Character degree graphs of solvable groups with diameter three

Theorem 1.1 ([15])

A graph is said to satisfy Pálfy’s Condition if given any three vertices, two of them are adjacent. If G is a solvable group, then Δ⁢(G) satisfies Pálfy’s Condition.

Theorem 1.2 ([17])

The graph in Figure 1 is not the character degree graph of any solvable group.

Theorem 1.3 ([9])

The graphs in Figure 2 are not the character degree graphs of any solvable group.

Theorem 1.4 ([8])

There exists a solvable group that has a character degree graph as shown in Figure 3.

1. Define ρ4⁢(G) to be the set of all vertices that are distance three from the vertex p1. As p4 is distance three from p1, the vertex p4∈ρ4.

2. Define ρ3⁢(G) to be the set of all vertices that are distance two from the vertex p1.

3. Define ρ2⁢(G) to be the set of all vertices that are adjacent to p1 and adjacent to some prime in ρ3⁢(G).

4. Define ρ1⁢(G) to be the vertices consisting of p1 and those that are adjacent to p1 and not adjacent to anything in ρ3⁢(G).

5. We relabel if necessary so that |ρ1∪ρ2|≤|ρ3∪ρ4|

The graphs in Figure 4 do not occur as Δ⁢(G) for any solvable group G. In particular, if G is a solvable group where Δ⁢(G) has diameter three and |ρ⁢(G)|=6, then Δ⁢(G) is isomorphic to the graph in Figure 3.

Let G be a solvable group where Δ⁢(G) has diameter three. Then |ρ3|≥3.

Let G be a solvable group where Δ⁢(G) has diameter three. Then G has a normal non-abelian Sylow p-subgroup for exactly one prime p and p∈ρ3.

Let G be a solvable group with Δ⁢(G) having diameter three. If n=|ρ1∪ρ2|, then |ρ3∪ρ4|≥2n.

Let G be a solvable group with Δ⁢(G) having diameter three. Then G has Fitting height 3.

Theorem 2.1 ([7])

Let G be a solvable group where Δ⁢(G) has two connected components. Then G is one of the following examples:

1. G has a normal non-abelian Sylow p-subgroup P and an abelian p-complement K for some prime p. The subgroup P′⊆ℂP⁡(K) and every non-linear irreducible character of P is fully ramified with respect to P/ℂP⁡(K).

2. G is the semi-direct product of a subgroup H acting on a subgroup P, where P is elementary abelian of order 9 and cd⁡(H)={1,2,3}. Let Z=ℂH⁡(P). We have Z⊆ℤ⁡(H), and H/Z≅SL2⁡(3), where the action of H on P is the natural action of SL2⁡(3) on P.

3. G is the semi-direct product of a subgroup H acting on a subgroup P, where P is elementary abelian of order 9 and cd⁡(H)={1,2,3,4}. Let Z=ℂH⁡(P). We have Z⊆ℤ⁡(H), and H/Z≅GL2⁡(3), where the action of H on P is the natural action of GL2⁡(3) on P.

4. G is the semi-direct product of a subgroup H acting on an elementary abelian p-group V for some prime p. Let Z=ℂH⁡(V) and K the Fitting subgroup of H. Write m=|H:K|>1, and |V|=qm, where q is a p-power. We have Z⊆ℤ⁡(H), K/Z is abelian, K acts irreducibly on V, m and |K:Z| are relatively prime, and (qm-1)/(q-1) divides |K:Z|.

5. G has a normal non-abelian 2 -subgroup Q, and an abelian 2 -complement K with the property that |G:KQ|=2 and the quotient G/Q is not abelian. Let Z=ℂK⁡(Q), and C=ℂQ⁡(K). The subgroup Q′⊆C, and Z is central in G. Every non-linear irreducible character of Q is fully ramified with respect to Q/C. Furthermore, Q/C is an elementary abelian 2 -group of order 22⁢a for some positive integer a, Q/C is irreducible under the action of K, and K/Z is abelian of order 2a+1.

