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Let $V$ be a vector space of finite dimentional $d$ over a field of prime order $p$. For what values of $d$ and $p$, one can find $d+1$ (pairwise distinct) hyperplanes (subspaces of dimension $d-1$) $W_1,\dots,W_{d+1}$ such that \par (1) $W_i+(W_{i+1}\cap\cdots \cap W_{d+1})=V$ for all $i=1,\dots,d$,\par (2) $V\not=W_1 \cup \cdots \cup W_{d+1}$?

I somehow believe the answer (for all values of $d$ and $p$) is negative!!

I would like to know if the question has any corresponding in the language of finite geometry?

The question comes from a problem about certain automorphisms of finite $p$-groups. Let $G$ be a finite non-abelian $p$-group and let $M_1,\dots,M_s$ be maximal subgroups of $G$ whose set-theoretic union is NOT the whole of $G$. Take any element $g\not\in M_1 \cup \cdots \cup M_{s}$ and any central element $z$ of order $p$. Then the maps $\theta_i:g\mapsto gz$ and $\theta_i: m_i \mapsto m_i$ for all $m_i\in M_i$ can be extended to automorphisms of $G$ of order $p$ for each $i=1,\dots,s$. If we may further assume that $Z(G)\leq \Phi(G)$, then the subgroup $\Theta:=\langle \theta_1,\dots,\theta_s\rangle$ is elementary abelian. The question is now as follows: What is the maximal rank (or dimension) of $\Theta$? The rank, of course, will depend on the choice of $M_i$. The ineteger $d$ in the former question corresponds to the minimal number of generators of $G$. We know that the rank is at least the minimal number of generators of $G$.

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    $\begingroup$ Something seems wrong with the question: (1) alone is impossible to fulfill! To see this, notice that, for every $i\in\left\lbrace 1,2,...d\right\rbrace$, the space $W_{d+1}\cap W_d\cap ...\cap W_i$ is a proper subspace of $W_{d+1}\cap W_d\cap ...\cap W_{i+1}$ (because otherwise, $W_i + W_{i+1}\cap ...\cap W_{d+1}$ would be $\subseteq W_i$ and thus couldn't be the whole $V$, contradicting (1)). This yields that $W_{d+1} \supset W_{d+1}\cap W_d \supset W_{d+1}\cap W_d\cap W_{d-1} \supset ... \supset W_{d+1}\cap W_d \cap ...\cap W_1$, where all inclusions are proper. But in ... $\endgroup$
    – darij grinberg
    Commented Jan 17, 2012 at 0:33
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    $\begingroup$ ... a strictly decreasing chain of subspaces of a finite-dimensional space, dimensions must go down by at least $1$ in each step. This is impossible for this chain, since $\dim W_{d+1} = d-1$ and $\dim\left(W_{d+1}\cap W_d\cap ...\cap W_1\right)\geq 0$. $\endgroup$
    – darij grinberg
    Commented Jan 17, 2012 at 0:34

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