The commuting graph of a solvable 𝐴-group

Let 𝐺 be a solvable 𝐴-group such that G / Z ⁢ ( G ) is neither a Frobenius nor 2-Frobenius group. Then the diameter of the commuting graph of 𝐺 is at most 6.

Lemma 2.1 ([13, Theorem 4.1])

Let 𝐺 be an 𝐴-group. Then G ′ ∩ Z ⁢ ( G ) = 1 .

Lemma 2.2 ([13, Corollary 4.5])

Let G = G ( 0 ) ⁢ ⊵ ⁢ G ′ = G ( 1 ) ⁢ ⊵ ⁢ ⋯ ⁢ ⊵ ⁢ G ( n ) = 1 be the derived series of 𝐺. Let M G ⁢ ( G ( i ) ) be a relative system normalizer for G ( i ) in 𝐺. Then G = M G ⁢ ( G ( i − 1 ) ) ⁢ G ( i ) , and M G ⁢ ( G ( i − 1 ) ) is a complement for G ( i ) in 𝐺. In particular, if 𝑀 is any absolute system normalizer of 𝐺, then 𝑀 is a complement of G ′ in 𝐺.

Lemma 2.3 ([13, Theorem 5.4])

If 𝐺 is a solvable 𝐴-group with derived length 𝑛, then F ⁢ ( G ) = Z ⁢ ( G ) × Z ⁢ ( G ′ ) × ⋯ × Z ⁢ ( G ( n − 1 ) ) .

Lemma 2.4 ([9, Theorem 6.4])

Let 𝑁 be a normal subgroup of a finite group 𝐺, and suppose that 𝐴 is a complement for 𝑁 in 𝐺. The following are equivalent.

1. 𝐺 is a Frobenius group with kernel 𝑁.

2. C G ⁢ ( a ) ≤ A for all nontrivial a ∈ A .

3. C G ⁢ ( n ) ≤ N for all nontrivial n ∈ N .

Lemma 2.5 ([1, Theorem 36.2])

Let 𝑝, 𝑞, and 𝑟 be distinct primes, 𝑋 a group of order 𝑟 that acts faithfully on a 𝑞-group 𝑄 and let 𝑉 be a faithful GF ⁢ ( p ) ⁢ X ⁢ Q -module. Additionally if q = 2 and 𝑟 is a Fermat prime, assume 𝑄 is abelian. Then C V ⁢ ( X ) ≠ 0 .

Let 𝐺 be a solvable, non-abelian 𝐴-group such that G / Z ⁢ ( G ) is neither a Frobenius nor 2-Frobenius group.

Let 𝐺 be a solvable 𝐴-group of derived length 2 with trivial center such that G ′ is not a Hall subgroup. Let 𝑔 be an element of 𝐺 and 𝑝 a prime dividing the order of 𝑔. If the 𝑝-part of 𝑔 does not lie in G ′ and does not lie in any absolute system normalizer of 𝐺, then C G ′ ⁢ ( g ) ≠ 1 and there exists an absolute system normalizer 𝑀 of 𝐺 such that C M ⁢ ( g ) ≠ 1 .

Proof

Let 𝐺 be a solvable 𝐴-group of derived length 2 with trivial center such that G ′ is not a Hall subgroup. From Lemma 2.2, we have that G = M ⁢ G ′ , where 𝑀 is an absolute system normalizer of 𝐺 and M ∩ G ′ = 1 . Since Z ⁢ ( G ) = 1 , note that F ⁢ ( G ) = G ′ by Lemma 2.3. Let 𝑔 be an element of 𝐺 such that the 𝑝-part of 𝑔 does not lie in G ′ for some prime 𝑝. Denote the 𝑝-part of 𝑔 by g p . If g p does not lie in any absolute system normalizer of 𝑀, then we have that 𝑝 divides both | G ′ | and | M | . Let 𝑃 be a Sylow 𝑝-subgroup of 𝐺 containing g p . We have that P = M p × O p ⁢ ( G ) , where M p is the Sylow 𝑝-subgroup of an appropriately chosen conjugate of 𝑀. Since g p ∉ M p , we have that g p = m p ⁢ f p , with m p ∈ M p and f p ∈ O p ⁢ ( G ) . Note that, because m p and f p lie in the same Sylow 𝑝-subgroup of 𝐺, they will commute. We will show that g ∼ m p and g ∼ f p .

