Let $G$ be a finite solvable group. We show that $G$ does not have a normal
nonabelian Sylow $p$-subgroup when its prime character degree graph $\Delta(G)$
satisfies a technical hypothesis.Comment: 6 pages, 1 figur

Bissler, Mark W.

Laubacher, Jacob

Lyons, Corey F.

English

arXiv

Let G be a finite solvable group. We show that G does not have a normal nonabelian Sylow p-subgroup when its prime character degree graph Δ(G) satisfies a technical hypothesis.  Δ(G) \u3ehttps://digitalcommons.snc.edu/faculty_staff_works/1024/thumbnail.jp

Bisslier, Mark W.

St. Norbert College

St. Norbert CollegeDigital Commons @ St. Norbert CollegeFaculty Creative and Scholarly Works12-17-2018On the absence of a normal nonabelian SylowsubgroupMark W. BisslierWestern Governors UniversityJacob LaubacherSt. Norbert CollegeCorey F. LyonsColorado State UniversityFollow this and additional works at: https://digitalcommons.snc.edu/faculty_staff_worksThis Article is brought to you for free and open access by Digital Commons @ St. Norbert College. It has been accepted for inclusion in FacultyCreative and Scholarly Works by an authorized administrator of Digital Commons @ St. Norbert College. For more information, please contactsarah.titus@snc.edu.Recommended CitationBisslier, Mark W.; Laubacher, Jacob; and Lyons, Corey F., "On the absence of a normal nonabelian Sylow subgroup" (2018). FacultyCreative and Scholarly Works. 23.https://digitalcommons.snc.edu/faculty_staff_works/23ON THE ABSENCE OF A NORMAL NONABELIAN SYLOWSUBGROUPMARK W. BISSLER, JACOB LAUBACHER, AND COREY F. LYONSAbstract. Let G be a finite solvable group. We show that G does not have a normal non-abelian Sylow p-subgroup when its prime character degree graph ∆(G) satisfies a technicalhypothesis.1. Introduction and PreliminariesIn this paper we fix G to be a finite solvable group. Set F(G) to be the Fitting subgroupof G, write Irr(G) for the set of irreducible characters of G, and let cd(G) denote the set ofirreducible character degrees of G. Denote ρ(G) to be the set of primes that divide degreesin cd(G). The prime character degree graph of G, written ∆(G), is the graph whose vertexset is ρ(G). Two vertices p and q of ∆(G) are adjacent if there exists d ∈ cd(G) such that pqdivides d. In this paper, we concern ourselves solely with graphs ∆(G) that are connectedwith ρ(G) ≥ 5.We fix the following notation for an arbitrary vertex p ∈ ρ(G):π := {q ∈ ρ(G) : q is adjacent to p in ∆(G)}andρ := {q ∈ ρ(G) : q is not adjacent to p in ∆(G)}.Observe that ρ induces a complete subgraph in ∆(G) due to Pálfy’s condition from [8], andthat ρ(G) is a disjoint union: {p} ∪ π ∪ ρ. We let π∗ and ρ∗ denote arbitrary nonemptysubsets of π and ρ, respectively. Finally, let π∗∪ρ∗ be an arbitrary vertex set which inducesa complete subgraph in ∆(G). Let β be a subset of ρ(G) that contains π∗ ∪ ρ∗ such that βalso induces a complete subgraph in ∆(G). Set B to be the union of all such β’s satisfyingthese properties for π∗ ∪ ρ∗. Consider τ := B \ (π∗ ∪ ρ∗), and denote τ∗ to be a subset of τ .Notice that τ could be empty, depending on the initial set π∗ ∪ ρ∗.In this note we build upon the work done in [2] and [6] which concerns a normal nonabelianSylow p-subgroup of a finite solvable group. We use tools from [2] which classify vertices asadmissible. Recall the definition and result directly related to admissible vertices below.Definition 1.1. ([2]) A vertex q of a graph Γ is admissible if:(i) the subgraph of Γ obtained by removing q and all edges incident to q does not occuras the prime character degree graph of any solvable group, and(ii) none of the subgraphs of Γ obtained by