On the number of isomorphism classes of transversals. (Q369297)

In this paper the authors show that there does not exist a subgroup \(H\) of a finite group \(G\) such that the number of isomorphism classes of normalized right transversals (NRT) of \(H\) in \(G\) is four. In p. 346 the authors prove the following Theorem: Let \(G\) be finite group and \(H\) be a subgroup of \(G\). Then \(|I(G,H)|\neq 4\). The proof of this Theorem is based on the method of contradiction. Assuming the falsity of the result, the authors can find a pair \((G,H)\), called a minimal counterexample, such that: a) \(|G|\) is minimal; b) The index \([G:H]\) is minimal; c) \(|I(G,H)|=4\). The authors study various properties of a minimal counterexample and come to the case of a finite non-Abelian simple group. Finally the authors derive a contradiction. But they do not have an alternate proof of this Theorem (called Main Theorem by the authors) where the use of the classification of finite simple groups could be avoided.