If g = g p , then 𝑔 is a 𝑝-element. Furthermore, 𝑔 lies in the same Sylow subgroup as m p and f p , so we have g ∼ m p and g ∼ f p . So suppose that the order of 𝑔 is divisible by at least 2 distinct primes. We can write g = g p ⁢ g q 1 ⁢ g q 2 ⁢ ⋯ ⁢ g q r , where g q i represents the q i -part of 𝑔 for distinct primes q 1 , q 2 , … , q r . To show that g ∼ m p and g ∼ f p , we will show that g q i ∼ m p and g q i ∼ f p for each 𝑖.

We can write g q i = m q i ⁢ f q i , where m q i lies in a conjugate of 𝑀 and f q i ∈ G ′ . If m q i = 1 , then g q i = f q i ∈ G ′ . As G ′ is abelian, we have f p ∼ f q i . Therefore, since g p ∼ g q i = f q i , we get that f q i ⁢ m p ⁢ f p = m p ⁢ f p ⁢ f q i = m p ⁢ f q i ⁢ f p . Hence, we have f q i ∼ m p , and we are done. Therefore, suppose that m q i ≠ 1 . We then get the following:

Note that we have not assumed that m p and m q i lie in the same conjugate of 𝑀, so we must show that these elements commute. To do so, let 𝐾 be a Hall { p , q i } -subgroup of 𝐺 containing g p ⁢ g q i . Let K 0 be the intersection of 𝐾 with an appropriately chosen conjugate of 𝑀 and let K 1 = K ∩ G ′ . Let M p and M q i be the Sylow 𝑝- and q i -subgroups of K 0 respectively, and let F p and F q i be the Sylow 𝑝- and q i -subgroups of 𝐾 respectively. Note that F p = O p ⁢ ( G ) and F q i = O q i ⁢ ( G ) . We will show that the number of conjugates of K 0 in 𝐾 is the number of conjugates of M p times the number of conjugates of M q i .

First, we will show that N K ⁢ ( M p ) = N F q i ⁢ ( M p ) ⁢ F p ⁢ K 0 . Note that, as K 0 is abelian, it will centralize M p . Similarly, as the elements of F p centralize every 𝑝-element in 𝐺, we have that F p centralizes M p . Thus, N F q i ⁢ ( M p ) ⁢ F p ⁢ K 0 ≤ N K ⁢ ( M p ) . Thus, we need to show the reverse containment. Let x ∈ N K ⁢ ( M p ) . We can write x = x 1 ⁢ x 0 for some x 1 ∈ K 1 , x 0 ∈ K 0 . Furthermore, note that x 1 = k q i ⁢ k p , where k q i ∈ F q i , k p ∈ F p . Then we have that

Hence, k q i ∈ N F q i ⁢ ( M p ) . Therefore, x = k q i ⁢ k p ⁢ x 0 ∈ N F q i ⁢ ( M p ) ⁢ F p ⁢ K 0 . Through a similar argument, we can show that

Next, we want to show that N K 1 ⁢ ( K 0 ) = N K 1 ⁢ ( M p ) ∩ N K 1 ⁢ ( M q i ) . To do this, first suppose x ∈ N K 1 ⁢ ( K 0 ) . Then we have that x ⁢ M p ⁢ x − 1 ⊆ K 0 , so 𝑥 will send M p to one of its conjugates in K 0 . However, as K 0 is abelian (as it is a subgroup of a conjugate of 𝑀, which is abelian), we have that M p is the only Sylow 𝑝-subgroup of K 0 . Hence, x ⁢ M p ⁢ x − 1 = M p . Similarly, we have that x ⁢ M q i ⁢ x − 1 = M q i . Therefore,

