Finite groups with a small proportion of vanishing elements ∗

Dongfang Yang†, Yu Zeng‡and Silvio Dolfi§

arXiv:2303.16391v2 [math.GR] 9 Jul 2023

Abstract

The function Pv (G), measuring the proportion of the elements of a finite group G that are
zeros of irreducible characters of G, takes (as proved in [12]) only values m−1
m
, for 1 ≤ m ≤ 6,
in the interval [0, Pv (A7 )). In this paper, we give a complete classification of the finite groups
G such that Pv (G) = m−1
m
for m = 1, 2, · · · , 6.

Keywords Character theory, zeros of irreducible characters.

2020 MR Subject Classification 20C15.
∗ Acknowledgement: The first and second authors gratefully acknowledge the support of China Scholarship
Council (CSC) and the first author was supported by the Natural Science Foundation for the Universities in Jiangsu
Province (No. 23KJB110002). The third author was partially supported by INDAM-GNSAGA.
† Department of Mathematics, Changshu Institute of Technology, Changshu, China. email: dfyang1228@163.com
‡ Department of Mathematics, Changshu Institute of Technology, Changshu, China. email: yuzeng2004@163.com
§ Dipartimento di Matematica e Informatica, Università di Firenze, Italy. email: silvio.dolfi@unifi.it

1
1 Introduction
We say that an element g of a finite group G is a vanishing element of G if there exists an
irreducible character χ of G such that χ(g) = 0, and we denote by V(G) the set of the vanishing
elements of G.
A natural way to measure the relative abundance of vanishing elements in a finite group G is
to consider the proportion
|V(G)|
Pv (G) =
|G|
which can be seen as the probability of getting a vanishing element by picking at random (with
uniform distribution) an element in G. Relevant questions in this context are those concerning
the possible values of the function Pv (G), and the conditions on the structure of a group G that
are related to specific values of Pv (G). We recall that by Burnside’s classical theorem on zeros
of irreducible characters, G is abelian if and only if Pv (G) = 0. Strengthening this result, in [11]
L. Morotti and H. Tong-Viet show that a finite group G is abelian if and only if Pv (G) < 12 , and
1
that Pv (G) = 2 if and only if G is an A-group (i.e. a solvable group whose Sylow subgroups are
all abelian) and G/Z(G) is a Frobenius group with complement of order 2. They also prove ([11,
Theorem 1.6]) that if Pv (G) ≤ 32 , then Pv (G) ∈ {0, 21 , 23 } and conjecture that, for G in the class
of finite groups, the only values of Pv (G) smaller than Pv (A7 ), where A7 is the alternating group
m−1
on 7 letters, are of the form m for some integer 1 ≤ m ≤ 6. This conjecture has been confirmed
in [12], after that A. Moretó and P.H. Tiep proved in [10] that a group G is necessarily a solvable
1067
if Pv (G) < Pv (A7 ) = 1260 ≃ 0.846.
In this paper, we give a complete classification of the finite groups G such that Pv (G) <
Pv (A7 ). Before stating the main result of this paper, we need to introduce some notation. As
usual, for an abelian group A and a group H, we say that A is an H-module if it is given a group
homomorphism from H to Aut(A). We say that A is a cyclic H-module if A is generated, as
H-module, by a single element of A.
For a group H ∼
= C6 (where Cn is the cyclic group of order n) and an abelian 2-group A, we
say that

(T1) A is of type (T1) if A is a cyclic H-module of rank 2, A ∼

= C8 × C8 and, for suitable x, y ∈ H
such that o(x) = 3 and o(y) = 2, CA (x) = 1 and aay = (a4 )x for all a ∈ A.

(T2) A is of type (T2) if A is a cyclic H-module of rank 4, A = A0 A1 , with A0 ∼= C2n × C2n ,

n ∼
2 ≥ 8, A1 = [A0 , y] = C4 × C4 and, for suitable x, y ∈ H such that o(x) = 3 and o(y) = 2,
n−1
CA (x) = 1 and [a2 , y]x = a2 for all a ∈ A.

We say that an abelian 2-group A is a homogeneous H-module of type (T1) (resp. (T2)),
where H ∼
= C6 , if A is a direct sum of isomorphic H-modules of type (T1) (resp. (T2)).

2
Theorem A. Let G be a finite group, Q ∈ Syl2 (G) and P ∈ Syl3 (G). Then Pv (G) < Pv (A7 ) if
and only if

(a) G is an A-group such that [G : F(G)] = m ≤ 6, and |F(G)/Z(G)| is not divisible by 6 if

m = 5.

(b) Q is nonabelian and G has an abelian normal subgroup A such that Q ∩ A = Z0 × D, where
Z0 = CQ∩A (P ) ≤ Z(G) and D = [Q ∩ A, P ], and one of the following holds:

(b1) |G/A| = 4 and Q ∩ A = Z(Q).

(b2) G/A ∼
= S3 , and, for some x ∈ P − A, 1 6= D = Z × Z x , where Z = CD (Q) is either
elementary abelian or the direct product of a cyclic group of order 4 and a possibly trivial
elementary abelian group.
(b3) G/A ∼
= C6 , 1 6= D = B × C with B and C are normal subgroups of G such that
[C, Q, Q] = 1 and if B 6= 1, then exp(B) > exp(C) and either every y ∈ Q − A acts
as the inversion on C and B is a homogeneous G/A-module of type (T1), or B is a
homogeneous G/A-module of type (T2).

In Theorem 4.5 and Corollary 4.6, for completeness, we describe in full detail the groups that
satisfy condition (a) of Theorem A.
The paper is organized as follows: in Section 2 and Section 3 we collect several preliminary
results, while in Section 4 we prove Theorem A and Theorem 4.5.
Our notation is standard and for character theory follows [7]. All groups considered in the
paper will be assumed to be finite groups.

2 Preliminaries
Let G be a finite group. As already mentioned, we write

V(G) = {g ∈ G | χ(g) = 0 for some χ ∈ Irr(G)},

and we denote by N(G) = G − V(G) its complement in G, i.e. the set of the non-vanishing
elements of G.
We remark that, given a normal subgroup N of a group G and g ∈ G, if gN ∈ V(G/N ) then
gN ⊆ V(G). This implies that Pv (G/N ) ≤ Pv (G), a fact that will be used freely in the rest of
the paper.
We collect some useful results about vanishing elements.

Lemma 2.1 ([9, Main Theorem]). Let m, n be a positive integers, α1 , α2 , . . . , αn m-th roots of
unity and let p1 , p2 , . . . , pr be the (distinct) prime divisors of m. Then α1 + · · · + αn = 0 only if
n = n1 p1 + · · · + nr pr for suitable nonnegative integers ni .

3
Lemma 2.2 ([12, Lemma 2.4]). Let G be a group and z ∈ Z(G). Then x ∈ V(G) if and only if
zx ∈ V(G).

Lemma 2.3 ([4, Corollary 1.3]). Let g ∈ H ≤ G be such that G = CG (g)H. Then g ∈ V(G) if
and only if g ∈ V(H).

Lemma 2.4 ([12, Lemma 2.5]). Let A and B be normal subgroups of the group G such that
A ∩ B = 1. Then, for a ∈ A, we have a ∈ V(G) if and only if aB ∈ V(G/B).

Lemma 2.5 ([8, Theorem A]). If P is a p-group, p a prime number, then N(P ) = Z(P ) and
m−1
hence Pv (P ) = m , where m = [P : Z(P )].

Lemma 2.6 ([1]). Let G be a finite group and let P be a Sylow p-subgroup of G, p a prime
number. Then Z(P ) ∩ Op (G) ⊆ N(G).

For short, in the rest of the paper we write a = Pv (A7 ). Although N(G) is in general quite
far from being a subgroup of the group G, it is indeed an abelian normal subgroup under the
assumption that Pv (G) < a, as shown in the following result (which is the key theorem of [12]).

Theorem 2.7 ([12, Theorem 4.3]). If G is a finite group such that Pv (G) < a, then N(G) is an
abelian normal subgroup of G.

3 Auxiliary results
Let A be an abelian group and let  = Irr(A) be the dual group of A. For B ≤ A, let B ⊥ =
{α ∈ Â | α(b) = 1, ∀ b ∈ B} ≤ Â, and for V ≤ Â, let V ⊥ = {a ∈ A | λ(a) = 1, ∀ λ ∈ V } ≤ A.
Then B ⊥ = [ and B ⊥⊥ = B (see for instance [6, V.6.4]). Finally, we recall that for every
∼ A/B
−1
x ∈ Aut(A), x acts on  via αx (a) = α(ax ), where α ∈  and a ∈ A.

Lemma 3.1 ([12, Lemma 3.1]). Let A be an abelian group, y ∈ Aut(A) such that o(y) = 2, and
let Q = A ⋊ hyi. Then

(1) The map γ : A → A, defined by γ(a) = [a, y] for a ∈ A, is a group homomorphism with
Im(γ) = [A, y] = Q′ and ker(γ) = CA (y) = Z(Q). Moreover, A/CA (y) and [A, y] are
isomorphic hyi-modules, on which y acts as the inversion.

