We know that a group $G$ cannot be written as the set theoretic union of two of its proper subgroups. Also $G$ can be written as the union of 3 of its proper subgroups if and only if $G$ has a homomorphic image, a non-cyclic group of order 4.
In this paper http://www.jstor.org/stable/2695649 by M.Bhargava, it is shown that a group $G$ is the union of its proper normal subgroups if and only if its has a quotient that is isomorphic to $C_{p} \times C_{p}$ for some prime $p$.
I would like to make the condition more stringent on the subgroups. We know that Characteristic subgroups are normal. So can we have a group $G$ such that , $$G = \bigcup\limits_{i} H_{i}$$ where each $H_{i}$'s are Characteristic subgroups of $G$?
Best Answer
One way to ensure this happens is to have every maximal subgroup be characteristic. To get every maximal subgroup normal, it is a good idea to check p-groups first. To make sure the maximal subgroups are characteristic, it makes sense to make sure they are simply not isomorphic. To make sure there are not too many maximal subgroups, it makes sense to take p=2 and choose a rank 2 group.
In fact the quasi-dihedral groups have this property. Their three maximal subgroups are cyclic, dihedral, and quaternion, so each must be fixed by any automorphism.
So a specific example is QD16, the Sylow 2-subgroup of GL(2,3).
Another small example is 4×S3. It has three subgroups of index 2, a cyclic, a dihedral, and a 4 acting on a 3 with kernel 2. Since these are pairwise non-isomorphic, they are characteristic too. It also just so happens (not surprisingly, by looking in the quotient 2×2) that every element is contained in one of these maximal subgroups.