Now, suppose x ∈ N K 1 ⁢ ( M p ) ∩ N K 1 ⁢ ( M q i ) . Then we have that x ⁢ M p ⁢ x − 1 = M p and x ⁢ M q i ⁢ x − 1 = M q i . Thus, we have that

Hence, x ∈ N K 1 ⁢ ( K 0 ) . Thus, we have that

Using this, we see that

Thus, the number of conjugates of K 0 in 𝐾 is exactly the number of conjugates of M p times the number of conjugates of M q i . This means that each conjugate of M p commutes with each conjugate of M q i to form all the distinct conjugates of K 0 , meaning that each conjugate of M p centralizes each conjugate of M q i . Thus, we have that m p ∼ m q i .

This gives us that m q i − 1 ⁢ m p ∼ f q i ⁢ f p . Therefore, ( m q i − 1 ⁢ m p ) n 1 ∼ ( f q i ⁢ f p ) n 2 for any n 1 , n 2 ∈ Z . Therefore, we have m q i ∼ f p and f q i ∼ m p . Therefore, g q i ∼ m p and g q i ∼ f p . Since this holds for all 𝑖, we therefore have that g ∼ m p and g ∼ f p . Hence, C M ⁢ ( g ) ≠ 1 and C G ′ ⁢ ( g ) ≠ 1 . ∎

Assume Hypothesis 2.6 and suppose that Z ⁢ ( G ) = 1 . Furthermore, suppose that the derived length of 𝐺 is 2. Then Γ ⁢ ( G ) is connected of diameter at most 4.

Proof

Let the derived length of 𝐺 be 2. Thus, 𝐺 is metabelian and G ′ is abelian. From Lemma 2.2, we have that G = M ⁢ G ′ , where 𝑀 is an absolute system normalizer of 𝐺, and M ∩ G ′ = 1 . As M ≅ G / G ′ , we have that 𝑀 is abelian. Additionally, since Z ⁢ ( G ) = 1 , we have that F ⁢ ( G ) = G ′ by Lemma 2.3.

Since 𝐺 is not a Frobenius group, there exists x * ∈ M # such that C G ′ ⁢ ( x * ) ≠ 1 . Consider ⟨ x * ⟩ ⁢ G ′ . Let g ∈ G # , and note that we can write g = m ⁢ g ′ , where m ∈ M and g ∈ G ′ . We have that g ⁢ ⟨ x * ⟩ ⁢ G ′ ⁢ g − 1 = ( m ⁢ g ′ ) ⁢ ⟨ x * ⟩ ⁢ G ′ ⁢ ( m ⁢ g ′ ) − 1 = m ⁢ ⟨ x * ⟩ ⁢ G ′ ⁢ m − 1 as g ′ ∈ G ′ ≤ ⟨ x * ⟩ ⁢ G ′ . Furthermore, as m ∈ M and G ′ ⁢ ⊴ ⁢ G , we have that

Therefore, ⟨ x * ⟩ ⁢ G ′ ⁢ ⊴ ⁢ G . Hence, as 𝑀 intersects ⟨ x * ⟩ ⁢ G ′ nontrivially, we have that every conjugate of 𝑀 intersects ⟨ x * ⟩ ⁢ G ′ nontrivially. Since C G ′ ⁢ ( x * ) ≠ 1 , we have that Z ⁢ ( ⟨ x * ⟩ ⁢ G ′ ) ≠ 1 . So fix c * ∈ Z ⁢ ( ⟨ x * ⟩ ⁢ G ′ ) # . We will show that d ⁢ ( g , c * ) ≤ 2 for all g ∈ G # .