(2) [A, y]⊥ = CÂ (y) and [Â, y]⊥ = CA (y).

(3) The groups [Â, y] and [A, y] are isomorphic.

In the following, we denote by c(Q) the nilpotency class of a nilpotent group Q and by Z2 (Q)
the second term of the upper central series of Q. If A is an abelian p-group, p a prime, we denote

4
by exp(A) the exponent of A (the largest order of an element of A, in this case) and we define, for
i ≥ 0, Ωi (A) = {a ∈ A | o(a) divides pi }, which is a characteristic subgroup of A. We now state
a couple of lemmas for later use.

Lemma 3.2 ([12, Lemma 2.6]). Let A be a normal subgroup of the group G. Then, for a ∈ A,
a ∈ V(G) if and only if there exists a character α ∈ Irr(A) such that αG (a) = 0.

Lemma 3.3 ([12, Lemma 3.5]). Let A be an abelian normal 2-subgroup of the group G such that
[G : A] = 6, and let Q ∈ Syl2 (G) and P ∈ Syl3 (G). Then

(1) There exists an element y ∈ NQ (P ) such that y 6∈ A and y 2 ∈ CA (P ).

(2) If CA (P ) = 1 and either exp(Q′ ) ≤ 2 or c(Q) ≤ 2, then A ⊆ N(G).

Assume further that P acts nontrivially on A and that Q is nonabelian. Then

(3) G − A ⊆ V(G).

(4) If Pv (G) ≤ a, then CA (P ) ≤ Z(G).

In the rest of the section, we will address the most complicated cases arising in the process of
proving Theorem A. In order to avoid repetitions, we introduce the following setting.

Setting 3.4. Let A be an abelian normal 2-subgroup of the group G such that [G : A] = 6 and
assume that A has a complement H in G. Let P be the Sylow 3-subgroup of H, y an involution of
H and Q = Ahyi a Sylow 2-subgroup of G. Assume that Q is nonabelian and that CA (P ) = 1.
5
We remark that, by part (3) of Lemma 3.3, if G satisfyies Setting 3.4, then Pv (G) = 6 if and
only if A ⊆ N(G).
Given an abelian p-group A, we denote by rank(A) its rank (i.e. the number of factors in a
decomposition of A as a direct product of cyclic groups) and, for a ∈ A, k ∈ Z and x ∈ Aut(A),
we will write, for short, akx instead of (ak )x ; in particular, we will write a−x for (ax )−1 .

Remark 3.5. If a cyclic group P = hxi of order 3 acts on an abelian 2-group A and CA (P ) = 1,
2 2
then aax ax = 1 for every a ∈ A, because aax ax ∈ CA (P ). Considering A as a P -module, if B 6= 1
is a P -submodule of A and rank(B) ≤ 2, then B is an indecomposable P -module and hence B is
homocyclic of rank 2. Thus A, being a direct sum of indecomposable P -modules, has even rank.
Moreover, B is a uniserial P -module whose only submodules are the subgroups Ωi (B), i ≥ 0, and
whose composition factors are all isomorphic to C2 × C2 (see for instance [5]). As a consequence,
for every a ∈ A, a 6= 1, the P -submodule of A generated by a is ha, ax i = hai × hax i.

If A is an H-module, we say that B is a cyclic submodule of A if there exists a single element

b ∈ A such that B = hbiH , i.e. B is generated by b as an H-module. Observe that this is equivalent
to B = hbiG , the normal closure of hbi in the semidirect product G = A ⋊ H.

5
Remark 3.6. In the situation of Setting 3.4, a nontrivial cyclic H-submodule of A has rank
either 2 or 4. In fact, writing H = hx, yi, where P = hxi, the condition CA (P ) = 1 implies that
2
bbx bx = 1 for every b ∈ A. So, for 1 6= b ∈ A, B = hbiH = hb, bx , by , byx i has rank at most 4
and hence rank(B) ∈ {2, 4} by the previous remark. We also note that rank(B) = 4 if and only if
by 6∈ hb, bx i, and that rank(B) = 2 if and only if B = hbi × hbx i.

We denote, for short, by (Cn )k the direct product of k ≥ 0 copies of the cyclic group Cn .

Lemma 3.7 ([12, Lemma 3.7]). Assume Setting 3.4 and that A ⊆ N(G). Let Z = Z(Q) and
P = hxi. Then

(1) If H ∼
= C6 , then Z E G and exp(A/Z) = 2c−1 , where c = c(Q) ≤ 3.

(2) If H ∼
= C6 and B is a P -submodule of A such that rank(B) = 2 and A0 = B × B y has index
at most 4 in A, then exp(B) = 2.

(3) If H ∼
= S3 , A = Z × Z x and rank(A) ≤ 4, then exp(A) ≤ 4 and Z is not isomorphic to (C4 )2 .

3.1 S3 case

In this subsection, we classify the groups G that satisfy Setting 3.4 such that H ∼
= S3 and
Pv (G) = 65 .

Proposition 3.8. Assume Setting 3.4 and that H is isomorphic to S3 . Let P = hxi and Z =
CA (y). Then A = Z × Z x . Moreover, A ⊆ N(G) if and only if Z is isomorphic to a nontrivial
subgroup of C4 × (C2 )t , for some integer t ≥ 0.

Proof. We start by observing that Z = CA (y) = Z(Q) is nontrivial. As H = hy, y x i, we have

Z ∩ Z x = CA (y) ∩ CA (y)x = CA (hy, y x i) = CA (H) ≤ CA (P ) = 1.

An application of part (1) of Lemma 3.1 to the action of y on  yields µy = µ−1 and hence
2
µxy = µ−x 6= µ−x for every µ ∈ [Â, y] − {1A }, so [Â, y] ∩ [Â, y]x = 1A . Since by Lemma 3.1
[Â, y]⊥ = Z, it follows that

A = (1A )⊥ = ([Â, y] ∩ [Â, y]x )⊥ = ZZ x = Z × Z x .

Write Z = C × D, with C = hci such that o(c) = exp(Z). Then A = V × W where V = C × C x

2 2
and W = D × Dx . As z xy = z yx = z x = (zz x )−1 for every z ∈ Z, we have both V E G and
W E G.

Let us suppose, first, that A ⊆ N(G). Take d ∈ D, and let Z0 = hci×hdi. Since o(c) = exp(Z),
in order to conclude that Z is isomorphic to a subgroup of C4 × (C2 )t for some t ≥ 0, it suffices
to show that Z0 is isomorphic to a subgroup of C4 × C2 . Let A0 = Z0 × Z0x , Q0 = A0 ⋊ hyi and

6
G0 = A0 ⋊ H. As G = CG (a0 )G0 for every a0 ∈ A0 , by Lemma 2.3 we deduce that A0 ⊆ N(G0 ).
2 2
Note that (cx )y = cyx = cx 6= cx , so Q0 is nonabelian and hence G0 satisfies Setting 3.4. Now, an
application of part (3) of Lemma 3.7 to G0 yields that Z0 is isomorphic to a subgroup of C4 × C2 .

Conversely, we assume that Z is isomorphic to a subgroup of C4 × (C2 )t for some t ≥ 0, and

we show that A ⊆ N(G). Recall that A = V × W , where V = hci × hcx i with o(c) = exp(Z). So,
o(c) ≤ 4 and exp(W ) ≤ 2. If o(c) ≤ 2, then exp(A) ≤ 2, and hence part (2) of Lemma 3.3 yields
A ⊆ N(G). So, we may assume that o(c) = 4. Let a ∈ A and α ∈ Irr(A), and write a = ck clx w,
where w ∈ W , k, l ∈ Z. By Lemma 3.2, it suffices to prove that αG (a) 6= 0. Assume the contrary,
i.e.
2 2
αG (a) = α(a) + α(ax ) + α(ax ) + α(ay ) + α(ayx ) + α(ayx ) = 0. (.)
2 2 2 2
where α(a)α(ax )α(ax ) = α(aax ax ) = 1 and α(ay )α(ayx )α(ayx ) = α(ay ayx ayx ) = 1. As α is
a linear character of A and exp(A) ≤ 4, the values of α (and of any of its conjugate characters)
are 4-th roots of unity. Thus, an application of part (1) of Lemma 2.3 of [12] to (.) yields that
i i
max{o(αx ([a, y])) | 0 ≤ i ≤ 2} = 4 and hence, as o(αx ([a, y])) divides o([a, y]), that o([a, y]) = 4.
i i i
Since at least one of the elements α([a, y]x ) = α(a−x )α(ayx ) has order 4 (in C× ), for 0 ≤ i ≤ 2,
it follows that max{o(α(ah )) | h ∈ H} = 4. Since αG = (αh )G , for h ∈ H, up to substituting α
with a suitable conjugate αh , we may assume that both α(a) and α(ax ) have order 4, so α(a2 ) =
i
α(a2x ) = −1. By (.), part (2) of Lemma 2.3 of [12] implies that α(ayx )2 = 1 for 0 ≤ i ≤ 2. In
particular, α(a2y ) = α(ay )2 = 1. Recalling that a = ck clx w, as [a, y] = [ck clx w, y] = [clx , y][w, y] =
2
c−lx clx [w, y] has order 4 and o([w, y]) ≤ 2 (as [w, y] ∈ W ), we deduce that l is odd. As o(c) = 4,
then (cl )2 = c2 and hence a2 = c2k c2x . If k is even, then c2k = 1 and a2 = c2x ; so, recalling that
2
cy = c and that xy = yx2 , we have 1 = α(a2y ) = α(c2xy ) = α(c2x ) = α(a2x ) = −1, a contradiction.
2 2 2
If k is odd, then a2 = c2 c2x and 1 = α(a2yx ) = α(a2xy ) = α(c2xy c2x y ) = α(c2x c2x ) = α(a2x ) =
−1, again a contradiction. Thus, αG (a) 6= 0, as desired.

