Groups whose vanishing class sizes are prime powers

Let 𝐺 be a finite group. If each vanishing class size of 𝐺 is a prime power, then one of the following situations occurs.

1. G / F ⁢ ( G ) is a nilpotent group with at most one non-abelian Sylow subgroup and | ρ ⁢ ( cs ⁡ ( G / F ⁢ ( G ) ) ) | ≤ 1 .

2. G / F ⁢ ( G ) is an 𝐴-group and there exist two distinct primes 𝑟 and 𝑠 such that G / F ⁢ ( G ) ≅ K × H , where 𝐾 is an abelian { r , s } ′ -group and H / Z ⁢ ( H ) is a Frobenius { r , s } -group. Moreover, | ρ ⁢ ( cs ⁡ ( G / F ⁢ ( G ) ) ) | = dl ⁡ ( G / F ⁢ ( G ) ) = 2 .

Each non-vanishing element of a finite solvable group 𝐺 lies in the Fitting subgroup F ⁢ ( G ) .

Let 𝐺 be a finite group whose vanishing class sizes are prime powers. If 𝑥 is a non-vanishing element, then x ∈ F ⁢ ( G ) .

Theorem 2.1 ([15, Theorem 3.9])

Let 𝐺 be a finite group and suppose that there exists x ∈ G of prime power index. Then 𝐺 is not simple.

Let 𝐺 be a non-abelian finite simple group. Then 𝐺 has some conjugacy class of size divisible by 2 ⁢ q for some odd prime number 𝑞.

Proof

As 𝐺 is non-abelian, the celebrated theorem of Feit and Thompson verifies that 𝐺 has even order. Now [14, Theorem 33.4] forces 𝐺 to have an element 𝑥 whose conjugacy class is of even size. At the same time, Theorem 2.1 indicates that | x G | is not a 2-power, so it is of size divisible by 2 ⁢ q for some odd prime number 𝑞. ∎

Lemma 2.3 ([6, Lemma 2.2])

Let 𝐺 be a finite group and 𝑁 a normal subgroup of 𝐺. If 𝑁 has an irreducible character of 𝑝-defect zero , then every element of 𝑁 of order divisible by 𝑝 is a vanishing element in 𝐺.

Theorem 2.4 ([13, Corollary 2])

Let 𝑆 be a non-abelian finite simple group, and assume there exists a prime 𝑝 such that 𝑆 does not have an irreducible character of 𝑝-defect zero. Then one of the following holds.

1. The prime 𝑝 is 2, and 𝑆 is isomorphic to either M 12 , M 22 , M 24 , J 2 , HS , Suz , Ru , Co 1 , Co 3 , BM , or Alt ⁢ ( n ) for some n ≥ 7 .

2. The prime 𝑝 is 3, and 𝑆 is isomorphic to either Suz , Co 3 , or Alt ⁢ ( n ) for some n ≥ 7 .

Let 𝑆 be a sporadic simple group or an alternating group Alt ⁢ ( n ) for some n ≥ 7 . Then 𝑆 has an irreducible character 𝜃 which extends to Aut ⁢ ( S ) and a conjugacy class x S of size divisible by 2 ⁢ q for some odd prime number 𝑞 such that θ ⁢ ( x ) = 0 .

Proof

Using [9], it is straightforward to prove the claim whenever 𝑆 is a sporadic simple group. Thus, we may assume that 𝑆 is an alternating group Alt(𝑛) for n ≥ 7 . As a consequence of [15, Corollary 5.17], θ = χ - 1 G , where 𝜒 is the permutation character, is an irreducible character of Alt ⁢ ( n ) which extends to Sym ⁢ ( n ) . Let x = ( 1 , … , n - 1 ) ⁢ ( n ) if 𝑛 is even and x = ( 1 , … , n - 3 ) ⁢ ( n - 2 , n - 1 ) ⁢ ( n ) if 𝑛 is odd. We can observe that the conjugacy class x Alt ⁢ ( n ) is of size divisible by 2 ⁢ q for some odd prime number 𝑞; furthermore, θ ⁢ ( x ) = 0 . ∎

Lemma 2.6 ([3, Lemma 5])

Let 𝐺 be a finite group and M = S 1 × ⋯ × S k a minimal normal subgroup of 𝐺, where every S i is isomorphic to a non-abelian simple group 𝑆. If θ ∈ Irr ⁡ ( S ) extends to Aut ⁢ ( S ) , then θ × ⋯ × θ ∈ Irr ⁡ ( M ) extends to 𝐺.

