Character codegrees in finite groups

Guohua QIAN

Front. Math. China ›› 2023, Vol. 18 ›› Issue (1) : 15 -32.

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Front. Math. China ›› 2023, Vol. 18 ›› Issue (1) : 15 -32. DOI: 10.3868/S140-DDD-023-006-X
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Character codegrees in finite groups

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Abstract

For an irreducible character χ of a finite group G, we define its codegree as cod(χ)=|G:kerχ|χ(1). In this paper, we introduce some known results and unsolved problems about character codegrees in finite groups.

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Finite group / character / character codegree

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Guohua QIAN. Character codegrees in finite groups. Front. Math. China, 2023, 18(1): 15-32 DOI:10.3868/S140-DDD-023-006-X

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1 Introduction

Throughout this paper, G always denotes a finite group, and the term “character” always refers to complex character unless otherwise stated. The standard notation and terminology are used, where Irr(G) and cd(G) represent the irreducible character set and the irreducible character degree set of G, respectively. For an irreducible character (or representation), the most important quantity is undoubtedly its degree. In the 1960s, Isaacs et al. initiated the theoretical research on character degrees, and since then, the study on character degrees has been a major topic in finite group theory with thousands of research results.

Twenty years ago, the author began investigating the dual arithmetic quantity of character degree, namely, character codegree. This topic has garnered increasing attention and is becoming a hot topic in the character theory of finite groups. In this paper, we present some known results and unsolved problems regarding the character codegrees of finite groups.

Definition 1.1 [25] For χIrr(G), we define its codegree as

cod(χ):=|G:kerχ|χ(1).

Let cod(G) be the set of codegrees of irreducible characters of G. For example, the degrees and codegrees of the five irreducible characters of the symmetry group Sym4 are as follows:

χ1(1)=1,cod(χ1)=1;χ2(1)=1,cod(χ2)=2;χ3(1)=2,cod(χ3)=3;

χ4(1)=χ5(1)=3;cod(χ4)=cod(χ5)=8.

Note that the codegree of χIrr(G) is an integer factor of |G/kerχ| and |G|. Since

χ(1)cod(χ)=|G:kerχ|,

the character degree and the character codegree are dual arithmetic quantities with respect to the character. It follows from the equation |G|=χIrr(G)χ(1)2 that

χ(1)cod(χ)

for all χIrr(G), with equality holding if and only if χ=1G. The following proposition is fundamental to character codegree and ensures that we can use induction in most cases.

Proposition 1.1 [25]  Let χIrr(G).

(1) If NG with Nkerχ, then χ may be viewed as an irreducible character of G/N, and the codegrees of χ in G and in G/N are the same.

(2) If M is subnormal in G and ψ is an irreducible constituent of χM, then cod(ψ)cod(χ).

Proof (1) This follows directly from the definition of character codegree.

(2) By induction we may assume that M is maximal normal in G. We note that kerχMkerψ. Suppose first that kerχM. Then kerχkerψ, and since χ(1)/ψ(1) divides |G/M|, we conclude

cod(ψ)=|M:kerψ|ψ(1)||G:kerχ|χ(1)=cod(χ)

as desired. Suppose now that kerχM. The maximality of M implies G=kerχM, so χM=ψ,

cod(ψ)=|M:kerψ|ψ(1)=|M:Mkerχ|ψ(1)=|G:kerχ|χ(1)=cod(χ),

and we are done.□

Note that the codegree of χIrr(G) is defined as |G|/χ(1) in some literature, such as [8, 9, 34]. The main drawback of this definition is that the codegree of χ in G is inconsistent with the codegree of χ in G/kerχ.

What are the natural problems for the character codegrees? We believe that at least the following three aspects can be considered. Firstly, examine the arithmetic properties of the character codegrees themselves, though this area is almost unexplored. Secondly, study the relationship between character codegree and other arithmetic quantities, such as character degree and element order. Thirdly, investigate the dual problems of the character codegree corresponding to those of the character degree, which is a focus of most current research.

The fundamental Ito-Michler theorem determines which primes can divide a member in cd(G), while the analogous question for cod(G) is which primes can divide a member in cod(G). Let us answer the question below.

Proposition 1.2 [27]  A prime p divides some member in cod(G) if and only if pπ(G), that is, p is a prime divisor of |G|.