3.2 C6 case

Setting 3.9. Assume Setting 3.4 with H isomorphic to C6 and c(Q) ≥ 3.

In this subsection, we are going to classify the groups G satisfying Setting 3.9 and Pv (G) = 56 .

Remark 3.10. If G satisfies Setting 3.9 and A ⊆ N(G), then part (1) of Lemma 3.7 yields that
c(Q) = 3 and that exp(A/Z(Q)) = 4.

First, we deal with the case of Setting 3.9 where A is a cyclic H-module: see Proposition 3.13.
2π
In the following, we denote by ζn the primitive n-th root of unity e n i ∈ C× . We start with two
lemmas.

7
 Lemma 3.11. Assume Setting 3.9. Suppose that rank(A) = 2 and that exp(A) = 8. Then
A ⊆ N(G) if and only if for a suitable generator x of P , aay = a4x for all a ∈ A.

Proof. We observe that Q E G and then both Q′ and Z = Z(Q) are P -submodules of the uniserial
P -module A ∼
= C8 × C8 (see Remark 3.5). Since c(Q) > 2, Q′ is not contained in Z and hence
Z = Ω1 (A) and Q′ = A2 . Since c(A2 hyi) = 2, part (2) of Lemma 3.3 applied to the subgroup
A2 H yields A2 ⊆ N(A2 H) and hence A2 ⊆ N(G) by Lemma 2.3. Let a0 ∈ A be such that
o(a0 ) = 8; by the Frobenius action of P = hxi on A, we have A = ha0 i × hax0 i. Note that
a0 ay0 ∈ CA (y) = Z = {1, a40 , a4x
2
4x
0 , a0 }.

Assume first that A ⊆ N(G). We claim that a0 ay0 ∈ {a4x y

2
4x 4
0 , a0 }. Otherwise, a0 a0 ∈ {1, a0 },

so ay0 = ak0 for a suitable k ∈ {−1, 3}. Let α ∈ Irr(A) be such that α(a0 ) = ζ8 and α(ax0 ) = ζ85 ; so,
2
α(ax0 ) = (α(a0 )α(ax0 ))−1 = ζ82 . Hence,
X i i X i i X
αG (a0 ) = (αx (a0 ) + αx y (a0 )) = (αx (a0 ) + αx (ak0 )) = ζ8j = 0,
0≤i≤2 0≤i≤2 j∈{1,2,3,5,6,7}

so a0 ∈ V(G) by Lemma 3.2, a contradiction. Hence, for a suitable generator x of P , a0 ay0 = a4x
0 .

We now show that the action of x on A is as claimed. In fact, for every a ∈ A, a = ai0 ajx
0 for some

integers i and j. As xy = yx,

2
aay = ai0 ajx i jx y y i y jx
0 (a0 a0 ) = (a0 a0 ) (a0 a0 ) = a4ix
0 a0
4jx
= a4x .

Assume conversely that for a suitable generator x of P , aay = a4x for every a ∈ G. In
particular, aay = a4x for every a ∈ A − A2 . Let a ∈ A − A2 . As A2 ⊆ N(G), it suffices to
show that a ∈ N(G). Arguing by contradiction, by Lemma 3.2 there exists a character α ∈ Irr(A)
such that αG (a) = 0. Let K = ker(α) and observe that Z 6≤ K: otherwise, writing G = G/Z,
we have α ∈ Irr(A) and αG (a) = 0, so a ∈ V(G) ∩ A, against part (2) of Lemma 3.3 since
2 i
c(Q) = 2. Hence, as a4 a4x a4x = 1 and {α(a4x ) | 0 ≤ i ≤ 2} = {1, −1}, up to possibly
i i
exchanging α with some conjugate character αx (observing that αG = (αx )G for every i), we
2
can assume α(a4x ) = 1 and α(a4 ) = α(a4x ) = −1. Recalling that ay = a4x a−1 , we have that
i i 1−i
αx (ay ) = αx (a4x )αxi (a) = α(a4x )αxi (a) for 0 ≤ i ≤ 2 and then
2 2
αG (a) = α(a) + αx (a) + αx (a) + α(ay ) + αx (ay ) + αx (ay )
2
= α(a) + αx (a) + αx (a) + α(a) − αx (a) − αx2 (a)

Hence, the assumption αG (a) = 0 gives

2
α(a) + α(a) = αx (a) − αx (a) + αx2 (a) − αx (a).

Observing that the first member of the above equality is real, while the second member is purely
imaginary, we deduce that α(a) = −α(a) and we get α(a)4 = α(a4 ) = 1, a contradiction as
α(a4 ) = −1. Hence, a ∈ N(G) and we conclude that A ⊆ N(G).

8
 Lemma 3.12. Assume Setting 3.9. Let A = CD, where C is a P -submodule of rank 2 of A,
D = [C, y] and C ∩ D = D2 = Ω1 (D). If 2n = exp(A) ≥ 8, then A ⊆ N(G) if and only if for a
n−1
suitable generator x of P , a2 = [a2 , y]x for every a ∈ A.

Proof. See [12, Lemma 3.9].

The next result is a key for the classification we are after.

Proposition 3.13. Assume Setting 3.9 and that A is a cyclic H-module, say A = ha0 iG for some
a0 ∈ A. Then A ⊆ N(G) if and only if, for a suitable generator x of P , one of the following holds.

(1) rank(A) = 2, A = ha0 , ax0 i ∼

= (C8 )2 and aay = a4x for all a ∈ A; i.e. A is of type (T1).

(2) rank(A) = 4, A = CD where C = ha0 , ax0 i ∼

= (C2n )2 , n ≥ 3, D = [C, y] ∼
= (C4 )2 and
n−1
a2 = [a2 , y]x for every a ∈ A; i.e. A is of type (T2).

Moreover, if A ⊆ N(G), [A, y, y] is the unique irreducible H-submodule of A.

Proof. If either (1) or (2) holds, then A ⊆ N(G) follows directly by Lemma 3.11 and Lemma 3.12.
Assume, conversely, that A ⊆ N(G). Recall that, by Remark 3.10 c(Q) = 3 and exp(A/Z(Q)) =
4. As Q = Ahyi and A = ha0 iG , a0 6= 1. So, rank(A) = 2 or rank(A) = 4 by Remark 3.6.
Suppose first that rank(A) = 2. Then A is a uniserial P -module (see Remark 3.5) and,
as Q = Ahyi has class 3, Z(Q) < Q′ . Recall that by part (1) of Lemma 3.1, Q′ = [A, y] and
A/Z(Q) ∼
= (C4 )2 are isomorphic hyi-modules, on which y acts as the inversion. It follows that
Z(Q) = Ω1 (Q′ ) and that exp(A) = 8. Hence, by Lemma 3.11 we have case (1). Moreover, since
A is a uniserial P -module of rank 2, [A, y, y] = Ω1 (A) is the unique irreducible P -submodule, and
hence H-submodule, of A.

Suppose now that rank(A) = 4. Let C = ha0 , ax0 i and D = [C, y]; so, by Remark 3.6,
rank(D) = 2. Observe that A = CC y = CD. As C ∩ C y ✂ G, we set G = G/(C ∩ C y ). Since
y
A ⊆ N(G) implies A ⊆ N(G), applying part (2) of Lemma 3.7 to A = C × C , we have that
C∼= (C2 )2 and hence A/C ∼
= (C2 )2 . Write exp(A) = o(a0 ) = 2n and let Z = Z(Q) = CA (y). By
Lemma 3.1 and part (1) of Lemma 3.7 D = [A, y] ∼ = A/Z ∼ = (C4 )2 , so n ≥ 2 and C ∩ D = D2 =
Ω1 (D). We claim that n ≥ 3; in fact, otherwise |A| = |C||D/C ∩ D| = 26 and from |A/Z| = 24 it
follows |Z| = 22 , so Z = Ω1 (D) = Ω1 (C) and A/Z ∼ = C/Z × D/Z is elementary abelian, against
A/Z ∼
= (C4 )2 . An application of Lemma 3.12 to A yields that for a suitable generator x of P ,
n−1
a2 = [a2 , y]x for all a ∈ A. Hence, we have case (2).
Finally, by Lemma 3.1 y acts as the inversion on [A, y] = D ∼
= (C4 )2 , so [A, y, y] = D ∩ Z is
an irreducible H-module. In order to prove that D ∩ Z is the only irreducible H-submodule of A,
it is enough to show that rank(Z) = 2 as any irreducible H-submodule of A is contained in Z. We
observe that if rank(Z) > 2, then rank(Z) = rank(A) = 4 by Remark 3.6 and all the involutions

9
 n−2
of A would be fixed by y. Write w = [a0 , y]x and e = a20 w. Observe that e 6= 1 as w ∈
/ C, and
n−1 n−1
2
e = a20 w2 = a20 w−2 = 1, so e is an involution. Moreover,
n−2 n−2 n−2
x−1 x−1
[e, y] = [a20 w, y] = w2 [w, y] = w2 w−2 6= 1,

because CD (x) = 1. Hence, rank(Z) = 2, concluding the proof.