Let 𝐺 be a finite group. If all vanishing class sizes of 𝐺 are either odd or powers of 2, then 𝐺 is solvable.

Proof

We progress by induction on | G | .

Let 𝑀 be a minimal normal subgroup of 𝐺. If 𝑀 is non-abelian, then we have M = S 1 × ⋯ × S n , in which, for each i ∈ { 1 , … , n } , S i is isomorphic to a non-abelian simple group 𝑆. If, for each prime p ∈ π ⁢ ( M ) , the simple group 𝑆 has an irreducible character 𝜃 of 𝑝-defect zero, we can observe that θ × ⋯ × θ is an irreducible character of 𝑝-defect zero of 𝑀. As Lemma 2.3 implies that every non-trivial element of 𝑀 is a vanishing element of 𝐺, we conclude that the index of each element of 𝑀 in 𝐺 is either odd or a power of 2. However, Lemma 2.2 gives rise to the existence of some conjugacy class of 𝑆 of size divisible by 2 ⁢ q for some odd prime number 𝑞, which is a contradiction.

Hence, there exists a prime number p ∈ π ⁢ ( M ) for which 𝑆 does not have any irreducible character of 𝑝-defect zero. Now Proposition 2.5 and Theorem 2.4 imply that there exists θ ∈ Irr ⁡ ( S ) which extends to Aut ⁢ ( G ) , an element x ∈ S with | x S | = 2 n ⁢ m for some odd number 𝑚 and a positive integer 𝑛 such that 𝜃 vanishes on x S . Therefore, by Proposition 2.6, θ × ⋯ × θ extends to an irreducible character of 𝐺 which vanishes on x G . As | x G | is divisible by | x S | , we obtain a contradiction.

Therefore, 𝑀 must be abelian. Since G / M is solvable, by the inductive hypothesis, we deduce that 𝐺 is solvable. ∎

Lemma 3.1 ([17, Theorem D])

Let 𝑥 be a non-vanishing element of the solvable group 𝐺. Then the image of 𝑥 in G / F ⁢ ( G ) has a 2-power order, and in particular, if 𝑥 has odd order, then x ∈ F ⁢ ( G ) . In any case, if 𝐺 is not nilpotent, then 𝑥 lies in the penultimate term of the ascending Fitting series.

Lemma 3.2 ([17, Theorem B])

If 𝐺 is supersolvable, then the non-vanishing elements of 𝐺 all lie in Z ⁢ ( F ⁢ ( G ) ) . In particular, if 𝐺 is nilpotent, then the non-vanishing elements of 𝐺 are central.

Lemma 3.3 ([12, Lemma 2.6])

Let 𝐺 be a solvable group and 𝑁 a normal subgroup of 𝐺. If N / F ⁢ ( N ) is abelian, then N ∖ F ⁢ ( N ) ⊆ Van ⁡ ( G ) .

Let 𝐺 be a solvable group. Denote the image of a subgroup 𝐴 of 𝐺 in G / F ⁢ ( G ) by A ¯ . If A ¯ is an abelian normal subgroup, then each non-trivial element of A ¯ is the image of a vanishing element of 𝐺 in G / F ⁢ ( G ) .

Proof

Since F ⁢ ( G ) ⊆ F ⁢ ( A ⁢ F ⁢ ( G ) ) which is a nilpotent characteristic subgroup of the normal subgroup A ⁢ F ⁢ ( G ) of 𝐺, we observe that F ⁢ ( A ⁢ F ⁢ ( G ) ) = F ⁢ ( G ) . As long as A ¯ is abelian, Lemma 3.3 shows that A ⁢ F ⁢ ( G ) ∖ F ⁢ ( G ) is a subset of Van ⁡ ( G ) , which is the desired result. ∎

Lemma 3.5 ([18, Corollary 2])

Let 𝑥 be an element of 𝐺 for which | x G | = p α . Then x ∈ O σ ⁢ ( G ) ∩ S ⁢ ( G ) , where σ = π ⁢ ( o ⁢ ( x ) ) ∪ { p } and S ⁢ ( G ) is the solvable radical of 𝐺 .

