Correction of a theorem on the symmetric group generated by transvections

Let V denote a vector space over two-element field F₂ with finite positive dimension and endowed with a symplectic form B. Let SL(V) denote the special linear group of V. Let S denote a subset of V. Define Tv(S) as the subgroup of SL(V) generated by the transvections with direction α for all α∈S. Define G(S) as the graph whose vertex set is S and where α, β∈S are connected whenever B(α, β)=1. A well-known theorem states that under the assumption that S spans V, the following (i) and (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 also 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 (i) and (ii) are equivalent.