Character degree graphs of automorphism groups of characteristically simple groups

Abstract

Throughout this note, G will be a finite group, IrrðGÞ will be the set of irreducible characters of G, cdðGÞ will be the set of character degrees of G, and rðGÞ will be the set of primes that divide degrees in cdðGÞ. The prime vertex degree graph of G, written DðGÞ, is the graph with rðGÞ as its vertex set, and with an edge between p and q if pq divides some degree a A cdðGÞ. An overview of the literature on these graphs can be found in [2]. It seems that most groups G have graphs DðGÞ that are complete graphs (although we do not want to be precise on what we mean by ‘most’). For solvable groups, it has been shown that if DðGÞ is not complete, then the structure of G is limited (see [4] and [6]). In this note, we show that a large class of non-solvable groups have degree graphs that are complete. To study the degree graphs of non-solvable groups, a standard starting point is with the degree graphs of non-abelian simple groups. In a series of papers, Don White has classified the character degree graphs of all non-abelian simple groups. Except for some examples of small rank, nearly all non-abelian simple groups have degree graphs that are complete graphs. These results are summarized in [5]. Extending this idea, this note looks at the characteristically simple groups, groups for which there are exactly two characteristic subgroups. Each such group is a direct product of copies of a fixed simple group. With this in mind, we let G denote G G (n copies); thus S is characteristically simple when S itself is a simple group. We now fix S to be a non-abelian simple group. Because the non-abelian simple groups have already been addressed, we restrict our attention to when n > 1. However, it is relatively easy to show that DðSÞ is complete for n > 1. More generally, the graph DðHÞ associated with any non-abelian group H is always complete for every integer n > 1. Thus we expand our scope and consider extending these groups. We recognize first that S is isomorphic to InnðSÞ. Identifying S with its set of inner automorphisms, our goal herein is to prove the following result.