We observe that in part (2) of Proposition 3.13, n can be any integer larger than 2.

Lemma 3.14. Assume Setting 3.9. If A ⊆ N(G) and there exists a character α ∈ Irr(A) such
that ker(αG ) = 1, then A is a cyclic H-module.

Proof. Write K = ker(α). Then K ⊥ ∼ [ ∼

= A/K = A/K is a cyclic group of order o(α) and, as
α ∈ K ⊥ , we see that K ⊥ = hαi. It follows that, for h ∈ H, (K ⊥ )h = (K h )⊥ = hαh i.
Let P = hxi. As CA (x) = 1, by Brauer’s permutation lemma CÂ (x) = 1A and hence
x x2 2
λλ λ = 1A for all λ ∈ Â. So, K x ≥ K ∩ K x and then

K ∩ K x ∩ K y ∩ K xy = KG = ker(αG ) = 1 .

Set B = K ∩ K x ; so, B is P -invariant and B ∩ B y = 1. By [6, V.6.4] we have that B ⊥ =

K ⊥ (K x )⊥ = hα, αx i has rank at most 2, so B ⊥ = hαi × hαx i, as CÂ (P ) = 1A (and rank(B ⊥ ) 6= 1
as a cyclic 2-group has no automorphism of order 3). Moreover,

 = 1⊥ = (B ∩ B y )⊥ = B ⊥ B ⊥y = hα, αx ihαy , αxy i = hα, αx ih[α, y], [αx , y]i = B ⊥ [Â, y].

In particular, rank(A) = rank(Â) ≤ 4 and exp(A) = exp(Â) = o(α). Write e = exp(A). As A 6= 1,

rank(A) is either 2 or 4 by Remark 3.5. If rank(A) = 2, then by Remark 3.5 A = hai × hax i = haiG ,
and we are done. Thus, we can also assume that B 6= 1, since if B = 1, then  = 1⊥ = B ⊥ =
hαihαx i, so rank(A) = rank(Â) = 2. Hence, rank(A) = 4 by Remark 3.6.
Let Z = Z(Q). As H ∼
= C6 and c(Q) > 2, Q E G and by Remark 3.10 we have exp(A/Z) = 4
and c(Q) = 3. In particular, e ≥ 4. Observe also that B ∩ Z = 1, since B ∩ Z E G and B ∩ B y = 1.
By Lemma 3.1, [Â, y] ∼
= [A, y] ∼
= A/Z and, as  = B ⊥ B ⊥y , we see that [Â, y] = [B ⊥ , y][B ⊥y , y] =
[B ⊥ , y] = h[α, y], [α, y]x i has rank two. Since [Â, y] is a nontrivial P -module, we conclude that
Q′ = [A, y] ∼= [Â, y] ∼
= A/Z ∼ = (C4 )2 .
By Lemma 3.1 we also know that y acts as the inversion on both Q′ and A/Z. It follows that
Q′ ∩ Z = Ω1 (Q′ ) and A/Z2 (Q) ∼
= (C2 )2 .
As B ∼
= BZ/Z and BZ/Z is a nontrivial P -submodule of A/Z ∼
= (C4 )2 , we have two cases:

(a): B ∼
= (C4 )2 and A = B × Z. Since B ∼
= (C4 )2 , B × B y ≤ A ⊆ N(G). Observe that
G = CG (BB y )(BB y H), we have that B × B y ⊆ N(BB y H) by Lemma 2.3. Now, applying part
(2) of Lemma 3.7 to BB y H, we get a contradiction.

10
 (b): B ∼
= (C2 )2 . Note that in this case BZ/Z = CA/Z (y), so BZ = Z2 (Q). We also observe
that Z 6≤ Q′ . In fact, if Z ≤ Q′ , then Z = Ω1 (Q′ ), so B ∩ Q′ = 1; this implies that Q′ /Z is a
complement of BZ/Z in A/Z = ∼ (C4 )2 , a contradiction.

Let A = A/B. Take a ∈ A − BZ. Recalling that A ∼ [∼

= A/B = B⊥ ∼
= (Ce )2 , where e = exp(A),
2
we see that a 6∈ Z = A and hence that o(a) = e. Let C = ha, ax i. Since [A : C] = |A|/|B ⊥ | =
|B| = 22 , C is a maximal P -submodule of A. We also observe that e ≥ 8, as e ≤ 4 implies |A| ≤ 82 ,
so |Z| = 22 and Z is a minimal P -submodule of A, which is impossible as Z 6≤ Q′ .
We claim that C y 6= C. Working by contradiction, assume that C y = C, so C E G. As
CA/C (y) is a nontrivial P -submodule of the irreducible P -module A/C, we deduce that Q/C is
abelian and hence that Q′ ≤ C. We have observed above that Z 6≤ Q′ ; moreover, Q′ 6≤ Z because
c(Q) > 2, so Z cannot be contained in the uniserial P -module C. By the maximality of C, it follows
that A = CZ. So, C/(Z ∩ C) ∼ = A/Z ∼ = (C4 )2 and, as Z ∩ C = Ω1 (Q′ ) = Ω1 (C), we conclude
that e = 8 and C ∼ = (C8 )2 . Note that c(Chyi) = 3, as A = CZ implies Q′ = [A, y] = [C, y].
Since by assumption C ⊆ A ⊆ N(G), an application of Lemma 2.3 and Lemma 3.11 to the
group G0 = CH yields that aay = a4x for a suitable generator x of P . As B ∩ Z = 1, then
rank(Z) < rank(A) = 4 and hence Z has rank 2 by Remark 3.6. Since |Z| = |A/C||C ∩ Z| = 42 ,
we deduce that Z ∼
−1
= (C4 )2 . Let now z ∈ Z be such that z 2 = a4 and consider g = ax a2 z.
−1 −1 −1
Then o(g) = 8 and gg y = ax a2 z(ax a2 z)y = (aay )x (aay )2 z 2 = a4 a4 = 1, so g y = g −1 . Let
λ ∈ Irr(A) be such that λ(g) = ζ8 and λ(g x ) = ζ85 : consider, for instance, an extension to A of a
2 2
suitable character of hgi × hg x i. Since gg x g x = 1, λ(g x ) = λ(gg x )−1 = ζ82 . As g y = g −1 , we have
X i i X i X X
λG (g) = (λx (g) + λx y (g)) = λx (g) + λxi (g) = ζ8j = 0.
0≤i≤2 0≤i≤2 0≤i≤2 j∈{1,2,3,5,6,7}

Hence, by Lemma 3.2 we conclude that g ∈ V(G), against the assumption A ⊆ N(G).
Hence C y 6= C, and then A = CC y = haiH by the maximality of C, so A is a cyclic H-module.

The next result is also key to classifying the finite groups G such that Pv (G) = 65 .

Proposition 3.15. Assume Setting 3.9 and that A ⊆ N(G). If A is an indecomposable H-module,
then A is a cyclic H-module.

Proof. By Remark 3.10, c(Q) = 3 and, since

ą
Q. Q/ ker(λG ),
λ∈Irr(A)−{1A }

there exists a nonprincipal α ∈ Irr(A) such that c(Q/K) = 3, where K = ker(αG ). As A ⊆ N(G),
A/K ⊆ N(G/K) and by Lemma 3.14, A/K is a cyclic H-module. So, A = A0 K where A0 =
ha0 iH is the cyclic H-module generated by a suitable element a0 ∈ A. Let N = [A0 , y, y]. Since

11
c(A0 hyi) = c(Q/K) = 3, an application of Lemma 2.3 and of Proposition 3.13 to G0 = A0 ⋊ H
yields that N is a unique irreducible H-submodule of A0 . As

N/(N ∩ K) ∼
= N K/K = [A0 K/K, y, y] = [A/K, y, y] > 1,

we deduce that N ∩K = 1, and by the uniqueness of N we conclude that A0 ∩K = 1, so A = A0 ×K.

Since A is an indecomposable H-module, then K = 1 and A = A0 is a cyclic H-module.

The next three results will conclude the description of the groups G satisfying Setting 3.9 and
such that Pv (G) = 65 .

Lemma 3.16. Assume Setting 3.9 and let Y = hyi. Suppose that A = B × C, where B = hbiH
and C = hciH , and let D = hbciH . If A ⊆ N(G), then

(1) c(DY ) = c(Q) = max{c(BY ), c(CY )} = 3 and exp(D) = max{exp(B), exp(C)}.

(2) exp(B) ≥ exp(C) if and only if c(BY ) ≥ c(CY ). In particular, exp(B) = exp(C) if and only
if c(BY ) = c(CY ).