Lemma 3.6 ([2, Theorem 2])

Suppose that P ∈ Syl p ⁢ ( G ) is non-abelian and, for any x ∈ P - Z ⁢ ( P ) , the index of 𝑥 is a power of a prime. Then 𝑃 is a direct factor of 𝐺.

Lemma 3.7 ([2, Remark 3])

Suppose that P ∈ Syl p ⁢ ( G ) is not a direct factor of 𝐺 and, for all x ∈ P , the index of 𝑥 is a power of a prime. Then 𝐺 is 𝑝-solvable, 𝑃 is abelian, and there exists q ∈ π ⁢ ( G ) - { p } such that the index of 𝑥 is a power of 𝑞 for all x ∈ P .

Let G = R ⋊ H , where 𝑅 is a non-abelian Sylow 𝑟-subgroup and 𝐻 is an 𝑟-complement of 𝐺 with the property that C H ⁢ ( R ) < H . If θ ∈ Irr ⁡ ( R ) is nonlinear, then 𝜃 vanishes on some element of 𝑅 that is not conjugate to any element in C R ⁢ ( H ) .

Proof

Let θ ∈ Irr ⁡ ( R ) be nonlinear. The fact that 𝑅 is an 𝑀-group implies the existence of a proper subgroup 𝑋 in 𝑅 with a linear character λ ∈ Irr ⁡ ( X ) such that λ R = θ . Moreover, θ ⁢ ( x ) = 0 for each x ∈ R ∖ X R . Whenever 𝜃 does not vanish on any element of 𝑅 which is not conjugate to any element in C R ⁢ ( H ) , we observe that R = X R ∪ C R ⁢ ( H ) R which is impossible as both of these subgroups are proper. Hence the claim holds. ∎

Proof

Let G ¯ = G / F ⁢ ( G ) . As 𝐺 is solvable by Theorem 2.7, non-vanishing elements of 𝐺 in G ¯ have 2-power orders by Theorem 3.1. Moreover, each vanishing element of 𝐺 has a prime power index in G ¯ . Let S ¯ be a Sylow 2-subgroup of G ¯ and x ¯ = x ⁢ F ⁢ ( G ) . We shall break the proof into the following steps.

Let 𝑝 be an odd prime for which G ¯ has a non-abelian Sylow 𝑝-subgroup P ¯ . As the index of each non-central element of P ¯ is a prime power, Lemma 3.6 implies that P ¯ is a direct factor of G ¯ . Let 𝑦 be a non-vanishing element of 𝐺. The fact that, for 1 ¯ ≠ x ¯ ∈ Z ⁢ ( P ¯ ) ⊆ Z ⁢ ( G ¯ ) , the element x ¯ ⁢ y ¯ has order 2 n y ⁢ p m x for some non-negative integers n y and m x > 0 shows that x ⁢ y ∈ Van ⁡ ( G ) . As

we deduce that y ¯ has a prime power index in G ¯ . It follows that each element of G ¯ has a prime power index and hence, by [8, Theorem 2], the claim holds.

Let P ¯ be a Sylow 𝑝-subgroup of G ¯ . As P ¯ is not a direct factor of G ¯ and as each element of P ¯ has a prime power index in G ¯ , Lemma 3.7 implies that there exists q ∈ π ⁢ ( G ¯ ) ∖ { 2 , p } such that | x ¯ G ¯ | is a 𝑞-power for each x ¯ ∈ P ¯ . Let F ¯ = O { p , q } ⁢ ( G ¯ ) , and consider the group G ¯ / F ¯ . While x ¯ ∈ P ¯ , it is straightforward to see that x ¯ ⁢ F ¯ centralizes some Hall { p , q } ′ -subgroup of G ¯ / F ¯ . As

