# 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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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
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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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## 3 Answers

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
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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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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 honest-to-goodness 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.

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There are other places as well as other questions; you have to start somewhere. Gerhard "Ask Me About System Design" Paseman, 2010.11.15 –  Gerhard Paseman Nov 16 '10 at 5:34
Kellen, unfortunately, your answer is not entirely correct. If $G$ is cyclic and $g\in G$ is a generator, then $g$ is not contained in a proper subgroup. Otherwise what you wrote is the solution. –  Sándor Kovács Nov 16 '10 at 5:45
@Gerhard: Thank you for your comment, but I believe I can handle where and when "to start." I am quite capable of being active here and "other places" at my own discretion - even if I'm new to MO and this is an "easy" question. If I didn't assume otherwise, I might have taken this comment as "go elsewhere for easier questions to answer." If you mean well, as I do assume, I would suggest more careful word choice. What sounds like vague, sage, or cryptic advice may - in fact - also sound like a discouraging comment depending on how much the reader intends on exercising the principle of charity. –  Kellen Myers Nov 16 '10 at 16:49
It is meant to be a self-discriminatory comment: if your character is such that you take from it the message "go away", then react to it that way; if you take from it "keep trying", then react to it that way. I am willing for your character to take either or both (or other) interpretations, but what you do is your responsibility. Thank you for your acknowledgment and your observation on my choice of words. Gerhard "Ask Me About System Design" Paseman, 2010.11.16 –  Gerhard Paseman Nov 16 '10 at 19:22
Thanks Gerhard, but in the future, I would ask you not make comments about my character, or worse, comments intended to test or build my character. I do not believe my character is an appropriate topic of conversation - implicitly or explicitly - in a public form for professional collaboration & discussion. –  Kellen Myers Nov 16 '10 at 20:21
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