6. G is the semi-direct product of an abelian group D acting coprimely on a group T so that [T,D] is a Frobenius group. The Frobenius kernel is A=T′=[T,D]′, A is a non-abelian p-group for some prime p, and a Frobenius complement is B with [B,D]⊆B. Every character in Irr⁡(T∣A′) is invariant under the action of D and A/A′ is irreducible under the action of B. If m=|D:ℂD(A)|, then |A:A′|=qm where q is a p-power, and (qm-1)/(q-1) divides |B|.

Lemma 2.2 ([7, Lemma 5.1])

Let m be a positive integer, and let q be a prime power such that (qm-1)/(q-1) is relatively prime to m. If r is the number of distinct prime divisors of m, then the quotient (qm-1)/(q-1) has at least 2r-1 prime divisors.

Theorem 2.3 ([16, Theorem 3])

Let G be a solvable group with a disconnected character degree graph Δ⁢(G). If the smaller component has n vertices, then the larger component has at least 2n-1 vertices.

Lemma 2.4 ([9, Lemma 3])

Suppose that G has a non-abelian normal Sylow p-subgroup P for a prime p. Then ρ⁢(G/P′)=ρ⁢(G)\{p}.

Theorem 2.5 ([18])

If G is a solvable group and Δ⁢(G) has diameter three, then G has at most one normal non-abelian Sylow p-subgroup for some prime p∈ρ⁢(G).

Theorem 2.6 ([9, Theorem 2])

Suppose G is a solvable group with Φ⁢(G)=1. Assume that for all nontrivial normal subgroups M, Δ⁢(G/M) has two connected components or the diameter of Δ⁢(G/M) is at most 2. Then either Δ⁢(G) has two connected components or the diameter of Δ⁢(G) is at most 2.

Proof.

Because there is a shortest path between p1 and p4, there exists primes p2 and p3 such that the path p1,p2,p3,p4 is a shortest path between p1 and p4. The prime p3 is necessarily in ρ3 as p3 is distance 2 from p1 otherwise it contradicts that p1,p2,p3,p4 is a shortest path. The prime p2 is necessarily in ρ2 as p2 is adjacent to something in ρ3 and also adjacent to p1. ∎

Proof.

Let r be a prime in ρ1 and s be a prime in ρ4. Because ρ1∪ρ2 determines a complete subgraph of Δ⁢(G) and ρ2 is non-empty, there is a prime p2∈ρ2 that is adjacent to the prime r. Because p2 is adjacent to some prime in ρ3 and ρ3 is non-empty, there exists a prime p3∈ρ3 where p2 and p3 are adjacent. If p2 were adjacent to the prime s, then in particular, s would be distance two from p1 and s would be in ρ3. This contradicts the fact that s∈ρ4. As ρ3∪ρ4 determines a complete subgraph of Δ⁢(G), the prime p3 is adjacent to s and we have found a shortest path r,p2,p3,s. ∎

Proof.

Suppose that r is a prime that is not in ρ1 or ρ4. Without loss, we can assume that r is in ρ2. Because ρ1∪ρ2 determines a complete subgraph, we see that s cannot be in ρ1. Thus s must be in either ρ3∪ρ4. Because r is adjacent to some prime in ρ3, there exists a p3∈ρ3 and r and p3 are adjacent. Because ρ3∪ρ4 determines a complete subgraph, we see that p3 is adjacent to s and the distance between r and s must be less than or equal to two. ∎

Theorem 3.1 ([9, Theorem 4])

If |ρ⁢(G)|=6, then

Let G be a solvable group with Δ⁢(G) having diameter three. Assume that Oρ2⁢(G)<G and that the graph Δ⁢(Oρ2⁢(G)) does not have diameter three. Then, Oρ2⁢(G) must have a normal non-abelian Sylow p-subgroup and the graph Δ⁢(Oρ2⁢(G)) is disconnected. In particular, the group Oρ2⁢(G) is not Example (2.4) or Example (2.5) from Theorem 2.1.