we conclude that x ¯ ⁢ F ¯ ∈ O { p , q } ′ ⁢ ( G ¯ / F ¯ ) , which is impossible unless x ¯ ∈ F ¯ . Therefore, P ¯ ≤ F ¯ . Let Q ¯ ∈ Syl q ⁢ ( G ¯ ) . If, for some (so for all) y ¯ ∈ Q ¯ - Z ⁢ ( G ¯ ) , 𝑝 does not divide the index of y ¯ in G ¯ , the fact that F ¯ ∩ Q ¯ is an abelian group with the property that each element has index not divisible by 𝑝 in F ¯ implies that F ¯ ∩ Q ¯ ≤ Z ⁢ ( F ¯ ) . It follows that F ¯ = P ¯ × ( F ¯ ∩ Q ¯ ) . As, for each y ¯ ∈ Q ¯ , p ∤ | y ¯ G ¯ | and P ¯ ⊲ G ¯ , we deduce that P ¯ ⁢ Q ¯ = P ¯ × Q ¯ , which is a contradiction. So we may assume that | y ¯ G ¯ | is a 𝑝-power for some (and so for all) y ¯ ∈ Q ¯ . As 𝑝 does not divide | G ¯ / F ¯ | , we conclude that Q ¯ ⁢ F ¯ / F ¯ ≤ Z ⁢ ( G ¯ / F ¯ ) , which implies that Q ¯ < F ¯ and so F ¯ = P ¯ ⁢ Q ¯ which is a Hall { p , q } -subgroup of G ¯ .

Moreover, let x ¯ be an element in P ¯ - Z ⁢ ( G ¯ ) and y ¯ an element in Q ¯ - Z ⁢ ( G ¯ ) such that x ¯ ⁢ y ¯ = y ¯ ⁢ x ¯ . We can note that C G ¯ ⁢ ( x ¯ ⁢ y ¯ ) = C G ¯ ⁢ ( x ¯ ) ∩ C G ¯ ⁢ ( y ¯ ) and so | ( x ¯ ⁢ y ¯ ) G ¯ | has two prime divisors 𝑝 and 𝑞, which is a contradiction. Therefore, no element in P ¯ - Z ⁢ ( G ¯ ) commutes with any element in Q ¯ - Z ⁢ ( G ¯ ) and so F ¯ / Z ⁢ ( F ¯ ) is a Frobenius group. On the other hand, as | a ¯ G ¯ | is a 𝑞-power for some a ¯ ∈ P ¯ , for each r ∈ π ⁢ ( G ¯ ) - { p , q } , there exists some Sylow 𝑟-subgroup R ¯ of G ¯ such that R ¯ ⊆ C G ¯ ⁢ ( a ¯ ) . As | b ¯ G ¯ | is a 𝑝-power for each b ¯ ∈ Q ¯ , we have G ¯ = C G ¯ ⁢ ( a ¯ ) ⁢ C G ¯ ⁢ ( b ¯ ) , which implies that R ¯ ⊆ C G ¯ ⁢ ( b ¯ ) for each b ¯ ∈ Q ¯ . Similarly, we can deduce that R ¯ ⊆ C G ¯ ⁢ ( a ¯ ) for each a ¯ ∈ P ¯ , and hence a Hall { p , q } ′ -subgroup of G ¯ is a subset of C G ¯ ⁢ ( F ¯ ) . It follows that F ¯ is a direct factor of G ¯ .

Let H ¯ be a complement of F ¯ in G ¯ . Given any non-central element of F ¯ and any element of H ¯ , say x ¯ and y ¯ , respectively, as ( x ¯ ⁢ y ¯ ) G ¯ = x ¯ F ¯ × y ¯ H ¯ , we conclude that x ⁢ y ∈ Van ⁡ ( G ) . By our hypothesis, this holds if and only if H ¯ is abelian, which completes the proof of this step. Indeed, we have part (2) of the theorem.

Let G ¯ = S ¯ ⁢ T ¯ , where S ¯ is a Sylow 2-subgroup of G ¯ and T ¯ is of odd order. Since the index of each 𝑝-element of G ¯ is a 2-power for each odd prime number 𝑝, it follows from Lemma 3.5 that P ¯ ⊆ O { 2 , p } ⁢ ( G ¯ ) for some Sylow 𝑝-subgroup P ¯ of G ¯ contained in T ¯ . Thus, we have that P ¯ = O { 2 , p } ⁢ ( G ¯ ) ∩ T ¯ is normal in T ¯ , which implies that each Sylow subgroup in T ¯ is normal. Moreover, since each Sylow subgroup of odd order in G ¯ is abelian, then T ¯ is abelian. We proceed by calculating the vanishing class sizes of G ¯ .

