1. Introduction
There are some remarkable connections between normal coverings of groups and algebraic number fields. In particular, various open conjectures regarding Kronecker classes of field extensions can be reformulated as questions about covering groups by conjugates of subgroups. The interplay between these two subjects is detailed in [Reference Klingen7] and summarised in [Reference Bubboloni, Spiga and Weigel1, Section 1.7]; we also refer the reader to the papers [Reference Jehne3, Reference Klingen6, Reference Praeger8, Reference Praeger9]. For applications to Kronecker classes of field extensions, the most important open problem is the following conjecture of Neumann and Praeger [Reference Praeger9, Conjecture 4.3] (see also the Kourovka Notebook [Reference Khukhro and Mazurov5, 11.71]).
Conjecture 1.1 (Neumann and Praeger, 1990).
There exists a function
$f\colon \mathbb {N} \to \mathbb {N}$
such that the following holds. Let G be a finite group, let
$U \leqslant G$
and let
$\operatorname {Inn}(G) \leqslant A \leqslant \operatorname {Aut}(G)$
with
$|A:\operatorname {Inn}(G)| = n$
. If
$G = \bigcup _{a \in A} U^a$
, then
$|G:U| \leqslant f(n)$
.
This conjecture remains open. Establishing the veracity of this conjecture has important applications not only in number theory, but also in investigating Erdős–Ko–Rado-type theorems in algebraic combinatorics (see for instance [Reference Bubboloni, Spiga and Weigel1, Sections 1.4 and 1.6] or [Reference Fusari, Previtali and Spiga2, Section 1.1]). The conjecture can also be seen as a vast generalisation of the classical theorem of Jordan [Reference Jordan4] that a finite group cannot be covered by conjugates of a proper subgroup.
In 1994, Praeger gave two significant pieces of evidence for Conjecture 1.1. The first is a proof of Conjecture 1.1 when U is a maximal subgroup of G [Reference Praeger9, Theorem 4.4] and the second is Theorem 1.2 below, which, roughly speaking, establishes Conjecture 1.1 when G has bounded A-chief length [Reference Praeger9, Theorem 4.6]. To state this more precisely, we need some notation.
Let G be a finite group and let A be a subgroup of the automorphism group
$\operatorname {Aut}(G)$
. Then, the A-core of U, written
$U_A$
, and the A-cocore of U, written
$U^A$
, are defined by
$$ \begin{align*} U_A = \bigcap_{a \in A} U^a \quad \text{and} \quad U^A = \bigcup_{a \in A} U^a. \end{align*} $$
An A-chief series for G is an unrefinable A-invariant series
$G = G_0> \cdots > G_c = 1$
. By the Jordan–Hölder theorem, every A-chief series for G has the same length, called the A-chief length of G, written
$\ell _A(G)$
.
We are now in a position to state [Reference Praeger9, Theorem 4.6].
Theorem 1.2. There exists a function
$g\colon \mathbb {N} \times \mathbb {N} \to \mathbb {N}$
such that the following holds. Let G be a finite group, let
$U \leqslant G$
and let
$\operatorname {Inn}(G) \leqslant A \leqslant \operatorname {Aut}(G)$
with
$|A:\operatorname {Inn}(G)| = n$
. Write
$c = \ell _A(G/U_A)$
. If
$G = \bigcup _{a \in A} U^a$
, then
$|G:U| \leqslant g(n,c)$
.
Unfortunately, the proof of [Reference Praeger9, Theorem 4.6] is not correct. There is an error on page 29 (on line 7 counting from the bottom of the page). The mistake is subtle. Using the notation from the proof of [Reference Praeger9, Theorem 4.6], it occurs in the inductive step under the assumption that
$\overline {G} = Z_p^k$
. In the degenerate case when
$\overline {G}=\overline {U}$
, the inequality
$p \leqslant |\overline {G}:\overline {U}|$
is false and no upper bound on p is derived in this case. To the best of our knowledge, this error was first observed by the third author in the preparation of [Reference Fusari, Previtali and Spiga2].