Proof.

Suppose that G is a solvable group and Δ⁢(G) has diameter three. Let K=Oρ2⁢(G) and suppose K<G and so Δ⁢(K) is not diameter three. We assume that K has no normal non-abelian Sylow p-subgroups and derive a contradiction.

Thus K is Example (2.4) or Example (2.5) from Theorem 2.1 with the larger component being ρ3∪ρ4. By [7, Theorem 5.2], cd⁡(K) contains only one degree that is divisible by primes in ρ3∪ρ4. Pick primes p2∈ρ2, p3∈ρ3, and p4∈ρ4 such that p2 and p3 are adjacent in Δ⁢(G). Then there exists a character χ∈Irr⁡(G) such that p2⁢p3 divides χ⁢(1). Let θ be an irreducible constituent of χK in Irr⁡(K). By [3, Corollary 11.29], we know χ⁢(1)/θ⁢(1) divides |G:K|, and hence, p3 divides θ⁢(1). Because K has only one degree divisible by primes in ρ3∪ρ4 and that degree is divisible by p3 and p4, we conclude that p4 also divides θ⁢(1). Thus, p2 and p4 are adjacent in Δ⁢(G), which is a contradiction. ∎

Let G be a solvable group with a normal Sylow p-subgroup P for some prime p∈ρ⁢(G). Suppose ρ⁢(G/M)=ρ⁢(G) implies that M=1 whenever M is a normal subgroup of G. Then P′ is a minimal normal subgroup in G and in particular, P′ is a central subgroup in P.

Proof.

Let G be a solvable group with a normal Sylow p-subgroup P for some prime p∈ρ⁢(G). Assume ρ⁢(G/M)=ρ⁢(G) implies M=1. By Lemma 2.4, we have ρ⁢(G/P′)=ρ⁢(G)\{p}. Let X be a normal subgroup of G that is contained in P′ so that P′/X is a chief factor for G. Because ρ⁢(G/X) contains p and ρ⁢(G/P′), we see that ρ⁢(G/X)=ρ⁢(G). Thus X=1 by our hypothesis and P′ is a minimal normal subgroup of G. Because ℤ⁡(P) is a characteristic subgroup of P, it is normal in G. It follows that P′∩ℤ⁡(P) is a normal subgroup of G. Because P is a nilpotent subgroup of G, P′∩ℤ⁡(P) is not a trivial subgroup, and so P′⊆ℤ⁡(P). ∎

Let G be a solvable group and suppose G has a normal Sylow p-subgroup for some prime p∈ρ⁢(G). Suppose there is a normal subgroup N in G such that p does not divide |N|. Then ρ⁢(G/N) contains every prime in ρ⁢(G) that is not adjacent to p in Δ⁢(G).

Proof.

Let G be a solvable group with a normal Sylow p-subgroup P for some prime p∈ρ⁢(G). Let N be a normal subgroup of G such that p does not divide |N|. The subgroup P⁢N/N is a normal non-abelian Sylow p-subgroup of G/N and so p∈ρ⁢(G/N). Let q be a prime in ρ⁢(G)\ρ⁢(G/N), and let Q be a Sylow q-subgroup of G. Since q divides no degree in cd⁡(G/N), we use Itô’s Theorem [3, Corollary 12.34] to see that Q⁢N/N is abelian and QN is normal in G. Thus, the direct product P×Q⁢N is normal in G. Let χ∈Irr⁡(G) with q dividing χ⁢(1) and let θ∈Irr⁡(Q⁢N) be an irreducible constituent of χQ⁢N. We know that χ⁢(1)/θ⁢(1) divides |G:QN| by [3, Corollary 11.29]. Since q does not divide |G:QN|, we see q must divide θ⁢(1). Thus, q∈ρ⁢(Q⁢N), and so p and q will be adjacent in Δ⁢(P×Q⁢N). Because Δ⁢(P×Q⁢N) is a subgraph of Δ⁢(G), the primes p and q are adjacent in Δ⁢(G). In particular, ρ⁢(G/N) contains every prime in ρ⁢(G) that is not adjacent to p in Δ⁢(G). ∎