removing one or more of the edges incident toq occur as the prime character degree graph of any solvable group.Date: November 9, 2018.2010 Mathematics Subject Classification. Primary 20D10; Secondary 20D20.Key words and phrases. finite solvable groups, normal nonabelian subgroup, character degree graphs.12 MARK W. BISSLER, JACOB LAUBACHER, AND COREY F. LYONSLemma 1.2 (Lemma 2.1 from [2]). Let q be an admissible vertex of ∆(G). For everyproper normal subgroup N of G, suppose that ∆(N) is a proper subgraph of ∆(G). ThenOq(G) = G.Most notably, our results pertain to groups G and graphs ∆(G) which do not satisfyLemmas 2.3 and 2.4 from [2]. They also conclude in those lemmas that there is no normalnonabelian Sylow p-subgroup. This is a typical approach en route to showing that a graphcannot occur as the prime character degree graph of any solvable group.Next we recall several lemmas due to Lewis which pertain to groups having a normalnonabelian Sylow subgroup.Lemma 1.3 (Lemma 4.2 from [6]). Let G be a solvable group with a normal nonabelianSylow p-subgroup P . Let H be a p-complement for G. Assume that H acts faithfully on P ,that ρ ⊆ ρ(H), and that the Fitting subgroup of H is a π-group. Then ρ(H) = ρ.Lemma 1.4 (Lemma 4.3 from [6]). Let G be a solvable group with a normal nonabelianSylow p-subgroup P . Let H be a p-complement for G. Suppose that P ′ is minimal normalin G, and ρ(H) = ρ is not empty. Then there is a nonempty subset π∗ of π so that ∆(P ′H)has two connected components: ρ and π∗. Furthermore, if |ρ| = n, then |π∗| ≥ 2n − 1.Lemma 1.5 (Lemma 3 from [5]). Let G be a solvable group and let p ∈ ρ(G). If P is anormal Sylow p-subgroup, then ρ(G/P ′) = ρ(G) \ {p}.To conclude this section, we recall two powerful results concerning when the prime char-acter degree graph is disconnected.Theorem 1.6 (Theorem 5.5 from [4]). Let G be a solvable group and suppose that ∆(G) hastwo connected components. Then there is precisely one prime p so that the Sylow p-subgroupof the Fitting subgroup of G is not central in G.Theorem 1.7 (Pálfy’s inequality from [9]). Let G be a solvable group and ∆(G) its primecharacter degree graph. Suppose that ∆(G) is disconnected with two components having sizea and b, where a ≤ b. Then b ≥ 2a − 1.2. Main ResultsIn this section we prove that G has no normal nonabelian Sylow p-subgroup. Hypothesis2.1 below is strictly about the prime character degree graph ∆(G), whereas Hypothesis 2.2is about the group G itself. Further, we always assume |G| is minimal.Hypothesis 2.1. Concerning ∆(G), we assume the following:(i) for every vertex in ρ, there exists a nonadjacent vertex in π,(ii) for every vertex in π, there exists a nonadjacent vertex in ρ,(iii) all the vertices in π are admissible. Moreover, no proper connected subgraph withvertex set {p} ∪ π∗ ∪ ρ occurs as the prime character degree graph of any solvablegroup,(iv) for each vertex set π∗∪ρ∗ which induces a complete subgraph in ∆(G), all the verticesin the corresponding set τ are admissible. Moreover, no proper connected subgraphwith vertex set ρ(G) \ τ∗ occurs as the prime character degree graph of any solvablegroup, and(v) if a disconnected subgraph with vertex set ρ(G) does not occur, then it must specificallyviolate Pálfy’s inequality from [9]. Finally, if a disconnected subgraph with vertex setρ(G) does occur, then the sizes of the connected components must be n > 1 and 2n−1.ON THE ABSENCE OF A NORMAL NONABELIAN SYLOW SUBGROUP 3Observe the following consequences of Hypothesis 2.1. For (i) and (ii), since ∆(G) isassumed to be connected, it is required that |π| ≥ 2 and |ρ| ≥ 2. As another consequence of(ii), all vertices