Suppose $G$ is finite and $G=H\cup K\cup L$ for proper subgroups $H,K,L$. Show that $|G:H|=|G:K|=|G:L|=2$.
What I did: so if some of $H,K,L$ is contained in another, then we have $G$ being a union of two proper subgroups, which is impossible due to another result. So none of $H,K,L$ is contained is each other. If some element $h$ belongs to only $H$ and $k$ belongs to only $K$, then its product $hk$ cannot be in $H$ or $K$, so must belong to only $L$.
At least one of $H, K$ and $L$ must have index $2$ in $G$ by counting. Let’s say $|G:H| = 2$. Then, $H$ is normal in $G$. Then, $|K:K\cap H| = 2 = |L:L\cap H|$.
Now, for $G = H \cup (K – K\cap H) \cup (L – L\cap H)$, this union must be disjoint by counting and we must have $|G:K| = 2 = |G:L|$.