Let G be a solvable group where Δ⁢(G) has diameter three. Let M be a minimal normal subgroup of G such that Δ⁢(G/M) has diameter three. If G has a normal Sylow p-subgroup for p∈ρ⁢(G), then G/M does not have a normal Sylow q-subgroup for q∈ρ⁢(G)\{p}.

Proof.

Let G be a solvable group with Δ⁢(G) having diameter three. Suppose G has a normal Sylow p-subgroup P for p∈ρ⁢(G), and let M be a minimal normal subgroup of G and suppose Δ⁢(G/M) has diameter three. Suppose that the subgroup G/M has a normal non-abelian Sylow q-subgroup Q/M for a prime q∈ρ⁢(G)\{p}. Because the subgroup P⁢M/M is a normal Sylow p-subgroup of G/M, the Sylow subgroup P⁢M/M must be abelian or we have a contradiction as G/M can have at most one normal non-abelian Sylow p-subgroup. Thus, P′⊆M. Let R be a Sylow q-subgroup of G. We see that RM is normal in G and so RP is a normal subgroup in G. Further, [R,P]⊆M and so R centralizes P/P′ by [5, Corollary 3.28]. Hence, P=ℂP⁡(R)⁡P′=ℂP⁡(R)⁡Φ⁢(P). As P′⊆Φ⁢(P), by the Frattini Argument [4], P=ℂP⁡(R) and so RP is a direct product. Thus R is normal in RP, so it is characteristic, and R is normal in G. This contradicts the fact that G can have at most one normal non-abelian Sylow p-subgroup. ∎

Let G be a solvable group with Δ⁢(G) having diameter three. Assume for all proper nontrivial normal subgroups M that Δ⁢(G/M) and Δ⁢(M) do not have diameter three, and assume |ρ3|<2n-1 where n=|ρ1∪ρ2|. Then G does not have a normal Sylow p3-subgroup for any prime p3∈ρ3.

Proof.

Suppose G has a normal non-abelian Sylow p3-subgroup P for a prime p3∈ρ3. Then, since ρ⁢(G/P′)=ρ⁢(G)∖{p3} by Lemma 2.4 and the group G/P′ is a nontrivial proper factor group of G, the graph Δ⁢(G/P′) cannot have diameter three by the hypothesis. Because ρ⁢(G/P′) contains primes from ρ1 and ρ4, we have that the graph Δ⁢(G/P′) must be disconnected with components ρ1∪ρ2 and (ρ3∪ρ4)∖{p3}. Further, G/P′ cannot have any normal non-abelian Sylow subgroups or G would have more than one normal non-abelian Sylow subgroup, and that violates Theorem 2.5. So G/P′ is Example (2.4) from Theorem 2.1 as ρ1∪ρ2⊆ρ⁢(G/P′) and |ρ1∪ρ2|≥2.

Theorem 5.3 in [7] gives us that Op2⁢(G/P′)<G/P′, and so Op2⁢(G)<G for every prime p2∈ρ2. Fix the prime p2∈ρ2 and let K=Op2⁢(G). Because ρ⁢(K) contains ρ1 and ρ4, and K is a proper subgroup of G, the graph Δ⁢(K) is a subgraph of Δ⁢(G). By the hypothesis, the graph Δ⁢(K) must be disconnected. As K contains the subgroup P, the subgroup K has a normal non-abelian Sylow p3-subgroup and p3 is in the larger component. So K must satisfy the hypotheses of Example (2.6) from Theorem 2.1.

