When can a group be written as the settheoretic union of its proper subgroups?

closed as too localized by Gjergji Zaimi, Victor Protsak, Gerry Myerson, Denis Serre, Andreas Thom Nov 16 '10 at 6:45
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
The question does not match the title... As has been noted, a group $G$ is the union of proper subgroups if and only if $G$ is not cyclic. No group is the union of two proper subgroups (simple exercise often assigned as homework). A more interesting question is: when is a group a union of $n$ proper subgroups, $n\gt 2$, but no fewer? Theorem (Scorza) A group $G$ is the union of three proper subgroups if and only if $G$ has a quotient isomorphic to $C_2\times C_2$. Theorem (Cohn) A group $G$ is the union of four proper subgroups and no fewer if and only if $G$ has a quotient isomorphic to $S_3$, or a quotient isomorphic to $C_3\times C_3$. A group $G$ is a union of five proper subgroups but no fewer if and only if it has a quotient isomorphic to $A_5$. A group is a union of six proper subgroups but no fewer if and only if it has a quotient isomorphic to the dihedral group of order $10$, a quotient isomorphic to $C_5\times C_5$, or a quotient isomorphic to $\langle x,y\mid x^5, y^4, x^2yx^{1}y^{1}\rangle$. Theorem (Tomkinson) There are no groups that are the union of seven proper subgroups but no fewer. I seem to remember it has been shown that for any $n\gt 2$, there is a finite set of groups $S(n)$ so that $G$ is the union of $n$ proper subgroups but no fewer if and only if $G$ has a quotient isomorphic to a group in $S(n)$. The minimal number of subgroups that cover the symmetric and alternating groups $S_k$ and $A_k$ have only been found for smallish values of $k$, though upper and lower bounds are known. 


If and only if $G$ is not cyclic. If $G$ is a noncyclic group, you can write $G = \bigcup_{H \le G} H$. The LHS clearly contains the RHS. If $g \in G$, then $g$ generates some proper subgroup $H$ of $G$, hence $g$ is in the RHS. If $G$ is cyclic, then some $g \in G$ generates the whole group, so it can't lie in a proper subgroup. 


Almost* always  simply write $\langle g \rangle$ for the cyclic subgroup generated by $g\in G$. Then: $$G \subseteq \bigcup_{g\in G}\langle g\rangle \subseteq \bigcup_{H<G} H.$$ I'm not sure this is a research level question, which makes me sad, since it's the first time I've seen a question on MO I felt like I could honesttogoodness provide a complete answer. *Edit: I neglected to say, as I was "shooting from the hip" that this requires $\langle g \rangle <G$ (as the notation above should indicate) for all $g\in G$. This is true iff $G$ is not cyclic, as others have now pointed out. Thanks to Sándor for pointing this out courteously. I will note that the distinction between < and ≤ is somewhat important, and has been elsewhere neglected. I also think I would like to make it clear that I edited this response. This has also been elsewhere neglected. 