The purpose of this paper is to give a complete proof of Theorem 1.2. We were unable to fix the precise error in Praeger’s argument, so we give an alternative proof.
As a consequence of how we structure our proof, we also verify another case of Conjecture 1.1, namely, when U supplements a minimal A-invariant subgroup of G (see Proposition 3.1 below).
2. Preliminaries
2.1. Diagonal subgroups
Let T be a finite group and let k be a positive integer. It will be useful to introduce notation for some particular subgroups of
$T^k$
. For
${j \in \{1,\ldots ,k\}}$
, we write
$$ \begin{align*} T_j = \{ (x_1, \ldots, x_k) \in T^k \mid x_i = 1 \text{ if } i \neq j \}, \end{align*} $$
and more generally, for
$\Delta \subseteq \{1,\ldots ,k\}$
, we write
$$ \begin{align} T_\Delta = \{ (x_1, \ldots x_k) \in T^k \mid x_i = 1 \text{ if } i \not\in \Delta \}, \end{align} $$
so
$T_\emptyset =1$
,
$T_{\{i\}} = T_i$
and
$T_{\{1,\ldots ,k\}} = T^k$
. A subgroup
$H \leqslant T^k$
is called a diagonal subgroup if there exist
$\phi _1, \ldots , \phi _k \in \operatorname {Aut}(T)$
such that
$H = \{ (x^{\phi _1}, \ldots , x^{\phi _k}) \mid x \in T \}$
.
Lemma 2.1. Let T be a finite nonabelian simple group, let k be a positive integer, let
$A \leqslant \operatorname {Aut}(T^k)$
and let
$U < T^k$
such that
$U^A = T^k$
. Then there is a partition
$\{ \Delta _1, \ldots , \Delta _t \}$
of
$\{1, \ldots , k\}$
such that
$U = D_1 \times \cdots \times D_t$
, where
$D_i$
is a diagonal subgroup of
$T_{\Delta _i}$
for
$1 \leqslant i \leqslant t$
.
Proof. Fix i with
$1 \leqslant i \leqslant k$
and let
$\pi _i\colon G \to T$
be the projection
$(x_1, \ldots , x_k) \mapsto x_i$
onto the ith direct factor of
$T^k$
. We claim that
$U\pi _i = T$
. To see this, let
$x \in T$
be arbitrary and consider
$g = (x, \ldots , x) \in T^k$
. Since
$U^A = T^k$
, there exists
$a \in A$
such that
$g^a \in U$
. Since
$\operatorname {Aut}(T^k) = \operatorname {Aut}(T) \wr \operatorname {Sym}_k$
, there exists
$b \in \operatorname {Aut}(T)$
such that the ith component of
$g^a$
is
$x^b$
. In particular, x is
$\operatorname {Aut}(T)$
-conjugate to an element of
$U\pi _i$
. Therefore,
$$ \begin{align*} T = \bigcup_{a\in \operatorname{Aut}(T)}(U\pi_i)^a= (U\pi_i)^{\operatorname{Aut}(T)}. \end{align*} $$
However, by [Reference Saxl11, Proposition 2], this forces
$U\pi _i = T$
, as claimed. The result now follows by Goursat’s lemma.
2.2. A bound on the order of a finite simple group
For the rest of the paper, fix a nondecreasing function
$h\colon \mathbb {N} \to \mathbb {N}$
such that for all finite nonabelian simple groups T, if the number of
$\operatorname {Aut}(T)$
-classes of elements of T is m, then
$|T| \leqslant h(m)$
. By [Reference Pyber10, Lemma 4.4], such a function exists and, as explained in the comments following [Reference Praeger9, Theorem 4.5], one can take
$h(m) = 2^{c \, (\log {m})^2 \log \log m}$
for an appropriate constant c.