in π are adjacent to each other. Otherwise, if there existed two nonadjacentvertices in π, then the complement graph would have an odd cycle. This is prohibited bythe main theorem from [1]. Concerning (iii), one gleans that no proper connected subgraphwith vertex set ρ(G) occurs as the prime character degree graph of any solvable group(taking π∗ = π). Depending on the initial subset π∗ ∪ ρ∗, (iv) may require that some ofthe vertices in ρ are admissible. Finally, we know from (v) that if a disconnected subgraphwith vertex set ρ(G) does indeed occur as the prime character degree graph of a solvablegroup, then it must be represented by Example 2.4 from [6].Hypothesis 2.2. Concerning the group G, we assume P is a normal nonabelian Sylowp-subgroup of G, and we let H be a p-complement for G. We set F := F(H), which isnecessarily nontrivial.Concerning F , it is known there exists a degree in cd(G) that is divisible by all the primedivisors of |F |. This corresponds to the vertices in π(|F |) inducing a complete subgraph in∆(G). Next, assumingG has a normal nonabelian Sylow p-subgroup, we can now see the fullforce of Hypothesis 2.1(v). As stated above, this hypothesis forces the disconnected graph(should it occur) to be represented by Example 2.4 from [4]. In that case, all normal Sylowsubgroups are necessarily abelian. Therefore, assumingN is a group such that ρ(N) = ρ(G),we in fact get ∆(N) = ∆(G) by Hypothesis 2.1(iii). Since |G| is minimal we conclude thatN = G.Lemma 2.3. Assume Hypotheses 2.1 and 2.2. Then CH(P ) = 1.Proof. Set C := CH(P ). We follow the argument on page 259 in [6] in a generalized formto show C = 1.First we observe that C is normal in G, and that PC = P ×C. Therefore, any prime inρ(C) must be adjacent to p in ∆(G). So we know that ρ(C) ⊆ π and that r divides |H : C|for every r ∈ ρ. Since we know that H/C acts faithfully on P , we see that the primes inπ(|H : C|) must lie in ρ(G/C). Thus, ρ ⊆ ρ(G/C). We then conclude that ρ(G/C) musthave one of the following vertex sets: (a) {p} ∪ ρ, (b) {p} ∪ π∗ ∪ ρ, or (c) {p} ∪ π ∪ ρ. Hereπ∗ is taken as a nonempty proper subset of π.Suppose (a). Note that none of the primes in π divide |H : C|. We fix a nonlinearcharacter γ ∈ Irr(P ) and then γ extends to γ × 1C in P × C. For χ ∈ Irr(G|γ × 1C) noticethat p divides χ(1) and χ(1)/γ(1) divides |H : C|. Since none of the primes in ρ divide χ(1)and the only primes that could possibly divide |H : C| are the primes in ρ, we are forced toconclude that χPC = γ× 1C , which implies χP = γ. Using Gallagher’s Theorem (Corollary6.17 from [3]) we have that χ(1)d ∈ Irr(G) for every degree d ∈ cd(H). Thus, ρ(H) ⊆ π.This means that H has a normal abelian Hall ρ-subgroup, and hence Oq(G) < G for someq ∈ π. This is a contradiction since every vertex in π is admissible.Suppose (b). By Hypothesis 2.1(iii), we know that any connected subgraph with vertexset {p}∪π∗∪ρ does not occur as the prime character degree graph of any solvable group. If∆(G/C) is a disconnected graph that does occur, then by Theorem 5.5 from [4] we concludethat G/C has a central Sylow q-subgroup for some q ∈ π \ π∗. This implies Oq(G) < G,which is a contradiction since every vertex in π is admissible.We are left with (c), in which case ρ(G/C) = {p} ∪ π ∪ ρ = ρ(G). By Hypothesis 2.1(v),we see that |G/C| = |G|, and hence C = 1. 