(3) rank(D) = max{rank(B), rank(C)}.

Proof. By Remark 3.10, c(Q) = 3.

(1) As A = B × C where B, C are G-invariant subgroups of A and Q = AY where Y = hyi,

γi+1 (Q) = [A, Y, · · · , Y ] = [B, Y, · · · , Y ] × [C, Y, · · · , Y ]

| {z } | {z } | {z }
i i i

for i ≥ 0. Therefore, max{c(BY ), c(CY )} = c(AY ) = c(Q) = 3. Without loss of generality, we

may assume that c(BY ) = 3. To show that c(DY ) = 3, since c(DY ) ≤ c(Q) = 3, it suffices
to show that [bc, y, y] 6= 1, so [D, Y, Y ] 6= 1. An application of part (1) of Lemma 3.1 yields
[bc, y, y] = [b, y, y][c, y, y]. As A ⊆ N(G), by Lemma 2.3 B ⊆ N(BH) and the last statement of
Proposition 3.13 applied to BH gives [b, y, y] 6= 1. Therefore, as [b, y, y] ∈ B, [c, y, y] ∈ C and
B ∩ C = 1, we deduce that [bc, y, y] 6= 1.
Finally, we observe that o(b) = exp(B) and o(c) = exp(C), so exp(D) = o(bc) = max{o(b), o(c)} =
max{exp(B), exp(C)}.
(2) Set e = max{exp(B), exp(C)}. Then by (1) c(DY ) = c(Q) = 3 and o(bc) = exp(D) = e.
For F = hf iH ≤ A, as y acts as the inversion of [F, y] (by Lemma 3.1), c(F Y ) = 3 if and only if
[f, y]2 6= 1.
Assume, working by contradiction, that e = exp(B) ≥ exp(C) and c(BY ) < c(CY ) = 3. So,
e
/ C. Since [bc, y]2 = [b, y]2 [c, y]2 = [c, y]2 6= 1 (as c(CY ) = 3) and
[b, y]2 = 1. Observe that (bc) 2 ∈
B ∩ C = 1,
e e e
/ h[c, y]2 , [c, y]2x i = h[bc, y]2 , [bc, y]2x i.
(bc) 2 = b 2 c 2 ∈

12
This implies, in particular, that rank(D) = 4 and, applying Proposition 3.13 to DH, we deduce
that bc ∈ V(DH). Hence, bc ∈ V(G) by Lemma 2.3, a contradiction.
Assume now, again working by contradiction, that 3 = c(BY ) ≥ c(CY ) and exp(B) < exp(C).
Since [bc, y]2 = [b, y]2 [c, y]2 and [b, y]2 6= 1 (as c(BY ) = 3), we have
e e e e
/ h[b, y]2 , [b, y]2x i = h[bc, y]2 , [bc, y]2x i
(bc) 2 = b 2 c 2 = c 2 ∈

which as above implies bc ∈ V(G), a contradiction.

(3) We first show that rank(D) = 4 provided that either B or C has rank 4. To see this, we
may assume that rank(B) = 4. If rank(D) = 2, then D = hbc, (bc)x i and, for suitable integers i
and j,
[b, y][c, y] = [bc, y] = (bc)i (bc)jx = bi bjx (ci cjx ).

As A = B × C, where B and C are G-invariant subgroups of A, [b, y] = bi bjx (and [c, y] = ci cjx ).
So by ∈ hb, bx i and hence B = hb, bx i has rank 2, a contradiction.
We next show that rank(D) = 2 provided that both B and C have rank 2. In fact, otherwise
rank(D) = 4 by Remark 3.6. Without loss of generality, we may also assume that c(CY ) ≤
c(BY ) = 3. Hence, by (2), exp(B) ≥ exp(C). As A ⊆ N(G), an application of Lemma 2.3 to
both HB and HC yields that B ⊆ N(HB) and C ⊆ N(HC). Since rank(B) = 2, it follows by
Proposition 3.13 that bby = b4x (for a suitable generator x of P ) and o(b) = exp(B) = 8. As
o(c) = exp(C) ≤ exp(B) = o(b), then o(bc) = exp(D) = 8. Now, applying Lemma 2.3 to HD,
i
we have that D ⊆ N(HD). Since rank(D) = 4, Proposition 3.13 implies that [(bc)2 , y] = (bc)4x
where i = 1 or −1. Observing that
i i i
[b2 , y][c2 , y] = [(bc)2 , y] = (bc)4x = b4x c4x
i
and that A = B × C where B and C are G-invariant, we deduce that b−2 (b2 )y = [b2 , y] = b4x .
As bby = b4x , then b2 (b2 )y = (bby )2 = b8x = 1 where the last equality holds as o(b) = 8. Thus,
i
[b2 , y] = [b, y]2 = (b−2 bby )2 = b−4 = b4 , so b4 = b4x with i ∈ {1, −1}, and hence 1 6= b4 ∈ CA (P ),
a contradiction.

Lemma 3.17. Assume Setting 3.9 and let Y = hyi. Suppose that A = B × C, where B and C are
cyclic H-modules such that exp(B) ≥ exp(C). If A ⊆ N(G), then one of the following happens.

(1) c(BY ) = c(CY ) = 3, and B and C are isomorphic H-modules.

(2) c(BY ) = 3 > c(CY ), rank(B) ≥ rank(C) and exp(B) > exp(C). In addition, if rank(B) = 2,
then y acts as the inversion on C.

Proof. Let B = hbiG , C = hciG and D = hbciG . So, e = exp(B) = exp(A), as exp(B) ≥ exp(C).
Observe that c(Q) = 3 by Remark 3.10. By Lemma 3.16, 3 = c(DY ) = c(BY ) ≥ c(CY ),

13
exp(D) = e and rank(D) = max{rank(B), rank(C)}. Also, since A ⊆ N(G), an application of
Lemma 2.3 to HB, HC and HD yields that B ⊆ N(HB), C ⊆ N(HC) and D ⊆ N(HD).
e i e
We show first that if c(BY ) = 3 and bby = b 2 x for i ∈ {−1, 1}, then [b2 , y] 6= b 2 x . In fact,
e i
bby = b 2 x implies that rank(B) = 2. As BY satisfies Setting 3.9 and B ⊆ N(HB), Proposition
i i
3.13 yields e = 8 and bby = b4x . So, [b, y] = b−2 b4x and [b2 , y] = [b, y]2 = b−4 = b4 6= b4x , as
1 6= b4 6∈ CA (P ).

(1) Suppose that c(BY ) = c(CY ) = 3. So, exp(D) = exp(C) = exp(B) = e by Lemma 3.16.
We first show that rank(B) = rank(C). Working by contradiction, we may assume that rank(B) =
4, so rank(D) = 4, and that rank(C) = 2. As D ⊆ N(HD) and C ⊆ N(HC) and both HD and
HC satisfy Setting 3.9, Proposition 3.13 yields that e = 8 and, for a suitable generator x of P ,
i
[b2 , y][c2 , y] = [(bc)2 , y] = (bc)4x = b4x c4x , so [c2 , y] = c4x , and also that ccy = c4x for i ∈ {−1, 1},
against the second paragraph of this proof.
Next, we show that B and C are isomorphic H-modules. Assume first that rank(B) =
rank(C) = 2. Then rank(D) = 2. Since D ⊆ N(HD), an application of Proposition 3.13 to HD
yields that e = 8 and, for a suitable generator x of P , we may assume that (bc)(bc)y = (bc)4x .
Thus, bby = b4x and ccy = c4x . Defining ϕ(b) = c, ϕ(bx ) = cx , it is routine to check that ϕ extends
to an isomorphism of H-modules from B to C.
Assume now that rank(B) = rank(C) = 4. Then rank(D) = 4. Since D ⊆ N(HD), an
application of Proposition 3.13 to HD yields that, for a suitable generator x of P , [(bc)2 , y] =
e e e e
(bc) 2 x . Hence, as above [b2 , y] = b 2 x and [c2 , y] = c 2 x . Let b0 = wb 4 x , where w = [b, y]; so, w =
e e e2 −1
b− 4 x b0 . Then o(b0 ) = 2, as b20 = 1 and b0 6= 1 by Remark 3.6. We show that [b0 , y] = b− 2 x b− 16 x
and that B = hb, bx i × hb0 , bx0 i.
In fact, we have
e e e e e2 2
[b0 , y] = [wb 4 x , y] = [w, y][b 4 x , y] = w−2 w 4 x = b− 2 x b− 16 x .

where the third equality holds as [w, y] = w−2 (as y inverts [B, y]), and the last equality holds as
e
4 ≥ 2 = o(b0 ). In particular, it follows that [b0 , y] 6= 1. As Ω1 (hb, bx i) = hw2 , w2x i, y acts trivially
on Ω1 (hb, bx i) by part (1) of Lemma 3.1 and hence, as [b0 , y] 6= 1 and hb0 , bx0 i is an irreducible
P -module, we deduce that hb0 , bx0 i ∩ Ω1 (hb, bx i) = 1. Since B = hb, bx , by , bxy i = hb, bx , b0 , bx0 i,
e
we get B = hb, bx i × hb0 , bx0 i. Similarly, C = hc, cx i × hc0 , cx0 i, with [c, y] = c− 4 x c0 and [c0 , y] =
e e2 −1 e
c− 2 x c− 16 x , where c0 = [c, y]c 4 x . Defining ϕ(b) = c, ϕ(bx ) = cx , ϕ(b0 ) = c0 and ϕ(bx0 ) = cx0 , one
readily checks that ϕ extends to an isomorphism of H-modules from B to C.

