# Union of two proper subgroups [closed]

When can a group be written as the set-theoretic union of its proper subgroups?

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## closed as too localized by Gjergji Zaimi, Victor Protsak, Gerry Myerson, Denis Serre, Andreas ThomNov 16 '10 at 6:45

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Whether this is an actual homework question or not, it's not the right level for MO. Voting to close. – Victor Protsak Nov 16 '10 at 5:56

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.

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i am confused by your notation in the last line of your theorem due to cohn: what are a and b? should they be x and y? – Sean Tilson Nov 16 '10 at 6:44
@Sean Tilson: Oops; yes. Sorry about that. – Arturo Magidin Nov 16 '10 at 14:38

If and only if $G$ is not cyclic. If $G$ is a non-cyclic 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.

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