10.48 (1986)
SolvedLet $V$ be a vector space of finite dimension over a field of prime order. A subset $R$ of $\text{GL}(V) \cup \{0\}$ is called regular if $|R| = |V|$, $0, 1 \in R$ and $vx \neq vy$ for any non-trivial vector $v \in V$ and any distinct elements $x, y \in R$. It is obvious that $\tau, \varepsilon, \mu_g$ transform a regular set into a regular one, where $x^\tau = x^{-1}$ for $x \neq 0$ and $0^\tau = 0$, $x^\varepsilon = 1 - x$, $x^{\mu_g} = xg^{-1}$ and $g$ is a non-zero element of the set being transformed. We say that two regular subsets are equivalent if one can be obtained from the other by a sequence of such transformations.
$\qquad$ a) Study the equivalence classes of regular subsets.
$\qquad$ b) Is every regular subset equivalent to a subgroup of $\text{GL}(V)$ together with 0?
Progress
a) They were studied (N. D. Podufalov, Algebra and Logic, 30, no. 1 (1991), 62–69).
b) No. A regular set is closed with respect to multiplication if and only if the corresponding $(\gamma, \gamma)$-transitive plane is defined over a near-field. (N. D. Podufalov, Letter of February, 13, 1989).
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