Conjugacy Class Size Conditions Which Imply solvability

Abstract

Let $G$ be a finite $p$-solvable group and let ${G}^{\ast } $ be the set of elements of primary and biprimary orders of $G$. Suppose that the conjugacy class sizes of ${G}^{\ast } $ are $\{ 1, {p}^{a} , n, {p}^{a} n\} $, where the prime $p$ divides the positive integer $n$ and ${p}^{a} $ does not divide $n$. Then $G$ is, up to central factors, a $\{ p, q\} $-group with $p$ and $q$ two distinct primes. In particular, $G$ is solvable.

‘Conjugacy Class Size Conditions Which Imply solvability’

“…It follows that y has index 1 or n in C G (z), and if there are p -elements of primary orders of both indexes 1 and n in C G (z), it follows that n = p a q b by [1,Lemma 2.1]. By [1,Lemma 2.2], we conclude that G is a {p, q}-group (or a p-group if b = 0) up to central factors. So the proof is finished in this case.…”

“…P  T C. The first two paragraphs of the original proof of Theorem C in [1] are correct. We assume then that there are p-elements of index p a in G and that p divides n.…”

“…The authors have realised that in the proof of [1,Theorem C] there is a mistake, although the conclusion of the theorem is correct. The proof of the theorem is divided into two cases: when there exist no p-elements of index p a in the group G and when they do exist.…”

“…The proof of the theorem is divided into two cases: when there exist no p-elements of index p a in the group G and when they do exist. In the second case of the proof, the authors apply [1,Lemma 2.3] to the abelian normal subgroup O p (G). However, the conclusion is false, so the final contradiction is not achieved as it is argued in the paper.…”

Solvability of Finite Groups With Four Conjugacy Class Sizes of Certain Elements

“…There are errors in Cases 1 and 2 of the proof in [9]. The error in Case 2 was corrected in [10]; our method of the proof of Theorem A corrects the error in Case 1. Furthermore, we prove a more general result.…”