H \not \subset K$ and $K \not \subset H$ because, otherwise, $H \cup K = K$ and $H \cup K = H$, respectively, which contradicts $K$ and $H$ being proper subgroups. So there exists $x \in G \setminus H$ and $y \in G \setminus K$. So $xy \in G$, so $xy \in H$ or $xy \in K$. If $xy \in H$, since $y$ and hence $y^{-1}$ are. elements of $H$, we have $(xy)y^{-1} = x \in H$, a contradiction. Similarly, if $xy \in K$, since $x$ and hence $x^{-1}$ are elements of $k$, we have $x^{-1} (xy) = y \in K$, a contradiction. So this construction is impossible.