Assume that x ¯ is an element of odd order and y ¯ ∈ Z ⁢ ( S ¯ ) such that x ¯ ⁢ y ¯ = y ¯ ⁢ x ¯ ; then x ⁢ y ∈ Van ⁡ ( G ) . Since C G ¯ ⁢ ( x ¯ ⁢ y ¯ ) = C G ¯ ⁢ ( x ¯ ) ∩ C G ¯ ⁢ ( y ¯ ) , for some odd number 𝑚 and positive integer n x , we get that | ( x ¯ ⁢ y ¯ ) G ¯ | = | x ¯ G ¯ | ⁢ | y ¯ G ¯ | = 2 n x ⁢ m , which is a contradiction unless x ¯ ∈ Z ⁢ ( G ¯ ) or y ¯ ∈ Z ⁢ ( G ¯ ) . While x ¯ ∈ Z ⁢ ( G ¯ ) , following a proof similar to Step 1, we can observe that G ¯ = S ¯ × T ¯ . Otherwise, y ¯ ∈ Z ⁢ ( G ¯ ) ; henceforth, we may assume that Z ⁢ ( S ¯ ) = Z ⁢ ( G ¯ ) ⊆ O 2 ⁢ ( G ¯ ) . We proceed by considering the following conditions.

Let b ¯ ∈ T ¯ ∖ { 1 } (observe that b ¯ ∉ Z ⁢ ( G ¯ ) ); we claim that C G ¯ ⁢ ( b ¯ ) = C S ¯ ⁢ ( b ¯ ) × T ¯ . First, we clearly have C G ¯ ⁢ ( b ¯ ) = C S ¯ ⁢ ( b ¯ ) ⁢ T ¯ . If a ¯ lies in C S ¯ ⁢ ( b ¯ ) , then a ¯ ⁢ b ¯ is not a 2-element of G ¯ and so a ⁢ b is a vanishing element of 𝐺. By our assumptions, | a ¯ ⁢ b ¯ G ¯ | is a prime power, and it is divisible by the 2-power | b ¯ G ¯ | > 1 . We conclude that | a ¯ ⁢ b ¯ G ¯ | is a 2-power, and so | a ¯ G ¯ | is a 2-power as well. Now a ¯ must centralize a Hall 2 ′ -subgroup of G ¯ contained in C G ¯ ⁢ ( b ¯ ) ; therefore, it centralizes T ¯ x ¯ for some x ¯ ∈ C S ¯ ⁢ ( b ¯ ) . It easily follows that C S ¯ ⁢ ( b ¯ ) = C S ¯ ⁢ ( T ¯ ) , and our claim is proved.

Now let g ¯ ∈ G ¯ be arbitrary. If 1 ≠ d ¯ ∈ T ¯ ∩ T ¯ g ¯ , then T ¯ and T ¯ g ¯ ⊆ C G ¯ ⁢ ( d ¯ ) and so T ¯ = T ¯ g ¯ . It follows that either T ¯ ∩ T ¯ g ¯ = { 1 } or T ¯ = T ¯ g ¯ for each g ¯ ∈ G ¯ . This implies that either O 2 ′ ⁢ ( G ¯ ) = 1 or T ¯ ⊲ G ¯ .

On the other hand, as long as a ¯ ∈ Z ⁢ ( S ¯ ) is a vanishing element of G ¯ , we have | a ¯ G ¯ | = | T ¯ | = p n for some prime number 𝑝 and positive number 𝑛. Therefore, we can assume that T ¯ is an abelian Sylow 𝑝-group for some odd prime number 𝑝. Whenever Z ⁢ ( S ¯ ) ∩ Van ⁡ ( G ¯ ) = ∅ , then the size of each vanishing class in G ¯ is a 2-power and so S ¯ is normal in G ¯ ; see [10, Theorem A].

Applying the previous arguments, we distinguish two cases for G ¯ .

Since S ¯ ≃ G ¯ / T ¯ , by the assumption and Lemma 3.2, we conclude that

and for any x ¯ ∈ S ¯ - Z ⁢ ( S ¯ ) , the index of x ¯ is a power of a prime. Whenever S ¯ is non-abelian, Lemma 3.6 yields that G ¯ = S ¯ × T ¯ , and the desired result follows. While S ¯ is abelian and non-normal, then Z ⁢ ( S ¯ ) ∩ Van ⁡ ( G ¯ ) ≠ ∅ and so the index of each element in G ¯ is a power of a prime. Now [8, Theorem 2] proves the claim.