Let g ∈ G # . If there exists a positive integer n < o ⁢ ( g ) such that g n ∈ G ′ , we have that g ∼ g n ∼ c * . Hence, d ⁢ ( g , c * ) ≤ 2 . Suppose that there does not exist such an integer 𝑛. In particular, this gives us that g ∉ G ′ . If 𝑔 lies in some conjugate of 𝑀, we have that g ∼ x ∼ c * , where 𝑥 lies in the intersection of ⟨ x * ⟩ ⁢ G ′ with an appropriately chosen conjugate of 𝑀. Hence, d ⁢ ( g , c * ) ≤ 2 . That leaves the case where 𝑔 is neither an element of G ′ nor of a conjugate of 𝑀. Hence, there exists a prime 𝑝 such that the 𝑝-part of 𝑔 is neither an element of G ′ nor of a conjugate of 𝑀. Note that, in this case, G ′ is not a Hall subgroup of 𝐺. Therefore, 𝑔 satisfies the hypothesis of Lemma 3.1, and so C G ′ ⁢ ( g ) ≠ 1 . Hence, we have that g ∼ f ∼ c * , where f ∈ C G ′ ⁢ ( g ) # . Therefore, d ⁢ ( g , c * ) ≤ 2 . Thus, we have that, for any g ∈ G , d ⁢ ( g , c * ) ≤ 2 . This means that the diameter of the commuting graph of 𝐺 will be at most 4. ∎

Let 𝐺 satisfy Hypothesis 2.6, and assume that 𝐺 has derived length 2. Then Γ ⁢ ( G ) is connected with diameter at most 4.

Proof

Let 𝐺 be a solvable 𝐴-group satisfying Hypothesis 2.6 and let Z = Z ⁢ ( G ) . We then have that G / Z is neither a Frobenius nor a 2-Frobenius group. Furthermore, the commuting graphs of 𝐺 and G / Z have the same diameter. Recall that Lemma 2.1 gives us that G ′ ∩ Z = 1 . Thus, we have that ( G / Z ) ′ = G ′ ⁢ Z / Z ≅ G ′ . Therefore, as 𝐺 is metabelian and thus G ′ abelian, we get that ( G / Z ) ′ is abelian. Therefore, the derived length of G / Z is 2. From the previous lemma, we get that the diameter of Γ ⁢ ( G / Z ) is at most 4. Hence, the diameter of Γ ⁢ ( G ) is at most 4. ∎

Suppose that 𝐺 satisfies Hypothesis 2.6 and that the order of 𝐺 is p a ⁢ q b for distinct primes 𝑝 and 𝑞. Then Γ ⁢ ( G ) is connected of diameter at most 4.

Proof

By [13, Theorem 8.3], we have that the derived length is at most 2. As 𝐺 is non-abelian, we have that the derived length is exactly 2. Thus, by Lemma 3.3, we have that Γ ⁢ ( G ) is connected of diameter at most 4. ∎

Suppose that 𝐺 is a group of cube-free odd order such that G / Z ⁢ ( G ) is neither a Frobenius nor 2-Frobenius group. Then Γ ⁢ ( G ) is connected of diameter at most 4.

Proof

Suppose that 𝐺 is a group of cube-free order and that the order of 𝐺 is odd. Note that as 𝐺 has cube-free order, 𝐺 is an 𝐴-group. By [13, Theorem 8.4], the derived length of 𝐺 is 2. Hence, by Lemma 3.3, we have that Γ ⁢ ( G ) is connected of diameter at most 4. ∎

Lemma 4.1 ([11, Lemma 3.4])

Let 𝐺 be a solvable group with trivial center and let 𝑉 be a minimal normal subgroup of 𝐺. If the diameter of Γ ⁢ ( G ) is at least 7, then F = C G ⁢ ( V ) .

Lemma 4.2 ([11, Lemma 3.5])

Let 𝐺 be any solvable group with trivial center. If the diameter of Γ ⁢ ( G ) is at least 7, then | J / F | is coprime to | F | .

Suppose 𝐺 satisfies Hypotheses 2.6 and that Z ⁢ ( G ) = 1 . Assume that the diameter of Γ ⁢ ( G ) ≥ 7 and that 𝑉 is a minimal normal subgroup of 𝐺. Let g ∈ G and n ∈ Z such that o ⁢ ( g n ) = p for some prime 𝑝. If C V ⁢ ( g n ) = 1 , then g n ∈ J .

