Correction of a theorem on groups generated by transvections over $\mathbb{F}_2$

Mathematics – Group Theory

Scientific paper

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4 pages, 1 figure

Scientific paper

Let $V$ denote a finite dimensional vector space over the two-element field $\mathbb{F}_2$ endowed with a symplectic form $B.$ Let ${\rm rad} V$ denote the radical of $V$ with respect to $B.$ Let ${\rm SL}(V)$ denote the special linear group of $V.$ Let $S$ denote a subset of $V.$ Define $Tv(S)$ as the subgroup of ${\rm SL}(V)$ generated by the transvections with direction $\alpha$ for all $\alpha\in S.$ Define $G(S)$ as the graph whose vertex set is $S$ and where $\alpha,\beta$ in $S$ are connected whenever $B(\alpha,\beta)=1.$ A well-known theorem states that if $S$ spans $V$ then $Tv(S)$ is isomorphic to a symmetric group if and only if $G(S)$ is a claw-free block graph. We give an example which shows that the necessity of the theorem is not true, and give a modification as follows. Let $S$ denote a subset of $V\setminus {\rm rad} V$ and assume that $S$ is a linearly independent set of $V.$ If $Tv(S)$ is isomorphic to a symmetric group then $G(S)$ is a claw-free block graph.

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