Proof We only prove the necessity of the statement. Let L/E be a chief factor of G with p dividing |L/E|. Suppose that L/E is an elementary abelian p-group. Then p divides the codegree of any nonprincipal irreducible character of L/E. Suppose that L/E is a direct product of some isomorphic nonabelian simple groups. By Thompson’s theorem [15, Corollary 12.2], we can take a nonprincipal character ψIrr(L/E) of p-degree. Therefore, in either case, there exists ψIrr(L) with pcod(ψ). Now let χ be an irreducible constituent of ψG. Then Proposition 1.1 yields pcod(χ), as wanted.□

We use the following notation in this paper: Irr(G) denotes the set of nonprincipal irreducible characters of G. For a character ψ, we use Irr(ψ) to represent the set of irreducible constituents of ψ. If NG, we write

Irr(G|N)={χIrr(G)Nkerχ},

cd(G|N)={χ(1)χIrr(G|N)},

cod(G|N)={cod(χ)χIrr(G|N)}.

Clearly, Irr(G|N)Irr(G/N) is a partition of Irr(G). For a positive integer m and a prime p, let π(m) be the set of prime divisors of m, and use mp or m|p to represent the p-part of m.

2 Character codegree and element order

The arithmetic quantities of character codegree and element order arise from group representation and group structure, respectively. At first glance, there appears to be no necessary connection between these two quantities. However, our research has revealed a strong connection between them, which has motivated us to further investigate the character codegree.

Proposition 2.1 [27]  Let G be a nonabelian simple group, and let p,qπ(G) with pq. Then pq divides some member in cod(G).

Proof Suppose pq does not divide any member in cod(G). Let a,bG be of order p and q respectively, and let χIrr(G). Then

χ(1)p=|G|porχ(1)q=|G|q.

That is, χ is of p-defect zero or q-defect zero. By [15, Theorem 8.17], we have χ(a)χ(b1)=0. This leads to 0=χIrr(G)χ(a)χ(b1)=1G(a)1G(b1)=1, a contradiction.□

Let πe(G) be the set of element orders of G.

Proposition 2.2 [27]  Let p,qπ(G) with pq. If pqπe(G), that is, G has an element of order pq, then pq divides some member in cod(G).

Proof Let E be a minimal normal subgroup of G. If {p,q}π(E), then E is a direct product of nonabelian simple groups, and Proposition 2.1 shows that pq divides some member in cod(E), thus Proposition 1.1 yields the required result. If pqπe(G/E) (in particular, if E is a {p,q}-group), then induction and Proposition 1.1 also imply the result. Furthermore, if G has two different minimal normal subgroups E1,E2 satisfying pπ(E1) and qπ(E2), by Proposition 1.2, we can take μIrr(E1) with pcod(μ) and νIrr(E2) with qcod(ν), so pq divides cod(μ×ν) and hence divides some member in cod(G). In summary, for any minimal normal subgroup E of G, we may assume that

pπ(E),qπ(E),andpqπe(G/E).

Suppose that G has two different minimal normal subgroups E1,E2. Let xG be of order pq. The above assumptions yield xqE1E2=1, a contradiction. Hence, we may also assume that G has a unique minimal normal subgroup E.

Suppose the result is false and let xG with o(x)=pq. By the uniqueness of E, every χIrr(G|E) is faithful, so χ(1)p=|G|p or χ(1)q=|G|q. Obviously, xGE. Note that χ(x)=0 for every χIrr(G|E) by [15, Theorem 8.17]. Now,

|CG(x)|=χIrr(G)|χ(x)|2=χIrr(G/E)|χ(x)|2=|CG/E(xE)|.

Since we assume pqπe(G/E), it follows pq|CG/E(xE)|. However, pq obviously divides |CG(x)|, a contradiction, and the proof is complete.□

Proposition 2.3 [28]  Let G be a solvable group and let m be a square-free member in πe(G). Then m divides some member in cod(G).

Furthermore, Isaacs removed the solvability assumption in Proposition 2.3 and proved the following nice result.

Theorem 2.1 [16]  Let m be a square-free member in πe(G). Then m divides some member in cod(G).

Proposition 2.4 [28]  Let mπe(G) be a prime power. Then m divides some member in cod(G).

Proposition 2.5  Let G be an abelian group. Then πe(G)=cod(G).

Proof For an abelian group G, it is well known that Irr(G) forms a group and is isomorphic to G under the usual linear character multiplication. For each λIrr(G), it is easy to see that cod(λ) is exactly the order of the element λ in the abelian group Irr(G). Therefore, πe(G)=cod(G).□

We have recently proven the following theorem, which is much stronger than Theorem 2.1 for solvable groups.