Assume that O 2 ′ ⁢ ( G ¯ ) = 1 and x ¯ ∈ Van ⁡ ( G ¯ ) ∩ S ¯ , which implies that x ∈ Van ⁡ ( G ) and | x ¯ G ¯ | is a prime power. It is easy to understand that, whenever x ¯ is a non-central element of S ¯ , for some positive integer n x , we have | x ¯ G ¯ | = 2 n x . Now Lemma 3.5 ensures that x ¯ ∈ O 2 ⁢ ( G ¯ ) . On the other hand, if x ¯ ∈ Z ⁢ ( S ¯ ) , the following relation guarantees that x ¯ is an element of O 2 ⁢ ( G ¯ ) :

Consequently, each element of S ¯ ∖ O 2 ⁢ ( G ¯ ) is a non-vanishing element of G ¯ , and as a result, we conclude that each element in S ¯ / O 2 ⁢ ( G ¯ ) is a non-vanishing element of G ¯ / O 2 ⁢ ( G ¯ ) . It follows from [11, Theorem A] that S ¯ / O 2 ⁢ ( G ¯ ) is normal in G ¯ / O 2 ⁢ ( G ¯ ) ; in particular, S ¯ ⊲ G ¯ .

Suppose S ¯ is non-abelian. We claim that S ¯ has no nonlinear T ¯ -invariant irreducible characters; if not, let θ ∈ Irr ⁡ ( S ¯ ) be nonlinear and T ¯ -invariant. Lemma 3.8 implies the existence of a zero of 𝜃, say x ¯ ∈ S ¯ , which is not conjugate in S ¯ to any element of C S ¯ ⁢ ( T ¯ ) . Let χ ∈ Irr ⁡ ( G ¯ | θ ) . It can be derived from the Clifford theorem that χ ⁢ ( x ¯ ) = 0 , which is a contradiction. For this reason, S ¯ has no nonlinear T ¯ -invariant irreducible characters. The Glauberman–Isaacs correspondence [15, Theorem 13.1] tells us that there exists a one-to-one correspondence between Irr T ¯ ⁡ ( S ¯ ) and Irr ⁡ ( C S ¯ ⁢ ( T ¯ ) ) , where Irr T ¯ ⁡ ( S ¯ ) denotes the set of all T ¯ -invariant irreducible characters of S ¯ . It now follows from [22, Lemma 5.4] that all irreducible characters of C S ¯ ⁢ ( T ¯ ) are linear, and this is why C S ¯ ⁢ ( T ¯ ) is abelian. Moreover, applying [15, Theorem 13.24] indicates that the number of T ¯ -invariant irreducible characters of S ¯ is given by | C S ¯ / S ¯ ′ ⁢ ( T ¯ ) | which is equal to | C S ¯ ⁢ ( T ¯ ) ⁢ S ¯ ′ / S ¯ ′ | ; see [16, Corollary 3.28]. The fact that

forces C S ¯ ⁢ ( T ¯ ) ∩ S ¯ ′ to be trivial.

Looking at this another way, Lemma 3.4 shows that the pre-image of each non-trivial element of Z ⁢ ( S ¯ ) belongs to Van ⁡ ( G ) . Consequently, we deduce that either Z ⁢ ( S ¯ ) = Z ⁢ ( G ¯ ) or T ¯ is an abelian Sylow 𝑝-group for some odd prime number 𝑝. As a consequence of the following relation, the case Z ⁢ ( S ¯ ) = Z ⁢ ( G ¯ ) is not possible:

Henceforth, we can assume that T ¯ is an abelian Sylow 𝑝-subgroup for some odd prime number 𝑝. Furthermore, since each element of G ¯ / S ¯ ′ ≅ S ¯ / S ¯ ′ ⋊ T ¯ has a prime power index, [8, Theorem 2] yields that G ¯ / S ¯ ′ is either nilpotent or a Frobenius group. Using [16, Problem 1D.17], it is easy to observe that the case where G ¯ / S ¯ ′ is nilpotent will not occur; thus G ¯ / S ¯ ′ is a Frobenius group. As a result, no element in S ¯ ∖ S ¯ ′ commutes with any element of odd order; in other words, C S ¯ ⁢ ( T ¯ ) ∩ ( S ¯ ∖ S ¯ ′ ) = ∅ . This implies that G ¯ = S ¯ ⋊ T ¯ is a Frobenius group with the property that each element of S ¯ - Z ⁢ ( S ¯ ) is a non-vanishing element of G ¯ due to C S ¯ ⁢ ( T ¯ ) = 1 . While A ¯ is a non-central abelian characteristic subgroup of S ¯ , Lemma 3.4 guarantees that the index of some element a ¯ ∈ A ¯ - Z ⁢ ( S ¯ ) is a prime power, which is impossible. Thereby, S ¯ is of nilpotency class at most two.