Proof

Let g ∈ G such that o ⁢ ( g n ) = p for some prime 𝑝 and that C V ⁢ ( g n ) = 1 . Suppose g n ∉ J . As G / F is solvable and J / F = F ⁢ ( G / F ) , we have that there exists a prime 𝑡 such that the Sylow 𝑡-subgroup of J / F is not centralized by g n . Since 𝐺 is an 𝐴-group (and therefore G / F is also an 𝐴-group), we have that p ≠ t . Choose T ∈ Syl t ⁢ ( J ) to be ⟨ g n ⟩ invariant. Note that T ⁢ ⟨ g n ⟩ acts faithfully on 𝑉 by Lemma 4.1. However, as C V ⁢ ( g n ) = 1 , Lemma 2.5 gives us a contradiction. Therefore, g n ∈ J . ∎

Let 𝐺 satisfy Hypotheses 2.6 and suppose that Z ⁢ ( G ) = 1 . Then Γ ⁢ ( G ) is connected of diameter at most 6.

Proof

Suppose that 𝐺 satisfies Hypotheses 2.6 and that Z ⁢ ( G ) = 1 . Since 𝐺 is neither Frobenius nor 2-Frobenius, we know that Γ ⁢ ( G ) is connected by the main result of [11]. Therefore, we need only prove that the upper bound of the diameter of the commuting graph is 6. We proceed by contradiction and assume that Γ ⁢ ( G ) has diameter at least 7. If we can show that every element in 𝐺 has a distance of at most 3 from some specific element c * ∈ F # , then we will get that the diameter of Γ ⁢ ( G ) is at most 6, giving us our desired contradiction.

First, assume that 𝐽 is not a Frobenius group. From Lemma 4.2, we have that | J / F | and | F | are coprime. Thus, 𝐹 is a normal Hall subgroup of 𝐽 and so has a complement in 𝐽. Let 𝐾 be a complement to 𝐹 in 𝐽. Note that, as K ≅ J / F , we have that 𝐾 is abelian. Since 𝐽 is not Frobenius, we have that there exists k * ∈ K # such that C * = C F ⁢ ( k * ) ≠ 1 . Consider N G ⁢ ( C * ) . As 𝐹 is abelian and C * ≤ F we have that 𝐹 normalizes C * . Since F ⁢ ⊴ ⁢ G , we have that x ⁢ c ⁢ x − 1 ∈ F for all c ∈ F , x ∈ K . Therefore, since ( x ⁢ c ⁢ x − 1 ) ⁢ k * = x ⁢ c ⁢ k * ⁢ x − 1 = x ⁢ k * ⁢ c ⁢ x − 1 = k * ⁢ ( x ⁢ c ⁢ x − 1 ) for all x ∈ K , c ∈ C * , we have that x ⁢ c ⁢ x − 1 ∈ C * , and so 𝐾 normalizes C * . Thus, N G ⁢ ( C * ) ≥ K ⁢ F = J , and so 𝐽 normalizes C * . Hence, every conjugate of 𝐾 lies in N G ⁢ ( C * ) . In particular, as 𝐾 intersects C G ⁢ ( C * ) nontrivially, every conjugate of 𝐾 intersects C G ⁢ ( C * ) nontrivially.

Fix c * ∈ ( C * ) # . Let g ∈ G . If we can find a positive integer n < o ⁢ ( g ) for which C F ⁢ ( g n ) ≠ 1 , then we have that

for some f ∈ C F ⁢ ( g n ) # . Thus, we must assume that C F ⁢ ( g n ) = 1 for every positive integer n < o ⁢ ( g ) . In particular, when g n has prime order, we have that g n ∈ J by Lemma 4.3. As C F ⁢ ( g n ) = 1 , we have that g n has order coprime to the order of 𝐹. Therefore, g n lies is some Hall π ⁢ ( J / F ) -subgroup of 𝐽, which is contained in some conjugate of 𝐾. Without loss of generality, we may assume g n lies in 𝐾. Therefore, g n ∼ k * and we have that

Hence, in either case, we get that d ⁢ ( g , c * ) ≤ 3 . Therefore, if 𝐽 is not a Frobenius group, we get that the diameter of the commuting graph is at most 6, which contradicts our initial assumption that the diameter is at least 7.