Theorem 2.2 [31]  Let G be a solvable group and let mπe(G). Then m divides some member in cod(G).

In view of the above results, we have the following conjecture.

Conjecture 2.1 If mπe(G), then m divides some member in cod(G).

We think that Conjecture 2.1 is meaningful. Before fully solving this conjecture, there are several related issues that are worth investigating.

Problem 2.1 Suppose mπe(G) with |π(m)|=2. Prove that m divides some member in cod(G).

Problem 2.2 Suppose mπe(G) and G is a nonabelian simple group. Prove that m divides some member in cod(G).

Problem 2.3 Let G be a finite group (or solvable group), and let λ be an irreducible character of some subgroup of G. Is there an irreducible component χ of λG such that cod(λ)cod(χ)?

Note that if Problem 2.3 has an affirmative answer for all finite groups, then Conjecture 2.1 holds directly.

Proposition 2.3 can be restated as follows: If a solvable group G contains an element x, then there exists χIrr(G) such that π(o(x))π(cod(χ)). In [24], Moretó asked whether the converse of Proposition 2.3 holds, namely, If a solvable group G has an irreducible character χ, is there an element xG such that π(cod(χ))π(o(x))? Unfortunately, there are examples showing that the answer to this question is negative.

Conjecture 2.2 If mcod(G) and G is solvable, then there exists dπe(G) such that |π(d)π(m)|12|π(m)|.

We have observed that, in addition to cd(G), πe(G) can also be used as a frame of reference of cod(G).

3 The p-part of character codegree

Let pπ(G), and let X be a set of positive integers associated with G, such as πe(G), cs(G) (the set of conjugacy class sizes of G), cd(G) and cod(G). For any member m of such X, we have 1mp|G|p. Typically, we study the p-part of X by estimating the largest or smallest p-part of its members and by giving a structural description of the group G under the assumption that the p-parts of the members in X are all “small” or all “large”. We define

ep(G)=max{logp(χ(1)p)|χIrr(G)},

cp(G)=max{logp(cod(χ)|p)|χIrr(G)}.

There are many results on the p-part of character degree (see [30, 41] and their references), including the well-known Ito-Michler theorem which states that ep(G)=0 if and only if G has a normal abelian Sylow p-subgroup. As to cp(G), the following one is fundamental.

Theorem 3.1 [27]  For a prime pπ(G), we have

(1) cp(G)1;

(2) for a nonidentity normal subgroup N of G, p does not divide any member in cod(G|N) if and only if PSylp(G) acts fixed-point-freely on N, that is, PN is a Frobenius group with P as its complement and N as its kernel.

Proof (1) This follows by Proposition 1.2.

(2) We only need to prove the necessity. We claim that pπ(N). Otherwise, by Proposition 1.2, we may take λIrr(N) of codegree divisible by p. Then, for any χIrr(λG), we have χIrr(G|N) and pcod(χ), which leads to a contradiction. Thus pπ(N) as claimed, so PSylp(G) acts coprimely on N. Now it suffices to show that P acts fixed-point-freely on N. Let E be a minimal normal subgroup of G.

Suppose pπ(E). Let μIrr(E) with pcod(μ), and let νIrr(N). Note that EN=E×N, and it follows that μνIrr(EN|N) with pcod(μν). For any χIrr((μν)G), we have χIrr(G|N) and pcod(χ), a contradiction. Therefore, all minimal normal subgroups E of G are p-groups. By replacing (G,N) with (G/E,NE/E) and noting that the condition is still valid for (G/E,NE/E), we conclude by induction that either PE/E acts fixed-point-freely on NE/E or NE=E.

Suppose that G has two different minimal normal subgroups E1,E2. Since PEi/Ei acts fixed-point-freely on NEi/Ei (or N=Ei) for i=1,2, it follows easily that P also acts fixed-point-freely on N, and we are done.

Suppose that G has a unique minimal normal subgroup E. Then EN and Irr(G|E)Irr(G|N). For every χIrr(G|E), since χ is faithful and of codegree not divisible by p, we have χ(1)p=|P|. This implies that ψPN is irreducible for every ψIrr(N|E). For ψIrr(N/E), since PE/E acts fixed-point-freely on N/E, ψPN is also irreducible. Therefore, every ψIrr(N) induces an irreducible character of PN, which forces PN to be a Frobenius group with kernel N and complement P.□

Regarding the p-part of the character codegree, there are two independent works [5] and [33]. It is proved in [5] that if G is p-solvable, then the p-length of G is at most cp(G). However, [33] provides a stronger result as stated in the following theorem.

