# Covering of a group by seven proper subgroups: Counterexample

We say that a group $G$ is union of $k$ proper subgroups $H_1,H_2,\cdots, H_k$, if $\cup_{i=1}^k H_i=G$, and union of any $k-1$ subgroups among $H_i$'s is proper subset of $G$.

It is a theorem of M. J. Tomkinson which says, "there is no group which is union of seven proper subgroups". The question, with which I troubled, is due to an example.

Let $G$ be an elementary abelian $3$-group of order $3^n$ ($n>2$). Then to cover $G$, we need at least $3+1$ proper subgroups. It is in fact possible; $G$ can be written as union of four maximal subgroups, say $M_1,M_2,M_3,M_4$. In this case, it can be shown that the intersection of these maximal subgroups, say $L$, has index $p^2$ in $G$. Now, consider the maximal subgroup $M_4$, and we try to cover $M_4$ by four maximal subgroups, say $H_1,H_2,H_3,H_4$. If $n$ is sufficiently large, then, it is possible to choose $H_i$'s diferent from $L$ (am I correct?). Then, $G$ is union of seven proper subgroups: $M_1,M_2,M_3,H_1,H_2,H_3,H_4$. Is this not a counterexample to the theorem of Tomkinson?

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If one wanted a group covered by seven subgroups, surely the elementary abelian group of order 8 is an easier example? – Padraig Ó Catháin May 29 '13 at 10:29

Tomkinson, M.J.. "Groups as the union of proper subgroups." Mathematica Scandinavica 81.2 (1998): 191-198.

In section 3, he proves that any group which is the union of seven proper subgroups can be given as the union of fewer than seven proper subgroups.

The main theorem of the paper is that if you define $\sigma(G)$ to be the smallest $n$ such that $G$ is the union of $n$ proper subgroups, then there is no finite $G$ satisfying $\sigma(G) = 7$.

In summary, I agree that your construction can yield seven proper subgroups whose union is $G$, but it is not a counterexample of the theorem, when the theorem is stated correctly.

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