Since S ¯ is non-abelian, we conclude that { 1 } ≠ S ¯ ′ ⊆ Z ⁢ ( S ¯ ) ⊊ S ¯ . Without loss of generality, we can assume that S ¯ ′ is a minimal normal subgroup of S ¯ . Therefore, | S ¯ ′ | = 2 and each non-central element of S ¯ has index 2 in G ¯ . Let 𝜃 be a nonlinear irreducible character of S ¯ , and let x ¯ ∈ S ¯ ∖ Z ⁢ ( S ¯ ) . As G ¯ is a Frobenius group, we have [ χ S ¯ , θ ] = [ χ , θ G ¯ ] = 1 for some nonlinear irreducible character 𝜒 of G ¯ . Therefore, χ ⁢ ( x ¯ ) = ∑ i θ i ⁢ ( x ¯ ) , where θ i are distinct conjugates of 𝜃 in G ¯ . As the centralizer of each non-central element of S ¯ in S ¯ has order equal to | S ¯ | / 2 , we deduce that θ ⁢ ( x ¯ g ¯ ) = 0 for every g ¯ ∈ G ¯ and in particular θ i ⁢ ( x ¯ ) = 0 for each 𝑖. As a result, χ ⁢ ( x ¯ ) = 0 and x ¯ ∈ Van ⁡ ( G ¯ ) , which is a contradiction. Hence S ¯ is abelian, and the claim obviously holds. ∎

Proof

By Theorem 1.1, a Sylow 2-subgroup S ¯ of G ¯ is either abelian or a non-abelian direct factor of G ¯ . Let 𝑥 be a non-trivial non-vanishing element of 𝐺. If x ∉ F ⁢ ( G ) , then Lemma 3.1 implies that x ¯ is a non-trivial 2-element of G ¯ .

First suppose that S ¯ is non-abelian; by Theorem 1.1, G ¯ = S ¯ × T ¯ . As G ¯ is nilpotent and x ¯ is a non-vanishing element of G ¯ , by [17, Theorem D], we observe that x ¯ ∈ Z ⁢ ( G ¯ ) . As x ¯ is a non-trivial 2-element, we deduce that x ¯ ∈ Z ⁢ ( S ¯ ) . By the correspondence theorem, there exists a normal subgroup 𝑍 of 𝐺 containing F ⁢ ( G ) such that Z / F ⁢ ( G ) ≅ Z ⁢ ( S ¯ ) . Since F ⁢ ( Z ) = F ⁢ ( G ) and Z / F ⁢ ( G ) is abelian, Lemma 3.3 assures that x ∈ F ⁢ ( G ) , which contradicts our hypothesis.

Now consider the second case where S ¯ is abelian. While S ¯ ⊲ G ¯ , x ¯ lies in S ¯ , and we deduce from Lemma 3.4 that 𝑥 is a vanishing element of 𝐺, which is impossible.

If S ¯ is a non-normal abelian Sylow subgroup of G ¯ , then Theorem 1.1 implies that 2 ∈ ρ ⁢ ( cs ⁡ ( G ¯ ) ) and G ¯ = A ¯ × H ¯ with the property that, for a prime 𝑝, ρ ⁢ ( cs ⁡ ( G ¯ ) ) = { 2 , p } , A ¯ is an abelian { 2 , p } ′ -subgroup of G ¯ , and H ¯ = P ¯ ⋊ S ¯ such that H ¯ / Z ⁢ ( H ¯ ) is a Frobenius group, where P ¯ ∈ Syl p ⁢ ( G ¯ ) . Since x ¯ is a 2-element, without loss of generality, we may assume that x ¯ ∈ S ¯ . Since F ⁢ ( H ¯ ) = P ¯ × Z ⁢ ( S ¯ ) and H ¯ / F ⁢ ( H ¯ ) is abelian, Lemma 3.3 shows that x ¯ ∈ Z ⁢ ( S ¯ ) , which is a contradiction, using a proof similar to the previous paragraphs. ∎