Theorem 3.2 [33]  For a finite group G and a prime p, there exists a normal subgroup N of G such that Op(G)NOpp(G), Opp(G/N)=Opp(G)/N, and G/N possesses a faithful irreducible character χ of p-degree. In particular, we have

logp(|G/Opp(G)|p)logp(|G/N|p)cp(G).

Corollary 3.1 [33]  Let G be a finite group and p be a prime. The following results hold:

(1) if Opp(G)=Op(G), then logp(|G|p)=cp(G);

(2) if G is p-solvable, then logp(|G/Opp(G)|p)<12cp(G).

The previous results show that cp(G) is “large”, prompting the natural question of whether ep(G) (or logp(|G/Op(G)|p)) can be bounded in terms of cp(G). In other words, does there exist a function f such that ep(G)f(cp(G)) for all finite groups G? Remark 3.1(1) below shows that the answer to this question is negative, and Remark 3.1(2) indicates that cp(G) cannot be bounded by ep(G).

Remark 3.1 (1) Let Zp act faithfully and irreducibly on an elementary abelian q-group V, where p and q are different primes. Let G be the direct product of n copies of ZpV. Then, it is easy to verify that

logp(|G/Op(G)|p)=n,ep(G)=nandcp(G)=1.

Since n can be arbitrarily large, neither logp(|G/Op(G)|p) nor ep(G) can be bounded by cp(G).

(2) Let a p-group A act faithfully and irreducibly on an elementary abelian p-group V of order pn, and let G=AV. Then, it is easy to verify that ep(G)=0 and cp(G)=n. This shows that cp(G) cannot be bounded by ep(G).

The part (1) of Theorem 3.1 indicates that cp(G)=0 if and only if p|G|. Now we consider the case when cp(G)=1, which means p2 does not divide any irreducible character codegree of G. Since all groups G with |G|p=p satisfy cp(G)=1, we only need to consider the case where |G|pp2. It is shown in [5] that if cp(G)=1 and |G|pp2, then G is p-solvable with an elementary abelian Sylow p-subgroup.

Theorem 3.3 [33]  Let p be a prime and G be a finite group with |G|pp2. Then, cp(G)=1 if and only if G has a normal series GOp(G)=PV>V1 satisfying the following conditions:

(1) PSylp(G) is elementary abelian, V=Opp(G) is a p-group;

(2) viewing P as a linear space over Fp, we have H/CH(P)Z(GL(P)), where HHallp(G/V);

(3) V is solvable and CV(P)=1;

(4) if in addition |P|p3, then V is nilpotent, and V=CV(a)[V,a] for all aP.

Let pπ(G) and assume that pcod(χ) for any nonlinear χIrr(G). By Theorem 3.1, we see that either G is abelian or PSylp(G) acts fixed-point-freely on G, and in particular the Fitting height of G is at most 2.

Problem 3.1 Let pπ(G) and assume that p2cod(χ) for any nonlinear χIrr(G). Describe the structure of G.

Problem 3.2 For a given prime p and for every irreducible p-Brauer character of G, we can similarly define its codegree. Investigate the structure of G in which p or p2 does not divide any irreducible p-Brauer character codegree of G.

Proposition 3.1 [4]  Let pπ(G) and e be a positive integer. If mp=pe for all 1<mcod(G), then e=1 and G is an elementary abelian p-group.

Problem 3.3 Let p be a prime and e be a positive integer. Study the structure of G in which mp{1,pe} for all mcod(G).

Problem 3.4 Describe the structure of nonsolvable groups G in which c2(G) is small; for related work, see [26].

4 The codegree graph

Let X be a set of positive integers associated with a finite group G, such as πe(G), cs(G), cd(G) and cod(G). For the convenience of expressing and studying the relationship between the members of X, we introduce an undirected simple graph Γ(X) as follows: the vertices are prime divisors of members of X, and two distinct vertices p and q are connected by an edge if and only if pq divides some member of X.

The graph Γ(X) has been extensively studied when X=πe(G),cs(G) or cd(G), and the research content is roughly as follows.

(1) Describing the graph-theoretical properties of Γ(X) for finite groups in general or for some large class of finite groups.

