2004
DOI: 10.1023/b:desi.0000029214.50635.17
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Remarks on Singer Cyclic Groups and Their Normalizers

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Cited by 21 publications
(55 citation statements)
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“…Let M := x i | 0 ≤ i < q n −1 q−1 . The first two statements in the remark below follow from results in [2,7,25]. The third statement follows from the second statement and the facts that Z G ≤ stab G ( 1 ) and M ∈ lt (H, Z G ).…”
Section: Mlss For G L N (Q) and P G L N (Q)mentioning
confidence: 81%
“…Let M := x i | 0 ≤ i < q n −1 q−1 . The first two statements in the remark below follow from results in [2,7,25]. The third statement follows from the second statement and the facts that Z G ≤ stab G ( 1 ) and M ∈ lt (H, Z G ).…”
Section: Mlss For G L N (Q) and P G L N (Q)mentioning
confidence: 81%
“…As described above, since the number of non-zero isotropic points in P (V ) with respect to the quadratic space (V, Q) are (q m + 1)(q m−1 − 1)/(q − 1), therefore, we need to construct a cyclic group A of order q m + 1, which must be sharply transitive on a partial spread S and a cyclic set B of cardinality (q m−1 − 1)/(q − 1), which must be sharply transitive on the projective subspace of P (V ). Then, we take advantage of the Lemma 3, Lemma 5 and Lemma 6 for proving the existence of MLSs for O [19] . Then b 1 ∈ GL 2m (q) can be well defined as follows: Also, we observe that the group A 1 is sharply transitive on S We at first consider O + 2m (q).…”
Section: Lemma 4 [62123] If G Is a Solvable Group Then G Has An Mlsmentioning
confidence: 99%
“…If in addition, the group G preserves a product decomposition, then the same decomposition is preserved by G (2) . Thus, in this case, the form of this group can be found by means of the classification of 3/2-transitive imprimitive linear groups given in [8].…”
Section: Theorem 32 Let G ≤ Sym(v ) Be a Uniprimitive 3/2-transitivementioning
confidence: 99%
“…Then C = Inv(Γ ) and so Aut(C) = Γ (2) . On the other hand, it is easy to see that the orbits of the group Γ v other than {v} all have the same size |K|.…”
Section: Lemma 33 If the Scheme C Is Nontrivial Then T Is A Charactmentioning
confidence: 99%
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