By [7, Lemma 3.6], K/P′ satisfies the hypotheses of Example (2.4). Let

Because S is a characteristic subgroup of K and K is normal in G, S is normal in G, and S⊆F∩K. Because F∩K is a normal subgroup of K, we have F∩K=S. Further, R is characteristic in K and so R is normal in G. By the Diamond Lemma in [5], we obtain R/S≅R⁢F/F is a normal subgroup of G/F. Because R⁢F/F is normal in G/F, we have R⁢F/F⊆E/F and R⁢F⊆E. Since E/F is a Hall (ρ3∪ρ4)∖{p}-subgroup, and R/S is a Hall (ρ3∪ρ4)∖{p3}-subgroup by [7, Lemma 3.4], R⁢F=E.

Because G/P′ satisfies the hypotheses of Example (2.4), G/P′ is the semi-direct product of a subgroup H⁢P′/P′ acting on an elementary abelian p3-group P/P′, the positive integer m is defined to be the index |H:𝔽(H)|, the order of P/P′ is qm where q is a p3-power, and the quotient (qm-1)/(q-1) divides |E:F|. Since there are |ρ1∪ρ2|=n primes that divide m, there must be 2n-1 distinct primes that divide the quotient (qm-1)/(q-1) by Lemma 2.2. Because p3 does not divide |E:F|, and there are less than 2n-1 primes in ρ3, there must be at least one prime p4∈ρ4 that divides (qm-1)/(q-1).

As the graph Δ⁢(G) is connected, there exist primes p2∈ρ2 and p3∈ρ3 where p2 and p3 are adjacent. Then there is a character χ∈Irr⁡(G) such that p2⁢p3 divides χ⁢(1). Let θ be an irreducible constituent of χK. By [3, Corollary 11.29], p3 divides θ⁢(1) and so P′ is not contained in ker⁡θ. The size of the smaller component of Δ⁢(K) is either |ρ1∪ρ2|=n or |(ρ1∪ρ2)\{p2}|=n-1=a.

Because K is Example (2.6), K contains a Frobenius group where P is the Frobenius kernel and we let B be a Frobenius complement. The positive integer m1 is defined to be the index |K:R|. There is a p3-power q1 such that |P:P′|=q1m1 and (q1m1-1)/(q1-1) divides |B|. By [7, Lemma 3.6], the character degrees cd⁡(K∣P′) are all divisible by p⁢|B|. Because θ∈Irr⁡(K∣P′), the quotient (q1m1-1)/(q1-1) divides θ⁢(1).

If the smaller component of Δ⁢(K) is ρ1∪ρ2, then

Since the prime p4 divides (qm-1)/(q-1), the prime p4 divides θ⁢(1). If the smaller component of Δ⁢(K) is ρ1∪ρ2\{p2}, then let m2=|G:K|. Then

so q1=qm2. Let r be a Zsigmondy prime divisor of qm. Then r divides qm-1 and r does not divide qs-1 for any s<m. Because q1m1-1=qm-1, and m2<m, the prime r divides the quotient (q1m1-1)/(q1-1). Thus, as p4 is a Zsigmondy prime divisor of qm-1, the prime p4 divides (q1m1-1)/(q1-1) and so divides θ⁢(1).

As p4 divides θ⁢(1), the prime p4 also divides χ⁢(1). Then the primes p2 and p4 are adjacent, but this is impossible as p2∈ρ2 and p4∈ρ4. Thus, G has no normal Sylow p3-subgroup. ∎

Let G be a solvable group such that Δ⁢(G) has diameter three. Assume for all proper nontrivial normal subgroups N that Δ⁢(G/N) and Δ⁢(N) do not have diameter three. Further, assume G has no normal Sylow p4-subgroup for p4∈ρ4, Oρ4⁢(G)=G, |ρ1∪ρ2|=n, and |ρ3|<2n-1. Then ρ⁢(G/M)=ρ⁢(G) implies M=1, whenever M is a normal subgroup of G.

Proof.

