Q: If $G = H \cup K \cup L$, then $[G:H] = [G:K] = [G:L]=2$ This is a problem from Isaacs's Algebra. Let $G$ be a finite group and $H, K, L$ be proper subgroups of $G$ such that $G= H \cup K \cup L$. He asks us to show that, in this case, $[G:H] = [G:K] = [G:L] =2$.
It seems simple enough, but I can't get started on it! Any hint would be greatly appreciated.
