{
  "id": "1108.2409",
  "version": "v2",
  "published": "2011-08-03T03:59:05.000Z",
  "updated": "2012-08-14T14:31:07.000Z",
  "title": "Correction of a theorem on the symmetric group generated by transvections",
  "authors": [
    "Hau-wen Huang"
  ],
  "comment": "7 pages",
  "categories": [
    "math.GR",
    "math.CO"
  ],
  "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.",
  "revisions": [
    {
      "version": "v2",
      "updated": "2012-08-14T14:31:07.000Z"
    }
  ],
  "analyses": {
    "keywords": [
      "symmetric group",
      "transvections",
      "correction",
      "claw-free block graph",
      "well-known theorem states"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 7,
      "language": "en",
      "license": "arXiv",
      "status": "editable",
      "adsabs": "2011arXiv1108.2409H"
    }
  }
}