Suppose G is a solvable group where Δ⁢(G) has diameter three, G has no normal Sylow p4-subgroups for any prime p4∈ρ4, the subgroup Oρ4⁢(G)=G, |ρ1∪ρ2|=n, and |ρ3∪ρ4|<2n-1. Assume that if N is a proper nontrivial normal subgroup of G that the graphs Δ⁢(N) and Δ⁢(G/N) do not have diameter three. Let M be a nontrivial normal subgroup of G, and suppose that ρ⁢(G/M)=ρ⁢(G). Then Δ⁢(G/M) does not have diameter three, and since Δ⁢(G/M) is a subgraph of Δ⁢(G), the graph Δ⁢(G/M) must be disconnected with components ρ1∪ρ2 and ρ3∪ρ4. The group G/M must be one of the examples from Theorem 2.1. Because both components have more than one vertex, we must be in Example (2.4) or Example (2.6). If G/M is Example (2.4), then it has no normal Sylow p-subgroups for any prime p∈ρ⁢(G/M) and in particular, G has no normal non-abelian Sylow p-subgroups either as ρ⁢(G/M)=ρ⁢(G). By [7, Theorem 5.3], we have Op2⁢(G/M)<G/M. Let H/M=Op2⁢(G/M). Then since |G/M:H/M| is a nontrivial power of p2, the index |G:H| is a nontrivial power of the prime p2. So Op2⁢(G)<G. This contradicts Lemma 4.1, hence G/M is Example (2.6). By [7, Lemma 3.6], the group G/M has a normal Sylow p-subgroup for some prime p∈ρ3∪ρ4. Let Q/M be that Sylow p-subgroup of G/M and P a Sylow p-subgroup of G that is contained in Q.

Consider K=Op2⁢(G) for a fixed prime p2∈ρ2. Because Op2⁢(G/M)<G/M by [7, Theorem 5.3], we have K is a proper subgroup of G, and so Δ⁢(K) must be disconnected. Also, notice P is contained in K.

Since K is Example (2.6), K has a normal non-abelian Sylow r-subgroup R for some prime r∈ρ3∪ρ4. Because R is a characteristic subgroup of K, R is normal in G. Further, as |G:K| is a power of p2, R is a Sylow subgroup of G. Because G/M has a normal non-abelian Sylow p-subgroup, and R⁢M/M is a normal non-abelian Sylow r-subgroup of G/M, p=r and R=P. As G has no normal non-abelian Sylow p-subgroups for p∈ρ4, the prime p must be in ρ3. But this contradicts Lemma 4.5. Thus ρ⁢(G/M) does not equal ρ⁢(G). ∎

Let G be a solvable group with Δ⁢(G) having diameter three. Assume for all proper nontrivial normal subgroups M that Δ⁢(G/M) and Δ⁢(M) do not have diameter three. Further, assume ρ⁢(G/M)=ρ⁢(G) implies M=1. Then G does not have a normal Sylow p2-subgroup for any prime p2∈ρ2.

Proof.

Suppose G has a normal Sylow p2-subgroup P for a fixed prime p2∈ρ2 and let H be a p2-complement of G. By [9, Lemma 3], ρ⁢(G/P′)=ρ⁢(G)\{p2}. Because ρ⁢(G/P′) contains the primes in ρ1 and ρ4, and Δ⁢(G/P′) is a subgraph of Δ⁢(G), the graph Δ⁢(G/P′) is disconnected. Because P is not central in G and H acts on P nontrivially, we see that P/P′ is not central in G/P′. From [7, Theorem 5.5], the Fitting subgroup of G/P′ has at most one non-central Sylow subgroup, which is P/P′. Let F=𝔽⁡(G). Since all of the other Sylow subgroups of 𝔽⁡(G/P′)=F/P′ are central, F/P′ is abelian, and thus G/P′ is as described in Example (2.4) from Theorem 2.1. This is because any solvable group having an abelian Fitting subgroup whose graph has two connected components, where at least one connected component has size larger than one, must satisfy the hypotheses of Example (2.4). Further, G/P′ has Fitting height 3.