(2) Describing the group structure of G under some graph-theoretical assumption of Γ(X). Notice that existing results have shown that Γ(X) usually has strong connectivity. This indicates that only when the connectivity of Γ(X) is assumed to be not too strong (at least not a complete graph), it is possible to obtain a description of the structure of the group.

We use Γ(G) to denote the codegree graph Γ(cod(G)) of G. We know that π(G) is the set of vertices of Γ(G) by Propositions 1.2, and that Γ(G) is a complete graph when G is a simple group by Proposition 2.1. Furthermore, Proposition 2.2 implies that Γ(πe(G)), often called the prime graph of G, is a subgraph of Γ(G), and the two graphs have the same set of vertices. This indicates that the codegree graph of a group has stronger connectivity than the prime graph of the group. The following theorem presents the basic graph-theoretic properties of the codegree graph.

Theorem 4.1 [27]  For distinct primes p,q,r in π(G), there exist distinct u and v in {p,q,r} such that uv divides some member of cod(G), meaning that there is at least one edge in the codegree graph Γ(G) connecting two points from p,q, and r. In particular,

(1) Γ(G) has at most two connected components;

(2) if Γ(G) is connected, diam(Γ(G)), the diameter of Γ(G), is at most 3.

We now discuss the connection between the graph-theoretical properties of Γ(G) and the structural properties of G, starting with the two classes of graphs with the least connectivity: triangle-free graphs and disconnected graphs.

Proposition 4.1 [3]  If Γ(G) does not contain triangles, then G is solvable and |π(G)|5.

Problem 4.1 Classify finite groups G with triangle-free Γ(G).

If all character codegrees of G are prime powers, then the codegree graph Γ(G) is disconnected unless G is of prime power order. Finite groups whose character codegrees are all or almost all primes have been studied in [25] and [3]; moreover, finite groups with all character codegrees being prime powers have been classified in [27] and [37].

Recall that a finite group G with normal subgroups M and N such that G/N is a Frobenius group with kernel M/N and M is a Frobenius group with kernel N is called a 2-Frobenius group. For finite groups with disconnected codegree graphs, we have the following theorem.

Theorem 4.2 [27]  The codegree graph Γ(G) is disconnected if and only if G is a Frobenius or 2-Frobenius group, and in this case, Γ(G) has exactly two connected components, each of which is a complete graph.

Let 1<NG, and let Γ(G|N) be the graph defined by the set cod(G|N). Clearly, Γ(G|N) is a subgraph of Γ(G). It is proved in [27] that Γ(G|N) has at most two connected components; if Γ(G|N) is disconnected and G is nonsolvable, then G is very close to being a Frobenius group; if Γ(G|N) is disconnected and G is solvable, then the Fitting height of G is at most 4.

We note that the diameter of codegree graph of a finite group can reach the upper bound 3. For example, consider G=(Z2×Z3)(Z5×Z7) satisfying the following conditions: Z2 acts fixed-point-freely on Z5×Z7, while Z3 acts trivially on Z5 and fixed-point-freely on Z7. Then cod(G)={2,3,6,35,5,15}, the distance between 2 and 7 in Γ(G) is equal to 3, and so diam(Γ(G))=3.

Problem 4.2 Classify the finite groups G with diam(Γ(G))=3.

Problem 4.3 Given p,qπ(G) such that p and q are not connected by an edge in the codegree graph Γ(G), investigate the {p,q}-structure of G.

Theorem 4.3 [1]  If F(G)=1, where F(G) is the Fitting subgroup of G, then Γ(G) is a complete graph.

The above theorem generalizes Proposition 2.1, but its proof relies on the classification of finite simple groups.

Problem 4.4 Can Theorem 4.3 be proven without using the classification of finite simple groups?

As previously pointed out, the prime graph of a finite group must be a subgraph of its codegree graph. Moreover, there exist examples showing that the prime graph of a solvable group can be a proper subgraph of its codegree graph.

Problem 4.5 Provide a characteristic description of finite solvable groups for which the prime graph and codegree graph differ.

Problem 4.6 Describe the common properties of prime graphs and codegree graphs of solvable groups.

Conjecture 4.1 For an undirected simple graph Γ, Γ is the prime graph of some solvable group G1 if and only if it is the codegree graph of some solvable group G2.

