On the sum of character degrees coprime to p and p-nilpotency of finite groups☆
Introduction
In the representation theory of finite groups, one of the main and important problems is to study the relationship between the character degrees of a finite group and the structure of the group. This has been a topic of interest for a long time, going back to the works in the 1960's of Isaacs-Passman [12], [13] and later in the 1980's of Huppert and his school [7].
Let G be a finite group and be the set of all ordinary irreducible characters of G. Let . This invariant , which is often referred to as the character degree sum of G, has been known to having influence on the structure of G. In particular, in [20], [16], it was essentially shown that, the smaller is, the closer to abelian G is.
In this paper we consider a prime p and the -version of , and prove that there is a connection between this new invariant and the p-nilpotency of the group.
We write and
Theorem A Let G be a finite group and p be a prime. Suppose that , where Then G is p-nilpotent.
Recall that G is said to be p-nilpotent if G has a normal subgroup H of order relatively prime to p and is a power of p. Thompson's theorem [19, Theorem 1] on character degrees states that if p divides the degree of every nonlinear irreducible character of G, then G is p-nilpotent. Our Theorem A improves Thompson's theorem because the degree of every nonlinear irreducible character of G is divisible by p if and only if .
Of course Theorem A is not remarkable if there are no groups with . Using GAP [4] to check all the small groups of order up to 500, we indeed found several of them. For example, there are 151 among those such that and 778 among those such that .
Theorem A implies a known result that a finite group G must be nilpotent if (see [1, Chapter 11] for instance). In fact, we can do a bit more.
Corollary B Let G be a finite group and p the smallest odd prime divisor of . Assume that . Then G is nilpotent. Proof It follows from the hypothesis that Note that, for every prime divisor q of , every of degree not coprime to q will have degree at least 2. Therefore we have Hence It follows that for every prime divisor q of . By Theorem A, we deduce that G is q-nilpotent for every , which means that G is nilpotent. □
We also find a bound for to ensure the solvability of G. The next result is actually needed in the proof of Theorem A.
Theorem C Let G be a group and p be a prime. Suppose that , where Then G is solvable.
The case of Theorem C implies the main result of [20].
We remark that all the bounds in Theorems A and C are as tight as possible, as seen in the groups , , , and . We also remark that by the Cauchy-Schwarz inequality. The invariant , which is often referred to as the average of -character degrees, has been studied in [5], [14]. There, it was shown that if then G is 2-nilpotent and if for odd p then G is p-nilpotent. Consequently, if then G is 2-nilpotent and if for odd p then G is p-nilpotent. Theorem A improves not only Thompson's theorem but this result as well.
Section snippets
Preliminaries
In this section we collect and prove some lemmas that will be needed later on.
We will write for . For a character χ of G, we write to denote the set of irreducible constituents of χ. For and , we write for the inertia subgroup of λ in G, and . Also, we denote by the number of irreducible characters of G of degree k. Finally we write to denote the semidirect
Proof of Theorem C
Theorem C will be proved in the next four propositions.
Proposition 3.1 If G is a group such that , then G is solvable.
Proof Suppose that G is a counterexample of minimal order. Let be minimal such that M is nonsolvable and let N be a minimal normal subgroup of G contained in M. Then and . In addition, if , then we choose N such that , where is the unique largest solvable normal subgroup of M, namely, the solvable radical of M. I) Assume that N is abelian. Then
Proof of Theorem A
Theorem A will be proved in the following two propositions.
Proposition 4.1 If G is a group such that , then G is 2-nilpotent.
Proof Suppose that . Then we have and so Since for with , we obtain that This implies that and so Now, it follows from [5, Theorem 1.1(i)] that G
Acknowledgements
The authors are grateful to the referee for his or her comments and suggestions.
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Project supported by NSF of China (Nos. 11801208, 12061011), the Jiangsu Government Scholarship for Overseas Studies (2018), the key program of NSF of Guangxi of China (No. 2020GXNSFDA238014), and a grant from the Simons Foundation (No. 499532).