3. Proof of the main result
Before proving Theorem 1.2, we begin with a special case that arises in the proof, namely, in the notation of Theorem 1.2, when the subgroup U supplements a minimal A-invariant subgroup of G. In this case, we can actually prove the more general Conjecture 1.1, so this may be of independent interest.
Proposition 3.1. There exists a function
$f\colon \mathbb {N}\to \mathbb {N}$
such that the following holds. Let G be a finite group, let
$U \leqslant G$
and let
$\operatorname {Inn}(G) \leqslant A \leqslant \operatorname {Aut}(G)$
with
$|A:\operatorname {Inn}(G)| = n$
. Assume that there exists a minimal A-invariant subgroup L of G such that
$G = UL$
. If
$G = \bigcup _{a \in A} U^a$
, then
$|G:U| \leqslant f(n)$
.
Proof. Let
$h\colon \mathbb {N} \to \mathbb {N}$
be the function from Section 2.2 and define
$f:\mathbb {N}\to \mathbb {N}$
by
$$ \begin{align*} f(n)=\max\{n, \, h(n)^{n^2}\}. \end{align*} $$
Assume that
$G = \bigcup _{a \in A} U^a$
. If
$L \leqslant U$
, then
$U=UL=G$
and
$|G:U|=1 \leqslant f(n)$
, so the result holds. Therefore, for the rest of the proof, we may assume that
$L\not \leqslant U$
. In particular,
$U\cap L < L$
, so the A-core
$(U\cap L)_A$
is trivial since L is a minimal A-invariant subgroup of G.
First, assume L is abelian. Certainly, 
and, since L is abelian, we also have 
, so 
. Therefore, the set
$\{ (U \cap L)^a \mid a \in A \}$
has size at most
$|A:\operatorname {Inn}(G)| = n$
. Since
$L = (U \cap L)^A$
, we deduce that
$|L| \leqslant n |U \cap L|$
, so
$$ \begin{align*} |G:U| = |UL:U| = |L:U \cap L| \leqslant n \leqslant f(n). \end{align*} $$
Next, assume L is nonabelian. Then,
$L = T^k$
for a positive integer k and a nonabelian simple group T. Fix a partition
$\mathcal {S} = \{\Sigma _1, \ldots , \Sigma _r\}$
of
$\{1, \ldots , k\}$
such that
$$ \begin{align*} \{T_i \mid i \in \Sigma_1\}, \ldots, \{ T_i \mid i \in \Sigma_r\} \end{align*} $$
are the G-orbits on the set
$$ \begin{align*} \Omega=\{T_1, \ldots, T_k\} \end{align*} $$
of simple factors of L. Since
$G = UL$
and L acts trivially on
$\Omega $
, the G-orbits are also U-orbits. Since L has no proper nontrivial A-invariant subgroups, A acts transitively on the set
$\Omega $
, and since 
, the G-orbits form a system of imprimitivity for A. This means that r divides k, and writing
$k=rs$
, we have
$|\Sigma _i| = s$
for
$1 \leqslant i \leqslant r$
. Moreover, r divides
$n = |A:\operatorname {Inn}(G)|$
, so, in particular,
$$ \begin{align} r \leqslant n. \end{align} $$
For ease of notation, we will assume that
$\Sigma _i = \{(i-1)s+1, \ldots , is \}$
for
$1 \leqslant i \leqslant r$
. Moreover, we will write elements of
$L = T^k$
in the form
$(x_{11}, \ldots , x_{1s} \mid \ldots \mid x_{r1}, \ldots , x_{rs})$
, where
$x_{ij} \in T_{(i-1)s+j}$
. We will refer to
$T_{\Sigma _1},\ldots,T_{\Sigma _r}$
as the blocks of L. (Recall from (2.1) that
$T_{\Sigma _i}=\prod _{j \in \Sigma _i}T_j$
.)
