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.)

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    $\begingroup$ While I haven't encountered this number studied anywhere, there are some elementary things to be said, e.g. a non-cyclic finite group is certainly equal to the union of all its maximal subgroups, so $m(G)$ is bounded above by the number of maximal subgroups, a much more well-known invariant. $\endgroup$ Oct 19, 2014 at 21:53
  • $\begingroup$ @AlastairLitterick It looks like one could replace the term "proper subgroup" by the seemingly more restrictive term "maximal proper subgroup" in defining $m(G),$ since in any union $H_1 \cup \cdots H_n=G$ of proper subgroups each non-maximal $H_k$ could be replaced by a maximal proper subgroup $E_k$ which contained $H_k.$ In other words I think the two definitions would wind up equivalent. But that still leaves open how to get $m(G)$ for specific $G.$ $\endgroup$
    – coffeemath
    Oct 20, 2014 at 2:10

2 Answers 2


This is a well-studied invariant, going back to G. Scorza who, in 1926, showed that the groups $G$ with $m(G) = 3$ are those with $C_2^2$ as a homomorphic image. But much has been done since then. Here are some nice slides from a talk by Martino Garonzi. It, and the references at the end, should get you started.

  • $\begingroup$ Nice reference of Garonzi! $\endgroup$ Oct 20, 2014 at 7:09

As you noted, a group cannot be the union of two of its proper subgroups. It might be interesting to know, that research has been done on how this might generalize.

Theorem (Bruckheimer, Bryan and Muir) A group is the union of three proper subgroups if and only if it has a quotient isomorphic to $C_2 \times C_2$.

The proof appeared in the American Math. Monthly $77$, no. $1 (1970)$. The theorem seems to be proved earlier by the Italian mathematician Gaetano Scorza, I gruppi che possono pensarsi come somma di tre loro sottogruppi, Boll. Un. Mat. Ital. $5 (1926), 216-218$.

For 4, 5 or 6 subgroups a similar theorem is true and the Klein 4-group is for each of the cases replaced by some finite set of groups. For 7 subgroups however, it is not true: no group can be written as a union of 7 of its proper subgroups. This was proved by Tomkinson in 1997.

There is a nice overview paper by Mira Bhargava, Groups as unions of subgroups, The American Mathematical Monthly, $116$, no. $5, (2009)$.


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