4 MARK W. BISSLER, JACOB LAUBACHER, AND COREY F. LYONSNotice that Lemma 2.3 implies that H acts faithfully on P . With this in hand, we nowinvestigate F , the Fitting subgroup of H ∼= G/P .Claim 2.4. Assume Hypotheses 2.1 and 2.2. Then F is not a ρ-subgroup.Proof. For the sake of contradiction, we assume that F is a ρ-subgroup. Let γ ∈ Irr(P ) bea nonlinear character and let χ ∈ Irr(G|γ). We see that p divides both γ(1) and χ(1), andthat r does not divide χ(1) or χ(1)/γ(1) for all r ∈ ρ. Since this is true for all charactersχ ∈ Irr(G|γ), we apply Theorem 12.9 of [7] to get that G/P ∼= H has an abelian Hallρ-subgroup.Since we assumed that F was in fact a ρ-subgroup, we now get that F is the abelian Hallρ-subgroup. Thus Oq(G) < G for some q ∈ π, which is a contradiction since all the verticesin π are admissible. Therefore F is not a ρ-subgroup. Claim 2.5. Assume Hypotheses 2.1 and 2.2. Then F is not a π∗ ∪ ρ∗-subgroup.Proof. For the sake of contradiction, we assume that F is a π∗∪ρ∗-subgroup. It is sufficientto suppose π(|F |) = π∗ ∪ ρ∗. Observe that in this scenario, Hypothesis 2.1(i) and (ii) forceπ∗ and ρ∗ to be proper subsets of π and ρ, respectively.We fix the following notation:η := {s ∈ ρ(G) \ ({p} ∪ π(|F |)) : s and q are not adjacent in ∆(G) for some q ∈ π(|F |)}.In fact, η = ρ(G) \ ({p} ∪ B) and η must contain at least one element from π \ π∗ and fromρ \ ρ∗. For otherwise this would contradict Hypothesis 2.1(i) and (ii) as π∗ ∪ ρ∗ forms acomplete subgraph of ∆(G). Thus η is nonempty.Let γ ∈ Irr(PF ) be such that q divides γ(1) for all q ∈ π∗ ∪ ρ∗. For any characterχ ∈ Irr(G|γ), we have that q divides χ(1) for all q ∈ π∗ ∪ ρ∗, and that r does not divideχ(1) or χ(1)/γ(1) for any r ∈ η. By Theorem 12.9 from [7], we see that G/PF ∼= H/F hasan abelian Hall η-subgroup. Let M = OB(H), and note that F ⊆ M . Set E/M to be theFitting subgroup of H/M . Since H/M has an abelian Hall η-subgroup, we see that E/Mis the abelian Hall η-subgroup.Next we have that ρ(PE) contains {p}∪π∗∪ρ∗∪η. However this may be a proper subsetof ρ(G). Notice that by construction we have that ρ(G) \ ({p} ∪ π∗ ∪ ρ∗ ∪ η) = τ . Henceρ(PE) must be one of the following: (a) ρ(G) \ τ∗ or (b) ρ(G).Suppose (a). Then by Hypothesis 2.1(iv) we see that no proper connected subgraph withvertex set ρ(G) \ τ∗ occurs as the prime character degree graph of any solvable group. If∆(PE) is a disconnected graph that occurs, then by Theorem 5.5 from [4] we observe thatPE has a central Sylow q-subgroup for some q ∈ τ∗. This implies Oq(G) < G, which is acontradiction since every vertex in τ is admissible.We are left with (b). So ρ(PE) = ρ(G), and by Hypothesis 2.1(v) we see that PE = G.Since E/M is abelian, and since η ∩ π 6= ∅, we have that Os(G) < G for some s ∈ π.This is a contradiction since all the vertices in π are admissible. Therefore F is not aπ∗ ∪ ρ∗-subgroup. Claim 2.6. Assume Hypotheses 2.1 and 2.2. Then F is not a π-subgroup.Proof. For the sake of contradiction, we suppose F is a π-subgroup ofH and hence ρ ⊆ ρ(H).Furthermore, H acts faithfully on P as consequence of Lemma 2.3. We apply Lemma 4.2from [6] which yields ρ(H) = ρ.Next we will see that P ′ is minimal normal in G. Suppose that X is normal in G suchthat 1 ≤ X ≤ P ′ and P ′/X is a chief factor for G. Our goal will be to show that X = 1.ON THE ABSENCE OF A NORMAL NONABELIAN SYLOW SUBGROUP 5We know ρ(G/X) ⊇ ρ(G/P ′) = ρ(G) \ {p} (by Theorem 3 from [5]). However, p ∈ ρ(G/X)since G/X has a normal nonabelian Sylow p-subgroup. It follows that ρ(G) = ρ(G/X), andhence |G/X| = |G| by Hypothesis 2.1(v). Therefore X = 1 and P ′ is minimal normal in G.The hypotheses of Lemma 4.3 from [6] are satisfied, and therefore ∆(P ′H) has discon-nected components with vertex sets ρ and π∗ such that(1) |π∗| ≥ 2|ρ| − 1.In particular, notice that(2) |ρ| ≤ |π∗|.Hypothesis 2.1(v) gives two cases concerning the possibility of a disconnected subgraphof ∆(G) with vertex set ρ(G). We will investigate these