Since
$(U \cap L)^A = L$
, by Lemma 2.1, we may fix a partition
$\mathcal {D} = \{ \Delta _1, \ldots , \Delta _t \}$
of
$\{1, \ldots , k\}$
such that
$$ \begin{align*} U \cap L = D_1 \times \cdots \times D_t, \end{align*} $$
where
$D_i$
is a diagonal subgroup of
$T_{\Delta _i}$
for
$1 \leqslant i \leqslant t$
. Since
$U\cap L<L$
, for at least one i with
$1 \leqslant i \leqslant t$
, we have
$|\Delta _i| \geqslant 2$
. By relabelling the parts of
$\mathcal {D}$
, without loss of generality, we may assume that
$|\Delta _1| \geqslant 2$
.
Let m be the number of
$\operatorname {Aut}(T)$
-classes in T. Suppose that
$m> r$
. Then, we may fix nontrivial elements
$x_1, \ldots , x_r \in T$
that represent distinct
$\operatorname {Aut}(T)$
-classes. Consider
$$ \begin{align*} g = (x_1,1,\ldots,1 \mid x_2,1,\ldots,1 \mid \ldots \mid x_r, 1, \ldots, 1) \in T_{\Sigma_1}\times T_{\Sigma_2}\times\cdots\times T_{\Sigma_r}= L. \end{align*} $$
Since
$(U \cap L)^A = L$
, there exists
$a \in A$
such that
$g^a \in U \cap L$
. Fix
$l \in \Delta _1$
. Since U acts transitively on the set of factors of each block of L (that is, U acts transitively by conjugation on the simple direct factors of
$T_{\Sigma _j}$
for each j), there exists
$u \in U$
such that the lth entry of
$g^{au}$
is nontrivial. Now,
$g^{au} \in U \cap L$
, since 
, so
$D_1$
contains an element with a nontrivial lth entry. However, all nontrivial entries of g, and hence
$g^{au}$
, are pairwise not
$\operatorname {Aut}(T)$
-conjugate, so
$|\Delta _1| = 1$
, which is a contradiction. This contradiction implies
$$ \begin{align} m \leqslant r. \end{align} $$
Suppose that
$s> r$
. Let
$x \in T$
be nontrivial and consider
$$ \begin{align*} g &= (x,1,\ldots,1 \mid \ldots \mid x, 1, \ldots, 1) \in L, \\ h &= (1,\ldots,1 \mid x,1,\ldots,1 \mid x,x,1,\ldots,1 \mid \ldots \mid x, \ldots x, 1, \ldots, 1) \in L. \end{align*} $$
Since
$(U \cap L)^A = L$
, there exist
$a,b \in A$
such that
$g^a, h^b \in U \cap L$
. To prove our claim, we will need to establish two secondary claims.
First, let
$1 \leqslant i \leqslant t$
. We claim that
$|\Delta _i \cap \Sigma _j| \leqslant 1$
for
$1 \leqslant j \leqslant r$
. Fix
$l \in \Delta _i$
. As before, there exists
$u \in U$
such that the lth entry of
$g^{au}$
is nontrivial and
$g^{au} \in U \cap L$
, so
$D_i$
contains an element with a nontrivial lth entry. However, in each block, exactly one entry of g, and hence
$g^{au}$
, is nontrivial, so
$D_i$
is supported on at most one factor of each block, or said otherwise,
$|\Delta _i \cap \Sigma _j| \leqslant 1$
for
$1 \leqslant j \leqslant r$
, as required.
In particular, the above claim implies that
$1, \ldots , s$
are contained in distinct parts of the partition
$\{\Delta _1, \ldots , \Delta _t\}$
. By relabelling the coordinates (in a manner that preserves the block system) if necessary, without loss of generality, we may assume that
$1 \in \Delta _1$
, and by relabelling
$\Delta _2, \ldots , \Delta _t$
if necessary, without loss of generality, we may assume that
$i \in \Delta _i$
for
$1 \leqslant i \leqslant s$
.
