On the Isomorphism Classes of Transversals-II

On the number of isomorphism classes of transversals

“…Thus [G : N] = 3. Hence G is isomorphic to a subgroup of Sym (6). By Lemma 2.9(ii), the choices for the pair (G, H) in this case are G ∼ = Alt(4) × C 2 , H ∼ = C 2 × C 2 or G ∼ = Alt(4), H ∼ = C 2 .…”

“…Proof. Since Core G (H) = {1}, we can identify G with a subgroup of Sym (6). Thus, there exist subgroups K and L of Sym (6) As argued in the proof of [6, Lemma 2.12, p.2030], S and S i (1 ≤ i ≤ 3) are non-isomorphic NRTs.…”

“…Let H be a non-normal, abelian, corefree subgroup of G and N be a normal subgroup of G containing H such that (ii) Assume that [G : N] = 3. We can identify G with a subgroup of Sym (6). Since the order of an abelian subgroups of Sym( 6 As argued in the second paragraph of the proof of [6, Lemma 2.12, p.2030], we can show that S 4 is not a subgroup of G and S 4 = T ′ L ′ for any T ′ ∈ T (N, H) and L ′ ∈ T (G, N).…”

“…Then |N| = 16. Hence N is a Sylow 2-subgroup of Sym (6). Since G ⊆ N Sym(6) (N) (the normalizer of N in Sym (6)) and a Sylow 2-subgroup of Sym (6) is self-normalizing ( [15,Corollary 1,p.…”

“…Hence f (l) = hk 2 l or f (l) = hk 3 l. Suppose that f (l) = hk 2 l. (ii) Assume that [G : N] = 3. We can identify G with a subgroup of Sym (6). Since the order of an abelian subgroups of Sym (6) is at most 9 ([1, Theorem 1, p. 70]), |H| ≤ 9.…”