Therefore, we must have that 𝐽 is a Frobenius group with complement 𝐾. Since J / F ≅ K , we get that J / F is cyclic. Thus, Aut ⁢ ( J / F ) is abelian, and we have that

is abelian. Since G / J is abelian, we have that the derived length of 𝐺 is at most 3. Since 𝐺 is not abelian, the derived length is at least 2. However, if the derived length is 2, then we have already shown in Lemma 3.2 that the diameter of Γ ⁢ ( G ) will be at most 4, which is a contradiction. So that gives us that the derived length of 𝐺 is 3, and we can write 𝐺 as G = H ⁢ G ′′ , where H = M G ⁢ ( G ′ ) is a relative system normalizer for G ′ in 𝐺. Note that H ∩ G ′ = M 1 is an absolute system normalizer of G ′ . Furthermore, for the appropriate choice of Sylow system of G ′ , we have that an absolute system normalizer M 0 of 𝐺 will be contained in 𝐻. As | H | = | M 0 | ⋅ | M 1 | , we can write 𝐻 as H = M 0 ⁢ M 1 . Therefore, we get G = M 0 ⁢ M 1 ⁢ G ′′ . Note that F = Z ⁢ ( G ′ ) × G ′′ by Lemma 2.3. As 𝐽 is a Frobenius group and G ′ ≤ J , we have that G ′ is Frobenius and therefore Z ⁢ ( G ′ ) = 1 . Hence, F = G ′′ .

Note that, as 𝐺 is not a Frobenius group and every element of M 1 # acts fixed-point freely on 𝐹, we have that there exists x * ∈ M 0 # such that C F ⁢ ( x * ) ≠ 1 . Consider ⟨ x * ⟩ ⁢ F . Note that there are

conjugates of ⟨ x * ⟩ ⁢ F in 𝐺 and that there are | H : N H ( ⟨ x * ⟩ ) | = | M 1 : N M 1 ( ⟨ x * ⟩ ) | conjugates of ⟨ x * ⟩ in 𝐻. Therefore, each conjugate of ⟨ x * ⟩ in 𝐻 intersects a distinct conjugate of ⟨ x * ⟩ ⁢ F in 𝐺. Furthermore, every conjugate of 𝐻 will intersect each conjugate of ⟨ x * ⟩ ⁢ F nontrivially. We have that Z ⁢ ( ⟨ x * ⟩ ⁢ F ) ≠ 1 . So choose x * ∈ M 0 # to be as above and fix c * ∈ Z ⁢ ( ⟨ x * ⟩ ⁢ F ) # .

Let g ∈ G . As in the previous case, if we can find an integer n < o ⁢ ( g ) for which C F ⁢ ( g n ) ≠ 1 , then g ∼ g n ∼ f ∼ c * , where f ∈ C F ⁢ ( g n ) # , and d ⁢ ( g , c * ) ≤ 3 . Hence, we must have that C F ⁢ ( g n ) = 1 for all n < o ⁢ ( g ) . In this case, the order of 𝑔 must be coprime to the order of the Fitting subgroup. Hence, 𝑔 lies in some conjugate of H; without loss of generality, we can say that g ∈ H .