Second, let
$S\! = \{ j \mid |\Delta _1 \cap \Sigma _j| \!=\! 1 \}$
and let
$1 \leqslant \! i\! \leqslant s$
. We claim that
$S\! = \{ j \mid |\Delta _i \cap \Sigma _j| \!=\! 1 \}$
. Since U acts transitively on the set of factors of each block of L, there exists
$u \in U$
such that
$T_1^u = T_i$
and hence
$D_1^u = D_i$
since
$1 \in \Delta _1$
and
$i \in \Delta _i$
. However, U acts trivially on the set of blocks of G, so
$D_1$
and
$D_i$
are supported on the same blocks, or said otherwise,
$S = \{ j \mid |\Delta _i \cap \Sigma _j| = 1 \}$
, as required.
Since
$|\Delta _1 \cap \Sigma _j| \leqslant 1$
for
$1 \leqslant j \leqslant r$
, we must have
$|\Delta _1| = |S|$
, so, in particular,
$|S| \geqslant 2$
. By construction,
$1 \in \Delta _1$
and
$1 \in \Sigma _1$
, so
$1 \in S$
. Now, fix
$l> 1$
such that
$l \in S$
. Then,
$|\Delta _i \cap \Sigma _1| = |\Delta _i \cap \Sigma _l| = 1$
for
$1 \leqslant i \leqslant s$
. In particular, every element of
$U \cap L$
has the same number of nontrivial entries in block
$T_{\Sigma _1}$
and
$T_{\Sigma _l}$
. However, by the shape of the element h, this contradicts
$h^b \in U \cap L$
. This contradiction has arisen from the two claims that we proved under the supposition that
$s>r$
, so we conclude that
$$ \begin{align} s \leqslant r. \end{align} $$
Using the bounds in (3.1), (3.2) and (3.3),
$$ \begin{align*} |G:U| = |UL:U| = |L:U \cap L| \leqslant |L| = |T|^k \leqslant h(m)^k = h(m)^{rs} \leqslant h(n)^{n^2} \leqslant f(n), \end{align*} $$
which completes the proof.
It will be convenient to introduce some further notation for automorphism groups. Let G be a finite group, let
$A \leqslant \operatorname {Aut}(G)$
and let N be an A-invariant subgroup of G. Write
$$ \begin{align*} A|_N = \{ a|_N \mid a \in A \} \leqslant \operatorname{Aut}(N), \end{align*} $$
where
$a|_N$
denotes the restriction of the automorphism a to N, and similarly, write
$$ \begin{align*} A|_{G/N} = \{ a|_{G/N} \mid a \in A \} \leqslant \operatorname{Aut}(G/N), \end{align*} $$
where
$a|_{G/N}$
denotes the automorphism of
$G/N$
induced by a. Since every A-invariant series of G can be refined to an A-chief series of G, we see that
$\ell _{A|_N}(N) \leqslant \ell _A(G)$
with equality if and only if
$N=G$
, and
$\ell _{A|_{G/N}}(G/N) \leqslant \ell _A(G)$
with equality if and only if
$N = 1$
. For ease of notation, we write
$\ell _A(N)$
for
$\ell _{A|_N}(N)$
and
$\ell _{A}(G/N)$
for
$\ell _{A|_{G/N}}(G/N)$
.
We are now in a position to prove the main result of this paper.
Proof of Theorem 1.2.
Let
$h \colon \mathbb {N} \times \mathbb {N}$
be the function from Section 2.2, let
${f \colon \mathbb {N} \times \mathbb {N}}$
be a function that satisfies Proposition 3.1 and let
$g\colon \mathbb {N} \times \mathbb {N} \to \mathbb {N}$
be a function such that:
-
(i)
$g(n,0) \geqslant 1$
; -
(ii)
$g(n,c) \geqslant \max \{ g(n,c-1) \cdot g((n \cdot g(n,c-1))!,c-1), \, f(n) \}$
if
$c \geqslant 1$
; -
(iii) g is nondecreasing in both variables.