(2) Suppose now that c(BY ) = 3 > c(CY ). Then by Lemma 3.16 e = exp(D) = exp(B) >
exp(C) and c(DY ) = 3. Next, we show that rank(B) ≥ rank(C). In fact, if rank(B) < rank(C)
then rank(B) = 2 and rank(C) = 4; so rank(D) = 4. Since HD satisfies Setting 3.9 and D =
e
hbciG ⊆ N(HD), Proposition 3.13 yields that, for a suitable generator x of P , [(bc)2 , y] = (bc) 2 x .

14
 e i
Similarly, applying Proposition 3.13 to HB, we have that (e = 8 and) bby = b 2 x for i ∈ {−1, 1}.
e e i
So, [b2 , y] = b 2 x and bby = b 2 x for i ∈ {−1, 1}, against the second paragraph of this proof.
Finally, we assume that rank(B) = 2; then rank(C) ≤ 2 and rank(D) = 2. Then, using again
Proposition 3.13, o(bc) = 8 and for a suitable generator x of P

bc(bc)y = (bc)4x .

Thus ccy = c4x and, since o(c) < o(b) = 8, we deduce that cy = c−1 . Then, it easily follows that y
acts as the inversion on C.

Proposition 3.18. Assume Setting 3.9. Then A ⊆ N(G) if and only if A = B × C where B and
C are normal subgroups of G such that exp(B) > exp(C), [C, y, y] = 1, and either y acts as the
inversion on C and B is a homogeneous H-module of type (T1), or B is a homogeneous H-module
of type (T2).

Proof. We write A = B × C where B is a direct product of indecomposable H-modules, say Bi

for 1 ≤ i ≤ k, such that c(Bi hyi) > 2 and C is a H-module such that [C, y, y] = 1. Note that by
Lemma 2.3 and Remark 3.10, c(Bi hyi) = 3 for every 1 ≤ i ≤ k. By Proposition 3.15, Lemma 3.17
and Proposition 3.13, we can assume that k ≥ 2.

Suppose first that A ⊆ N(G). For i ∈ {2, . . . , k}, let Si = H(B1 × Bi ). Observe that
G = CG (B1 × Bi )Si , and hence an application of Lemma 2.3 yields B1 × Bi ⊆ N(Si ). As Si
satisfies Setting 3.9 and B1 and Bi are indecomposable H-modules, by Proposition 3.15, Lemma
3.17 and Proposition 3.13 we deduce, using again Lemma 2.3, that the Bj = hbj iG , for 1 ≤ j ≤ k,
are pairwise isomorphic H-modules of type (T1) or (T2). In the same way, applying Lemma 3.16
to (B1 × hciG )H, for c ∈ C, we have that exp(B) > exp(C). Finally, we assume that B is direct
product of isomorphic H-modules of type (T1) and we consider an element c ∈ C. Then part (2)
of Lemma 3.17 applied to (B1 × hciG )H yields cy = c−1 . So, y acts as the inversion on C.

Conversely, we assume that A = B × C, where B = B1 × · · · × Bk with Bi isomorphic

H-modules of type (T1) or of type (T2), and in the first case y acts as the inversion on C, for
1 ≤ i ≤ k, exp(B) > exp(C) and [C, y, y] = 1. We prove that A ⊆ N(G). Let a0 ∈ A, A0 = ha0 iG
and T = A0 H. Write a0 = b0 c0 , where b0 ∈ B and c0 ∈ C, and set B0 = hb0 iG , C0 = hc0 iG .
Therefore, A0 ≤ B0 × C0 . By Lemma 2.3, it suffices to show a0 ∈ N(T ). Also, by part (2) of
Lemma 3.3 we can assume c(A0 hyi) = 3. As c(C0 hyi) ≤ c(Chyi) ≤ 2 and (B0 × C0 )H satisfies
Setting 3.9, by Lemma 3.16 c(B0 hyi) = 3.
Suppose first that the modules Bi are of type (T1). Then gg y = g 4x for all g ∈ Bi , 1 ≤ i ≤ k.
Qk
Let b ∈ B0 and write b = i=1 gi , where gi ∈ Bi . Then
k
Y k
Y k
Y k
Y k
Y
bby = gi ( gi )y = (gi giy ) = gi4x = ( gi )4x = b4x .
i=1 i=1 i=1 i=1 i=1

15
This implies that rank(B0 ) = 2 and, since c(B0 hyi) = 3 and exp(B) = 8, also that exp(B0 ) = 8.
Let now a ∈ A0 and write a = bc with b ∈ B0 and c ∈ C0 . Recalling that y acts as the inversion
on C and that exp(C) < exp(B) = 8, we have

aay = bc(bc)y = (bby )(ccy ) = b4x = b4x c4x = (bc)4x = a4x .

This implies that rank(A0 ) = 2 and, as c(A0 hyi) = 3, that exp(A0 ) ≥ 8; so exp(A0 ) = exp(B) = 8.
Since T = A0 H satisfies Setting 3.9, an application of Proposition 3.13 to T yields a0 ∈ N(T ), as
desired.
n−1
Suppose now that the modules Bi are of type (T2). Then exp(B) = 2n ≥ 8 and g 2 = [g 2 , y]x
Qk
for all g ∈ Bi . Let a ∈ A0 and write a = bc with b ∈ B0 , b = i=1 bi where bi ∈ Bi , and c ∈ C0 .
Observe that o(c) < exp(B) = 2n and that o([c, y]) ≤ exp([C, y]) ≤ 2 (recall that [C, y, y] = 1 and
that y acts as the inversion on [C, y] by Lemma 3.1 if [C, y] 6= 1). So
k
Y k
Y k
Y k
Y
n−1 n−1 n−1 n−1
[a2 , y]x = [(c bi )2 , y]x = [c2 , y]x [b2i , y]x = b2i = c2 ( bi )2 = a2 . (.)
i=1 i=1 i=1 i=1

Since y acts as the inversion on [A0 , y] (by part (1) of Lemma 3.1), then [A0 , y] = h[a0 , y], [ax0 , y]i has
rank 2 and, recalling that A0 is a P -module and that c(A0 hyi) = 3, we deduce that [A0 , y] ∼ = (C4 )2 .
Thus, (.) implies that exp(A0 ) = 2n = exp(A). We next show that rank(A0 ) = 4. Assuming the
contrary, then A0 has rank 2, so it is a uniserial P -module. As c(A0 hyi) = 3 (and both CA0 (y) and
[A0 , y] are P -submodules of A0 ), then CA (y) ≤ [A0 , y]. Since by Lemma 3.1 A0 /CA (y) ∼
0 = [A0 , y], 0

we deduce that A0 ∼= (C8 )2 . So, exp(A) = exp(A0 ) = 8. As above, we write a0 = b0 c0 , with

Qk ąk
b0 ∈ B, b0 = i=1 gi , gi ∈ Bi , and c0 ∈ C. As A = ( Bi ) × C and [a0 , y] ∈ ha0 , ax0 i, we have
i=1
[gi , y] ∈ hgi , gix i for 1 ≤ i ≤ k. Also, since c(A0 hyi) = 3, there exists an index d, 1 ≤ d ≤ k, such
that c(hgd iG hyi) = 3 by Lemma 3.16. Let D = hgd iG . Since c(Dhyi) = 3 and rank(D) = 2, the
= (C8 )2 and [D, y] ∼
arguments used above similarly yield that D ∼ = (C4 )2 . Since Bd is of type (T2)
and exponent 8, it easily follows that |B| = |D|·22 , so D is a maximal P - submodule of Bd . Writing
Bd = hbd iG , we observe that bd 6∈ D, as D is H-invariant, so Bd = Dhbd , bxd i. Thus, D ∩ hbd , bxd i is
a P -submodule of D of order 16, and hence D ∩ hbd , bx i = [D, y]. Since [D, y] ≤ [Bd , y] ∼
d = (C4 )2 ,
we conclude that [Bd , y] = [D, y] ≤ hbd , bxd i, giving rank(Bd ) = 2, a contradiction.
n−1
Thus, rank(A0 ) = 4. As A0 = ha0 , ax0 ih[a0 , y], [a0 , y]x i and for every a ∈ A0 [a2 , y]x = a2 ,
n
where 2 = exp(A0 ), Proposition 3.13 yields a0 ∈ N(T ).

Remark 3.19. We remark that, using the notation of Proposition 3.18, if y acts as the inversion
on C and [C, y, y] = 1, then exp(C) divides 4.

16
4 Proof of the main theorem
We recall that, given a prime number p, an element g of a finite group G can be uniquely
written in the form g = gp gp′ , where gp is a p-element, gp′ is a p′ -element and gp and gp′ commute.
The element gp is called the p-part of g; we will use this notation below.

