# 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

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.