Assume that
$G = \bigcup _{a \in A} U^a$
. For ease of notation, write
$c = \ell _A(G/U_A)$
. We proceed by induction on
$|G|$
. In the base case where
$|G| = 1$
, we have
$|G:U| = 1 \leqslant g(1,0) = g(n,c)$
. For the inductive step, assume that
$|G|> 1$
. First, assume that
$U_A \neq 1$
. Since
$G/U_A = (U/U_A)^A$
and
$|G/U_A| < |G|$
, by induction,
$$ \begin{align*} |G:U| = |G/U_A : U/U_A| \leqslant g(n,c), \end{align*} $$
noting that
$|A|_{G/U_A}:\operatorname {Inn}(G/U_A)| \leqslant |A:\operatorname {Inn}(G)| = n$
and
$\ell _A((G/U_A)/(U/U_A)_A) = \ell _A(G/U_A) = c$
. Therefore, for the remainder of the proof, we may assume that
$U_A = 1$
.
Let L be a minimal A-invariant subgroup of G (we allow the possibility that
$G = L$
). If
$G=UL$
, then Proposition 3.1 ensures that
$|G:U| \leqslant f(n) \leqslant g(n,c)$
. Therefore, for the remainder of the proof, we may assume that
$UL < G$
.
Since
$G/L = (UL/L)^A$
and
$|G/L| < |G|$
, by induction,
$$ \begin{align} |G:UL| = |G/L:UL/L| \leqslant g(n,c-1), \end{align} $$
noting that
$|A_{G/L}:\operatorname {Inn}(G/L)| \leqslant |A:\operatorname {Inn}(G)| \leqslant n$
and
$\ell _A(G/L) = c-1$
.
Consider the A-core
$K = (UL)_A$
. Certainly,
$K \leqslant UL$
, so
$UK \leqslant UL$
, and as L is an A-invariant subgroup of
$UL$
, we have
$L \leqslant K$
and hence
$UL \leqslant UK$
. Therefore,
${UK = UL < G}$
, so
$K < G$
and hence
$\ell _A(K) < \ell _A(G) = c$
, or said otherwise,
$$ \begin{align} \ell_A(K) \leqslant c-1. \end{align} $$
For
$H \leqslant G$
, write
$\overline {H} = HZ(G)/Z(G)$
considered as a subgroup of
$\operatorname {Inn}(G) = G/Z(G)$
. Then, using (3.4),
$$ \begin{align*} |A:\overline{UL}| = |A:\operatorname{Inn}(G)| \cdot |\operatorname{Inn}(G):\overline{UL}| \leqslant |A:\operatorname{Inn}(G)| \cdot |G:UL| \leqslant n \cdot g(n,c-1). \end{align*} $$
Therefore,
$$ \begin{align} |{A|_K}:\operatorname{Inn}(K)| \leqslant |A:\overline{K}| = |A:\overline{(UL)_A}| = |A:\overline{UL}_A| \leqslant |A:\overline{UL}|! \leqslant (n \cdot g(n,c-1))!. \end{align} $$
Since
$K = (U \cap K)^A$
and
$|K| < |G|$
, by induction,
$$ \begin{align} |K:U \cap K| \leqslant g(|{A|_K}:\operatorname{Inn}(K)|, \ell_A(K)) \leqslant g((n \cdot g(n,c-1))!, c-1), \end{align} $$
where the final inequality follows from (3.5) and (3.6).
Using the bounds in (3.4) and (3.7),
$$ \begin{align*} |G:U| &= |G:UK| \cdot |UK:U| = |G:UL| \cdot |K:U \cap K| \\ &\leqslant g(n,c-1) \cdot g((n \cdot g(n,c-1))!, c-1) \leqslant g(n,c).\\[-2.7pc] \end{align*} $$
Acknowledgement
The second author wishes to thank the University of Milano-Bicocca for its hospitality.
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