A noncyclic finite group $G$ may be expressed as a union of some of its proper subgroups. (Say the subgroups "cover" $G$ in this case.) A relatively simple exercise in some introductory algebra texts, e.g. Jacobson [Basic Algebra I, 2ed p 36 ex 14 ], is to show that no group is the union of two proper subgroups. So I thought what if one is allowed more proper subgroups? One cannot express a cyclic group as a union of proper subgroups, since one of the subgroups would containing the generator and hence fail to be proper. On the other hand, if $G$ is noncyclic it has such an expression, since one can form all the cyclic groups generated by the elements of $G$, and then their union is $G$ and each is proper since $G$ is not cyclic.

My question is about the minimal number of proper subgroups needed to cover a given noncyclic group $G.$ If this minimal number is denoted $m(G)$ then I found e.g. $m(S\_3)=4$ and also $m(K)=3$ for the symmetric group $S\_3$ and the Klein four-group $K.$
I would be interested if anything is known about $m(S\_n)$ generally, or in what $m(G)$ turns out to be for other families of groups. (Even some more examples of $m(S\_n)$ for small $n$ would be nice.) Another interesting case might be for the multiplicative groups of invertible elements mod $n$ (in cases where no primitive root exists.)