Lemma 4.1 ([12, Proposition 3.2].). Let A be an abelian normal subgroup of the group G such
that [G : A] ≤ 6 and [G : A] 6= 5. Then, for a ∈ A:

(1) if [G : A] 6= 3 and G has abelian Sylow 3-subgroups, then a ∈ V(G) if and only if a2 ∈ V(G);

(2) if [G : A] = 3, then a ∈ V(G) if and only if a3 ∈ V(G).

Lemma 4.2. Let G be a finite group such that Pv (G) < a. Then for every odd prime p, the Sylow
p-subgroups of G are abelian.

Proof. Let G be a minimal counterexample and P a nonabelian Sylow p-subgroup of G for some
odd prime p. Recalling that Pv (G) ≤ Pv (G) for every factor group G of G, the minimality of
G implies that Op′ (G) = 1. Since Pv (G) < a, by Theorem 2.7 A = N(G) is an abelian normal
1
subgroup of G. Hence, Pv (G) = 1 − [G:A] < a implies [G : A] = m ≤ 6. So, m = 6, p = 3 and
A ≤ P , so [G : P ] = 2 and [P : A] = 3. Using again the minimality of G, we deduce that P ′ is a
minimal normal subgroup of G. It follows that P ′ ≤ Z(P ) and hence (as [G : P ] = 2) |P ′ | = 3.
So, by Problem 2.13 and Problem 6.3 of [7] every nonlinear irreducible character of P vanishes on
P − Z(P ). Clifford theory hence yields that every χ ∈ Irr(G) such that P ′ 6≤ ker χ vanishes on
P − Z(P ), so A = Z(P ) and [P : A] ≥ 32 , a contradiction.

Lemma 4.3. Let G be an A-group; let Z = Z(G) and F = F(G). Then

(1) F(G/Z) = F/Z and Z(G/Z) = 1.

(2) If Pv (G) < a, then N(G) = F and Pv (G/Z) = Pv (G).

(3) If [G : F ] = 5, then Pv (G) < a if and only if [F : Z] is not divisible by 6.

Proof. (1) As G is a A-group, then G′ ∩ Z = 1 ([6, VI.14.3(b)]). Let W/Z = Z(G/Z); then
[W, G] ≤ Z ∩ G′ = 1, so W ≤ Z and hence W = Z, i.e. Z(G/Z) = 1. Let X/Z = F(G/Z); X/Z
is abelian, so [X, X] ≤ G′ ∩ Z = 1, thus X is a normal abelian subgroup of G and X ≤ F . Hence,
X = F.

(2) First, we show that N(G) = F . As Pv (G) < a, by Theorem 2.7, N(G) is an abelian normal
subgroup of G, so N(G) ≤ F . If G is nilpotent, than G (being an A-group) is abelian, and clearly
N(G) = G = F . If m = [G : N(G)] is a prime, then G is nonabelian and N(G) = F . Finally,

17
if m ∈ {4, 6}, then for every a ∈ F Lemma 4.1 yields a ∈ N(G), because a2 is a nonvanishing
element of G by Lemma 2.6. Hence, N(G) = F .
In order to show that Pv (G/Z) = Pv (G), we observe that G/Z is a A-group and Pv (G/Z) ≤
Pv (G) < a so, by what we have just proved and part (1), we have N(G/Z) = F/Z, hence
Pv (G/Z) = Pv (G).

(3) Assume [G : F ] = 5. By (1), G/Z is a Frobenius group with abelian kernel F/Z. If 6 divides
|F/Z|, then there is a normal subgroup N of G, with Z ≤ N , such that G = G/N is a Frobenius
group with kernel F = U × V , where U is a 2-group, V is a 3-group and U and V are minimal
normal subgroups of G. Then, by Lemma 3.3 of [12] N(G) = U ∪ V and Pv (G) ≥ Pv (G) > a.
Assume now that 6 does not divide [F : Z] and observe that F = G′ × Z ([6, VI.14.7(b)]).
For every a ∈ G′ and α ∈ Irr(F ), αG (a) is a sum of five |G′ |-th roots of unity, and |G′ | is coprime
both to 5 and to either 2 or 3. Hence, a ∈ N(G) by Lemma 2.1 and Lemma 3.2, so F ⊆ N(G) by
4
Lemma 2.2, proving that Pv (G) ≤ 5 < a.

We are now ready to prove Theorem A, which we state again.

Theorem 4.4. Let G be a finite group, Q ∈ Syl2 (G) and P ∈ Syl3 (G). Then Pv (G) < Pv (A7 ) = a
if and only if

(a) G is an A-group such that [G : F(G)] = m ≤ 6, and |F(G)/Z(G)| is not divisible by 6 if

m = 5.

(b) Q is nonabelian and G has an abelian normal subgroup A such that Q ∩ A = Z0 × D, where
Z0 = CQ∩A (P ) ≤ Z(G) and D = [Q ∩ A, P ], and one of the following holds:

(b1) |G/A| = 4 and Q ∩ A = Z(Q).

(b2) G/A ∼
= S3 , and, for some x ∈ P − A, 1 6= D = Z × Z x , where Z = CD (Q) is either
elementary abelian or isomorphic to C4 × (C2 )t , with t ≥ 0.

(b3) G/A ∼
= C6 , 1 6= D = B × C with B and C are normal subgroups of G such that
[C, Q, Q] = 1 and, if B 6= 1, then exp(B) > exp(C) and either every y ∈ Q − A acts
as the inversion on C and B is a homogeneous G/A-module of type (T1), or B is a
homogeneous G/A-module of type (T2).

Proof. Suppose that Pv (G) < a. By Theorem 2.7, A = N(G) is an abelian normal subgroup of
1
G, so Pv (G) = 1 − [G:A] and hence [G : A] = m ≤ 6. Also, for every odd prime number p, G has
abelian Sylow p-subgroups by Lemma 4.2. Let Q be a Sylow 2-subgroup of G. If Q is abelian then
G is an A-group, and A = F(G) by Lemma 4.3. Hence, [G : F(G)] = m ≤ 6 and, recalling also
part (3) of Lemma 4.3, we have case (a).

18
 We can hence assume that Q is nonabelian, so m is even. If m is a power of 2, then G has a
factor group isomorphic to Q and m = 4 by Lemma 2.5. Thus, m ∈ {4, 6}.
Clearly, Q ∩ A is the (unique) Sylow 2-subgroup of A; let Z0 = CQ∩A (P ) and D = [Q ∩ A, P ],
where P ∈ Syl3 (G). By coprime action, Q ∩ A = Z0 × D. As AP E G, by the Frattini argument we
get G = ANG (P ), and this implies that both Z0 and D are normal subgroups of G. Let K be the 2-
complement of A. By part (4) of Lemma 3.3, we have Z0 K/K ≤ Z(G/K), so [Z0 , G] ≤ Z0 ∩ K = 1
and hence Z0 ≤ Z(G).
If m = 4, then P ≤ A and Z0 = Q ∩ A = Z(Q) and we have case (b1).
Now, we assume that m = 6 and we first prove that Q′ ≤ D. In fact, setting L = DKP and
observing that L E G, if Q′ 6≤ D then G/L ∼
= Q/D is a nonabelian 2-group, so by Lemma 2.5
N(G/L)=X/L, where X/L = Z(G/L). Hence A = N(G) ≤ X ∩ A, so A ≤ X, which is a
contradiction because 4 divides [G : X]. Thus, Q′ ≤ D and, in particular, D 6= 1. Let now
G = G/Z0 K. Note that A ⊆ N(G) and that N(G) is a subgroup of G by Theorem 2.7, so
m = [G : N(G)] divides 6. Since Q ∼
= Q is nonabelian, the argument in the second paragraph of
5
this proof yields m = 6 and hence Pv (G) = 6. Therefore, since D and A are isomorphic G/A-
modules, without loss of generality we can assume Z0 K = 1. Thus, A = D 6= 1, |P | = 3 and
CA (P ) = 1. By part (1) of Lemma 3.3 there exists an involution y ∈ NQ (P ), such that Q = Ahyi
and such that H = P hyi is a complement of A in G; in particular, |H| = 6. If H ∼ = S3 , then by
Proposition 3.8 we have case (b2). In order to conclude this part of the proof, we can hence assume
H=∼ C6 and c(Q) ≥ 3, since if c(Q) ≤ 2, then [D, Q, Q] = 1, so we have case (b3) with D = C and
B = 1. Hence, G satisfies all the conditions of Setting 3.9, and Proposition 3.18 yields case (b3).

