Correction of a theorem on the symmetric group generated by transvections

TL;DR

This paper corrects a previous theorem about when a subgroup generated by transvections in a symplectic vector space over  and its associated graph is isomorphic to a symmetric group, providing a new valid condition.

Contribution

It identifies an error in the existing theorem and offers a corrected version with additional assumptions on the generating set.

Findings

The original theorem is not universally true.

A corrected theorem is established under linear independence and radical exclusion conditions.

An explicit counterexample demonstrates the failure of the original statement.

Abstract

Let $V$ denote a vector space over two-element field $\mathbb F_2$ with finite positive dimension and endowed with a symplectic form $B.$ 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 under the assumption that $S$ spans $V,$ the following (i), (ii) are equivalent: (i) $Tv(S)$ is isomorphic to a symmetric group. (ii) $G(S)$ is a claw-free block graph. We give an example which shows that this theorem is not true. We give a modification of this theorem as follows. Assume that $S$ is a linearly independent set of $V$ and no element of $S$ is in the radical of $V.$ Then the above (i), (ii) are equivalent.