below, but first notice that by ourassumptions, the two connected components of a possibly disconnected subgraph of ∆(G)with vertex set ρ(G) are forced to have vertex sets ρ and {p} ∪ π.For Case 1, we consider if no disconnected subgraph of ∆(G) with vertex set ρ(G) occurs.By Hypothesis 2.1(v) we know that for component sizes a and b (with a ≤ b), we must havethe situation where b < 2a − 1. This gives two subcases:Case 1(a): |ρ| = a and |{p} ∪ π| = b. For this scenario we have|π∗| ≤ |π| < |{p} ∪ π| = b < 2a − 1 = 2|ρ| − 1,which contradicts (1).Case 1(b): |{p} ∪ π| = a and |ρ| = b. Thus|ρ| = b ≥ a = |{p} ∪ π| > |π| ≥ |π∗|,which contradicts (2).For Case 2, we consider if a disconnected subgraph of ∆(G) with vertex set ρ(G) occurs.By Hypothesis 2.1(v) we know the components must be of sizes n > 1 and 2n − 1, whichgives two subcases:Case 2(a): |ρ| = n > 1 and |{p} ∪ π| = 2n − 1. Observe that|π∗| ≤ |π| < |{p} ∪ π| = 2n − 1 = 2|ρ| − 1,which contradicts (1).Case 2(b): |{p} ∪ π| = n > 1 and |ρ| = 2n − 1. Similarly notice|ρ| = 2n − 1 > n = |{p} ∪ π| > |π| ≥ |π∗|,which contradicts (2). Therefore F is not a π-subgroup. We have shown that no primes in ρ(G) \ {p} will divide |F |. This leads to the followingresult:Theorem 2.7. Assume Hypothesis 2.1. Then G has no normal nonabelian Sylow p-subgroup.Proof. For the sake of contradiction, suppose Hypothesis 2.2. That is, suppose G has anormal nonabelian Sylow p-subgroup P , and let H be a p-complement for G with Fittingsubgroup F .By Claims 2.4, 2.5, and 2.6, we see that F must be trivial since no primes divide |F |. AsH is solvable it must have a nontrivial Fitting subgroup, a contradiction. 6 MARK W. BISSLER, JACOB LAUBACHER, AND COREY F. LYONSConsider the following graphs with the appropriate vertex p labeled in Figure 1 for severalexamples and applications of Theorem 2.7.••••p•p••••p••••••p•••••••p•••••••Figure 1. Graphs satisfying Hypothesis 2.1References[1] Zeinab Akhlaghi et al. On the character degree graph of solvable groups. Proc. Amer. Math. Soc.,146(4):1505–1513, 2018.[2] Mark W. Bissler and Mark L. Lewis. A family of graphs that cannot occur as character degree graphsof solvable groups. arXiv:1707.03020, 2017.[3] I. Martin Isaacs. Character theory of finite groups. Dover Publications, Inc., New York, 1994. Correctedreprint of the 1976 original [Academic Press, New York; MR0460423 (57 #417)].[4] Mark L. Lewis. Solvable groups whose degree graphs have two connected components. J. Group Theory,4(3):255–275, 2001.[5] Mark L. Lewis. Solvable groups with character degree graphs having 5 vertices and diameter 3. Comm.Algebra, 30(11):5485–5503, 2002.[6] Mark L. Lewis. Classifying character degree graphs with 5 vertices. In Finite groups 2003, pages 247–265.Walter de Gruyter, Berlin, 2004.[7] Olaf Manz and Thomas R. Wolf. Representations of solvable groups. Cambridge University Press: Cam-bridge, 1993.[8] Péter Pál Pálfy. On the character degree graph of solvable groups. I. Three primes. Period. Math. Hungar.,36(1):61–65, 1998.[9] Péter Pál Pálfy. On the character degree graph of solvable groups. II. Disconnected graphs. Studia Sci.Math. Hungar., 38:339–355, 2001.Department of General Education - College of Business, Western Governors University,Salt Lake City, Utah 84107E-mail address: mark.bissler@wgu.eduDepartment of Mathematics, St. Norbert College, De Pere, Wisconsin 54115E-mail address: jacob.laubacher@snc.eduDepartment of Mathematics and Physics, Colorado State University - Pueblo, Pueblo,Colorado 81001E-mail address: coreylyons314@gmail.com

On the Absence of a Normal Nonabelian Sylow Subgroup

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