5 Character codegree and character degree

In this section, we describe the group structure by comparing the size relationship, divisibility relationship, or coprimelity relationship between character degrees and character codegrees of the group. For a nonprincipal irreducible character χ, χ(1) must be strictly less than cod(χ). However, it is possible for a prime p that χ(1)p>cod(χ)|p or χ(1)p<cod(χ)|p.

Theorem 5.1 [12]  A finite group G is nilpotent if and only if χ(1)cod(χ) for all χIrr(G).

Localizing the above theorem, we obtain the following Theorem 5.2, which is a generalization of the Ito-Michler theorem.

Theorem 5.2 [29]  Let p be a prime and assume that χ(1)pcod(χ)|p for all χIrr(G). Then one of the following holds:

(1) G is p-closed, i.e., G has a normal Sylow p-subgroup;

(2) p=3, and G has a composition factor isomorphic to the alternating group Alt7;

(3) p=2, and G has a composition factor isomorphic to Alt7, Alt11, or Mathieu group M22.

Corollary 5.1 [29]  Let p be a prime and assume that either p3 or G is p-solvable. Then G is p-closed if and only if χ(1)p<cod(χ)|p for all χIrr(G) satisfying pχ(1).

Corollary 5.2 [29]  For a prime p5, the following statements are equivalent:

(1) G is p-closed;

(2) χ(1)p<cod(χ)|p for all χIrr(G) satisfying pχ(1);

(3) χ(1)pcod(χ)|p for all χIrr(G).

The following result (Theorem 5.3) due to Tong-Viet is actually the Brauer character version of Theorem 5.2.

Theorem 5.3 [36]  Let p be a prime and assume that [ϕ(1)p]μpcod(ϕ)|p for all irreducible p-Brauer characters ϕ of G, where μ2=8, μ3=2, and μp=1 for p5. Then G is p-closed.

Problem 5.1 Study the non-p-solvable groups G in which [χ(1)p]2cod(χ)|p for every χIrr(G), or χ(1)p[cod(χ)]2 for every χIrr(G), where p is a given prime.

Problem 5.2 Suppose that π(χ(1))π(cod(χ)) for every irreducible character χ of G. Is G solvable? If so, describe the structure of G.

We introduce some sufficient conditions for the solvability of finite groups by examining the size relationship between character degrees and character codegrees. The author once conjectured that if χ(1)2cod(χ) for all nonlinear χIrr(G), then G is solvable. Although the conjecture is not true, Gao and Liu obtained the following result.

Proposition 5.1 [13]  If χ(1)e>cod(χ) for all nonlinear χIrr(G), where

e=log(460815505920)log(10944)11.8876,

then G is solvable.

Problem 5.3 Study the finite groups G in which χ(1)2cod(χ) for all nonlinear χIrr(G). Given the classification for the nonsolvable case, estimate the upper bound of the derived length and the Fitting height of G for the solvable case.

Proposition 5.2 [38]  If cod(χ)p(χ)χ(1) for all nonlinear χIrr(G), where p(χ) is the largest prime divisor of |G:kerχ|, then G is solvable.

Problem 5.4 Describe the structure of the group in Proposition 5.2.

Conjecture 5.1 Suppose that cod(χ)<d(χ)χ(1) for all nonlinear χIrr(G), where d(χ) is the order of the largest Sylow subgroup of |G/kerχ|. Then G is solvable (If this is not true, classify the nonsolvable groups satisfying the assumption).

A finite group G is called an H-group if gcd(χ(1),cod(χ))=1 for every χIrr(G). In [19], we gave a complete description of all H-groups (see also [20]), here we only describe the nonsolvable H-groups.

Theorem 5.4 [19]  A nonsolvable group G is an H-group if and only if it satisfies the following conditions:

(1) G=D(L×N), where D, L and N are Hall subgroups of G;

(2) L=L, L has a solvable normal subgroup E such that L/EPSL(2,2f), where f2;

(3) E=1, or |E|=22f and cd(L|E)=22f1;

(4) D is cyclic, |D| is a square-free divisor of f;

(5) G/LDN is a solvable H-group.

Problem 5.5 For a given prime p, study the p-structure of finite groups G in which p does not divide gcd(χ(1),cod(χ)) for any χIrr(G).

Problem 5.6 Let p be a prime and k be a positive integer. Study the p-solvable groups G in which pk+1 does not divide gcd(χ(1),cod(χ)) for any χIrr(G).