First, suppose that J 0 = J ∩ M 0 ≠ 1 . Therefore, we have G ′ ⪇ J . Note that, as J / F = F ⁢ ( G / F ) , we have from Lemma 2.3 that

Therefore, Z ⁢ ( G / F ) ≠ 1 . Note that as H ≅ G / F , we therefore have Z ⁢ ( H ) ≠ 1 . Let z ∈ Z ⁢ ( H ) # . Therefore, we have that

Hence, d ⁢ ( g , c * ) ≤ 3 . In both cases, we get that d ⁢ ( g , c * ) ≤ 3 , and so the diameter of Γ ⁢ ( G ) is at most 6, a contradiction.

That leaves the case when J 0 = 1 and so J = G ′ . However, from Lemma 2.3, we have that

which means that Z ⁢ ( G / F ) = 1 . Hence, we have that Z ⁢ ( H ) = 1 . Suppose that x ∈ M 0 has prime order. By Lemma 4.3, we have that if C F ⁢ ( x ) = 1 , then x ∈ J . However, M 0 ∩ J = 1 , which gives us a contradiction. Thus, every element of prime order in M 0 centralizes some element in F # .

Therefore, we can choose any such element to act as x * as in the previous case. To ensure shortest path between M 0 and M 1 , we will therefore choose x * ∈ M 0 so that C M 1 ⁢ ( x * ) ≠ 1 . As before, note that each conjugate of ⟨ x * ⟩ in 𝐻 corresponds to a unique conjugate of ⟨ x * ⟩ ⁢ F in 𝐺. Fix c * ∈ Z ⁢ ( ⟨ x * ⟩ ⁢ F ) # .

By Lemma 4.3, as C F ⁢ ( g n ) = 1 for all n < o ⁢ ( g ) , we have that g n ∈ J = G ′ whenever g n has prime order. More specifically, we get that g n lies in some conjugate of M 1 . Without loss of generality, g n ∈ M 1 . Since this holds for all integers 𝑛 for which o ⁢ ( g n ) is prime, we get that every prime dividing o ⁢ ( g n ) also divides | M 1 | . Hence, 𝑔 is a π ⁢ ( M 1 ) -element, and so g ∈ H . If g ∈ M 1 , we have that

where y ∈ C M 1 ⁢ ( x * ) # . Thus, d ⁢ ( g , c * ) ≤ 3 . Thus, we must have that 𝑔 is a π ⁢ ( M 1 ) -element that does not lie in M 1 . Thus, there exists a prime 𝑝 dividing the order of 𝑔 such that the 𝑝-part of 𝑔 does not lie in M 1 . In particular, this gives us that M 1 is not a Hall subgroup of 𝐻. Note that the absolute system normalizers of 𝐻 are the conjugates of M 0 contained in 𝐻. Thus, we have that 𝐻 is a solvable 𝐴-group of derived length 2 and that M 1 = H ′ is not a Hall subgroup of 𝐻; furthermore, we have that 𝑔 is an element of 𝐻 such that the 𝑝-part of 𝑔 does not lie in M 1 = H ′ or any absolute system normalizer of 𝐻. Hence, by Lemma 3.1, we have that C M 0 ⁢ ( g ) ≠ 1 . Thus, we can find an element 𝑥 of prime order in C M 0 ⁢ ( g ) # so that

where f ∈ C F ⁢ ( x ) # . Therefore, we have d ⁢ ( g , c * ) ≤ 3 . This means that, in all cases, d ⁢ ( g , c * ) ≤ 3 . Hence, we have that the diameter of the commuting graph is at most 6, contradicting our original hypothesis that the diameter was at least 7. Therefore, we have that the diameter of the commuting graph will be at most 6. ∎

Proof of Theorem 1.1

Let 𝐺 be a solvable 𝐴-group such that G / Z is neither a Frobenius nor 2-Frobenius group. Since G / Z is neither a Frobenius nor 2-Frobenius group, we have that Γ ⁢ ( G / Z ) is connected. Theorem 4.4 gives us that the diameter of Γ ⁢ ( G / Z ) is at most 6. From [2], we have that the diameter of Γ ⁢ ( G ) is the same as the diameter of Γ ⁢ ( G / Z ) . Thus, Γ ⁢ ( G ) is connected with diameter at most 6. ∎