We now prove the other implication. In case (a), G is an A-group with m = [G : F(G)] ≤ 6.
1
If m 6= 5, then F(G) ⊆ N(G) by Lemma 4.1 and Lemma 2.6. Thus, Pv (G) ≤ 1 − m < a. If m = 5,
then Pv (G) < a by part (3) of Lemma 4.3.
5
So, we assume (b) and we prove that A ⊆ N(G), which implies Pv (G) ≤ 6 < a. Let K be
the 2-complement of A. We observe that A = D × Z0 × K, with Z0 ≤ Z(G) and D E G (as
G = ANG (P )). For a ∈ A, we write a = dzk with d ∈ D, z ∈ Z0 and k ∈ K, and we observe that
by Lemma 4.1 a ∈ N(G) if and only if a2 = dz ∈ N(G). If we are in case (b1), then Q ∩ A ⊆ N(G)
by Lemma 2.6. For the remaining two cases, we observe that by Lemma 2.2 dz ∈ N(G) if and only
if d ∈ N(G) and that, setting G = G/Z0 K, d ∈ N(G) if and only d ∈ N(G) by Lemma 2.4. Since
G satisfies Setting 3.4, in case (b2) we have D ⊆ N(G) by Proposition 3.8, hence A ⊆ N(G). So,
we are left with case (b3) and G/A ∼ = C6 . If B = 1, then c(Q) ≤ c(Q) = 2 and hence D ⊆ N(G)
by part (2) of Lemma 3.3. If B 6= 1, then the assumptions on B and C imply that c(Q) = 3.
Therefore, G satisfies all the conditions of Setting 3.9, so D ⊆ N(G) by Proposition 3.18 and hence
A ⊆ N(G), concluding the proof.

19
 Finally, we give a description of the groups of case (a) of Theorem A.

Theorem 4.5. Let G be an A-group with trivial center. Then 2 ≤ [G : F(G)] ≤ 6 if and only if
G has an abelian normal subgroup A such that one of the following occurs.

(1) G is a Frobenius group with kernel A and complement H ∼

= Cm , 2 ≤ m ≤ 6.

(2) G = A ⋊ Q, with A of odd order and either

(2i) Q = hyi ∼
= C4 and A = C × D, with C = CA (hy 2 i), D = [A, hy 2 i] and both C ⋊ Q/hy 2 i
and D ⋊ Q are Frobenius groups; or

(2ii) Q = {1, y1 , y2 , y3 }, with yi involutions, and A = A1 × A2 × A3 , Ai = CA (yi ), and

Q
( j6=i Aj ) ⋊ hyi i are Frobenius groups, for i = 1, 2, 3.

(3) G = AP0 Q0 , for P0 , Q0 ≤ G, |P0 | = 3, |Q0 | = 2, [P0 , Q0 ] = 1, A = B × C × D with

B = CA (P0 ), C = CA (Q0 ), and both BD ⋊ Q0 and CD ⋊ P0 are Frobenius groups.

(4) G = AP Q, with P ∈ Syl3 (G) and Q ∈ Syl2 (G), |Q| = 2, |P/P ∩ A| = 3, [P, Q] = P ,
A = C × D, with C = CA (P ) and D = [P, A] = CA (Q) normal subgroups of G, and both
CP ⋊ Q and D ⋊ P/P ∩ A are Frobenius groups.

Proof. Let G be an A-group such that Z = Z(G) = 1. If G satisfies one of the conditions (1) − (4),
then G is nonabelian, A ≤ F(G) (actually, A = F(G)) and [G : A] ≤ 6.

Conversely, writing A = F(G), we assume that 2 ≤ [G : A] ≤ 6 and show that one of (1) − (4)
follows. If [G : A] = r ∈ {2, 3, 5} and R ∈ Sylr (G), then CA (R) = 1 and A ∩ R = 1, because Z = 1
and R is abelian. So, G is a Frobenius groups with kernel A and cyclic complement of order r.
Next, we assume [G : A] = 4. Then G = AQ where Q ∈ Syl2 (G) and, as above, CA (Q) = 1
and A ∩ Q = 1, so |Q| = 4. If Q is cyclic, then either G is a Frobenius group with complement Q
or, denoting by Y the subgroup of order 2 of Q, by coprimality A = C × D, where C = CA (Y ) and
D = [A, Y ] are nontrivial normal subgroups of G, and both CQ/Y and DQ are Frobenius groups.
If instead Q is elementary abelian, say Q = {1, y1 , y2 , y3 }, with yi involutions, it is well known that
A is the product of the normal subgroups Ai = CA (yi ), i = 1, 2, 3, of G. Note that, for distinct
indices i, j and k, Ai ∩ Aj Ak = CAj (yi ) ∩ CAk (yi ) = 1 because any two involutions of Q generate
Q and CA (Q) = 1; hence, A = A1 × A2 × A3 . Observe also that at most one of the subgroups Ai
can be trivial, and that when this happens G is the direct product of two Frobenius groups with
Q
complements of order 2. In any case, ( j6=i Aj )hyi i are Frobenius groups, for i = 1, 2, 3.
So, we assume [G : A] = 6. Let P ∈ Syl3 (G), Q ∈ Syl2 (G) and observe that CA (P )∩CA (Q) ≤
Z = 1. As AP E G, then G = ANG (P ) and hence AQ ∩ NG (P ) contains a Sylow 2-subgroup Q0
of NG (P ) (because [NG (P ) : AQ ∩ NG (P )] = [G : AQ] = 3). But [Q0 ∩ A, P ] ≤ O2 (G) ∩ P = 1,
so Q0 ∩ A ≤ CA (Q) ∩ CA (P ) = 1 and hence Q = (Q ∩ A) ⋊ Q0 , with |Q0 | = 2.

20
 If G/A ∼
= C6 , by a similar argument we get P = (P ∩ A) ⋊ P0 , with |P0 | = 3 and P0 ≤ NG (Q);
so [P0 , Q0 ] = 1 and P0 Q0 , a cyclic group of order 6, is a complement of A in G. Since both P
and Q are abelian, coprimality considerations give A = B × C × D with B = CA (P ) = CA (P0 ),
C = CA (Q) = CA (Q0 ) and by the three subgroups lemma D = [A, P0 , Q0 ] = [A, Q0 , P0 ] (note also
that CA (P0 ) ∩ CA (Q0 ) = 1 implies CA (Q0 ) = [CA (Q0 ), P0 ] ≤ [A, P0 ], so C[A,P0 ] (Q0 ) = CA (Q0 )).
Since Z = 1, we have that both BDQ0 and CDP0 are Frobenius groups with complements,
respectively, of order 2 and of order 3.
Finally, we assume G/A ∼
= S3 . Observe that CG (A ∩ Q) is a normal subgroup of G that
contains AQ; but AQ/A is a non-normal maximal subgroup of G/A, hence A ∩ Q ≤ Z = 1 and
Q = Q0 has order 2. Let C = CA (P ) and D = [P, A]. Since P is abelian, A = C × D. We also
have P ∩ A = [P ∩ A, Q], because CP ∩A (Q) ≤ Z = 1, so [P, Q] = P (as G/A is nonabelian). Hence,
CCP (Q) = CC (Q)CP (Q) = 1, so CP Q is a Frobenius group with kernel CP and complement Q.
Since CD (P ) = C ∩ D = 1, also the semidirect product D ⋊ P/P ∩ A is a Frobenius group with
complement P/P ∩ A of order 3. We finally observe that D ≤ CA (Q), because if [D, Q] 6= 1 then
P Q/P ∩A would act fixed point freely on [D, Q], giving G/A ∼
= P Q/(P ∩A) cyclic, a contradiction.
So, by Dedekind’s law CA (Q) = D(CA (Q) ∩ C) = D, concluding the proof.

From part (1) of Lemma 4.3 we have the following.

Corollary 4.6. Let G be a nonabelian A-group. Then [G : F(G)] ≤ 6 if and only if G/Z(G)
satisfies one of the conditions (1) to (4) of Theorem 4.5.

References
[1] J. Brough, Non-vanishing elements in finite groups, J. Algebra 460 (2016), 387–391.

[2] S. Dolfi, E. Pacifici and L. Sanus, On zeros of characters of finite groups, Group Theory and
Computation, Indian Stat. Inst. Ser., Springer, Singapore, 2018, pp. 41–58.

[3] S. Dolfi, E. Pacifici, L. Sanus and P. Spiga, On the vanishing prime graph of solvable groups,
J. Group Theory 13 (2010), 189–206.

[4] M. Grüninger, Two remarks about non-vanishing elements in finite groups, J. Algebra 460
(2016), 366–369.

[5] M. Harris, On p′ -automorphisms of abelian p-groups, Rocky Mountain J. Math. 7 (1977),

751–752.

[6] B. Huppert, Endliche gruppen I, Springer-Verlag, Berlin, 1983.

[7] I.M. Isaacs, Character theory of finite groups, AMS Chelsea Publishing, Providence, RI, 2006.

21
 [8] I.M. Isaacs, G. Navarro and T.R. Wolf, Finite group elements where no irreducible character
vanishes, J. Algebra 222 (1999), 413–423.

[9] T.Y. Lam and K.H. Leung, On vanishing sums of roots of unity, J. Algebra 224 (2000),
91–109.

[10] A. Moretó and P.H. Tiep, Nonsolvable groups have a large proportion of vanishing elements,
Israel J. Math. 254 (2023), 229–242.

[11] L. Morotti and H.P. Tong-Viet, Proportions of vanishing elements in finite groups, Israel J.
Math. 246 (2021), 441–457.

[12] Y. Zeng, D. Yang and S. Dolfi, On the proportion of vanishing elements in finite groups, J.
Algebra 614 (2023), 825–847.

22