Problem 5.7 Study the finite groups G in which gcd(χ(1),cod(χ)) is square-free for all χIrr(G). If G is solvable, estimate the upper bounds for the Fitting height of G and the rank of G/F(G); if G is nonsolvable, provide a complete description of it.

Problem 5.8 Study the finite groups G in which χ(1)pcod(χ)|p for every χIrr(G) and every prime divisor p of cod(χ).

Problem 5.9 Study the finite groups G in which χ(1)pcod(χ)|p for every χIrr(G) and every prime divisor p of χ(1).

Obviously, each H-group satisfies the conditions given in above two problems.

Problem 5.10 If G is solvable and gcd(χ(1),cod(χ))>1 for all nonlinear χIrr(G), is there an upper bound for the Fitting height of G?

6 ρ-σ conjuncture

Let X be a set of positive integers associated with a finite group G, such as πe(G), cs(G), cd(G) and cod(G). We define ρ(X) as the set of primes that divide some member of X, and let

σ(X)=max{|π(m)|:mX}.

The ρ-σ conjecture for cd(G) was proposed by Huppert, and it is one of the central problems in character theory. The ρ-σ conjecture for general X is as follows: there exists a constant c such that |ρ(X)|cσ(X) for all finite groups. Obviously,

ρ(πe(G))=ρ(cod(G))=π(G).

For simplicity, we write

σe(G)=σ(πe(G)),σco(G)=σ(cod(G)).

Note that by Isaacs’ result (Theorem 2.1), we have

σe(G)σco(G).

For the ρ-σ conjecture on element orders, Keller proved the following well-known result for solvable groups.

Theorem 6.1 [18]  For every solvable group G, we have

|π(G)|C(|π(G)|)σe(G),

where C(n) is a function satisfying limnC(n)=4. In particular, there exists a constant C such that |π(G)|Cσe(G) for every solvable group G.

Using equalities (1), (2), inequality (3), and Keller's theorem, we immediately get the affirmative answer to the ρ-σ conjecture on character codegrees for solvable groups, and then extend it to all finite groups.

Theorem 6.2 [40]  There exists a constant c such that |π(G)|cσco(G) for all finite groups G.

Yang [39] improved the constant c in Theorem 6.2 to an absolute constant 233. For the solvable case, we directly applied Keller's result without essential study on σco(G). Therefore, the absolute constant 233 still has room for improvement.

Conjecture 6.1 For every solvable group G, we have |π(G)|3σco(G).

By Theorem 6.3(3) below, the absolute constant 3 in Conjecture 6.1 cannot be improved.

Conjecture 6.2 For every finite group G, we have |π(G)|4σco(G).

Theorem 6.3 [24, 17, 42]  For solvable groups G, the following results hold:

(1) if σe(G)=2 or σco(G)=2, then the exact upper bound of |π(G)| is 5;

(2) if σe(G)=3 or σco(G)=3, then the exact upper bound of |π(G)| is 8;

(3) if σe(G)=4 or σco(G)=4, then the exact upper bound of |π(G)| is 12.

Problem 6.1 Consider only solvable groups G. For a positive integer n, let f1(n) denote the exactly upper bound of |π(G)| when σe(G)=n, and let f2(n) denote the exact upper bound of |π(G)| when σco(G)=n. Is there a positive integer m such that f1(m)f2(m)?

Problem 6.2 Study finite groups G with σco(G)=2. Note that some special cases are discussed in [35].

Suppose that G is a solvable group. It is easy to prove that σe(G)=1 if and only if σco(G)=1. However, we do not know whether σe(G)=2 if and only if σco(G)=2? There are some more questions worth discussing in this direction.

Problem 6.3 Study the finite groups in which there is no prime dividing three character codegrees. See [32] and its references for the dual problem of the degree version.

Problem 6.4 Let k be a positive integer and suppose that there is no prime dividing k members in cod(G). Is |π(G)| or |cod(G)| bounded by some function of k?

7 Others

In this section, we continue to present our results and problems regarding character codegrees. Note that these results and problems also have corresponding versions with respect to character degrees as shown in [14, 15].

We start by discussing the nilpotent class of p-groups. In [11], an upper bound for the nilpotent class of a p-group in terms of its character codegrees was established, and the character codegrees of a maximal class p-group were investigated in [10]. For a dihedral group G of order 2n, its nilpotent class is n1, and cd(G)={1,2}, so e2(G)=1. This implies that the nilpotent class of a p-group cannot be bounded by ep(G). Interestingly, however, the nilpotent class of a p-group can be bounded by cp(G).

Theorem 7.1 [11]  Let G be a p-group with cp(G)=a2. Then the nilpotent class of G does not exceed 2a2. Furthermore, if p=2 and a3, or if p=3 and a4, then the nilpotent class of G does not exceed 2a3.

Problem 7.1 Is the nilpotent class of a p-group G bounded by |cod(G)|?

Problem 7.2 Is the nilpotent class of a Sylow p-subgroup of a finite group G bounded by cp(G)?

Next, we discuss the relationship between |cod(G)| and the structure of G. Obviously |cod(G)|=1 if and only if G=1; and if |cod(G)|=2, then it is easy to verify that G is an elementary abelian p-group for some prime p.

Proposition 7.1 [2]  If |cod(G)|=3, then either G is a p-group of class 2, or |π(G)|=2, and G is a Frobenius group with a complement of prime order.

The following proposition describes the nonsolvable groups with few character codegrees. Note that Proposition 7.1 implies that |cod(G)|4 for all nonsolvable groups G.

Proposition 7.2 [22, 23]  For a nonsolvable group G, the following hold.

(1) If |cod(G)|=4, then GPSL(2,2f) for some integer f2.

(2) If |cod(G)|=5, then one of the following holds:

   1) G has a minimal normal subgroup M of order 22f, with exactly two G-conjugacy classes, and G/MPSL(2,2f);

  2) GPSL(2,q), where q7 is an odd prime power;

  3) GPGL(2,q), where q5 is an odd prime power;

  4) GM10.

Conjecture 7.1 The Fitting height of a solvable group G does not exceed |cod(G)|.

Conjecture 7.2 The derived length of a solvable group G does not exceed |cod(G)|.

Since there is currently no efficient method to estimate the number of character codegrees of a finite group, Conjecture 7.2 seems difficult to verify even for p-groups. It is unclear what kind of number |cod(G)| is.

Problem 7.3 Is there any relationship between |cd(G)| and |cod(G)|?

Problem 7.4 Study the finite groups G in which χ1(1)=χ2(1) if and only if cod(χ1)=cod(χ2) for every pair of nonlinear irreducible characters χ1,χ2 of G.

Now we discuss how cod(G) determines the structure of G. A well-known conjecture proposed by Huppert states: If cd(G)=cd(S), where G is a finite group and S is a nonabelian simple group, then GS×A for some abelian group A. Similarly, the author proposed the dual conjecture as follows.

Conjecture 7.3 Let G be a finite group and S be a nonabelian simple group. If cod(G)=cod(S), then GS.

It is proved in [6], the first published article on the conjecture, that Conjecture 7.3 holds when SPSL(2,q).

Problem 7.5 Let H be a finite group and G a solvable group. If cod(G)=cod(H), is H also solvable? Alternatively, is there a solvable section of H whose order is close to that of |G|?

We acknowledge that Problem 7.5 may be challenging. However, we propose a more specific conjecture worth studying.

Conjecture 7.4 Let H be a finite group and G an abelian group. If cod(G)=cod(H), then H is solvable.

Next, we introduce the dual object of complex group algebra. For a given mcod(G), we define nG(m) as the number of irreducible characters χIrr(G) with codegree m, and we call it the multiplicity of m. Set CC(G)={(m,nG(m))|mcod(G)}. Although CC(G) lacks an algebraic structure, it is really dual to the complex group algebra C(G) of G.

We know that the structure of G is almost determined by its group algebra C(G), now it's natural to inquire about the extent to which CC(G) can determine the structure of G. Unfortunately, there is currently no method to calculate the multiplicity of a codegree, and this also leads us to know almost nothing about CC(G). Here we list only two questions regarding CC(G).

Problem 7.6 Let G1 and G2 be two finite groups with CC(G1)=CC(G2). Does it follow that |G1|=|G2|?

Problem 7.7 Let G1 and G2 be two finite groups with CC(G1)=CC(G2). If G1 is abelian, does it follow that G2 is also abelian?

It is worth noting that examining the average codegree 1|Irr(G)|χIrr(G)cod(χ) of G may yield interesting results. Although we only discuss the codegrees for irreducible complex characters here, we believe that there are parallel results or questions for irreducible Brauer characters in many situations. Additionally, investigating the codegrees for only a subset of the irreducible characters can also lead to outcomes, as demonstrated in [21, Theorem 1.4] and [7].

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