ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P1.09
https://doi.org/10.26493/1855-3974.2975.1b2
(Also available at http://amc-journal.eu)
Products of subgroups, subnormality,
and relative orders of elements
Luca Sabatini *
Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences
Reáltanoda utca 13-15, H-1053, Budapest, Hungary
Received 6 October 2022, accepted 22 February 2023, published online 31 August 2023
Abstract
Let G be a group. We give an explicit description of the set of elements x ∈ G such
that x|G:H| ∈ H for every subgroup of finite index H ⩽ G. This is related to the following
problem: given two subgroups H and K, with H of finite index, when does |HK : H|
divide |G : H|?
Keywords: Relative order, product of subgroups, subnormal subgroup.
Math. Subj. Class. (2020): 20D40, 20D25, 20F99.
1 Introduction
Let G be an arbitrary group, and let us write H ⩽f G to say that H is a subgroup of G
of finite index. Let x ∈ G and H ⩽f G. If H is a normal subgroup of G, then it is
easy to see that x|G:H| ∈ H . The same is not true in general: fixed H ⩽f G, the set
{x ∈ G : x|G:H| ∈ H} may not even be closed under multiplication (take G = Sym(3)
and H = ⟨(1 2)⟩). The goal of this paper is to understand this phenomenom and its
implications. As far as we can see, this has not been dealt with before in the literature.
Definition 1.1. Let x ∈ G and H ⩽ G. The relative order of x with respect to H is
oH(x) := |⟨x⟩ : ⟨x⟩ ∩H|.
The following result is proved in Section 2.
Lemma 1.2. Let n ≥ 1. Then xn ∈ H if and only if oH(x) is finite and divides n.
*The author thanks Bob Guralnick and Orazio Puglisi for useful conversations.
E-mail address: sabatini.math@gmail.com (Luca Sabatini)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 24 (2024) #P1.09
Given H,K ⩽ G, |HK : H| is the cardinality of the set of all cosets of H which are
intersected by K (we refer to Section 2 for more details). Since oH(x) = |H⟨x⟩ : H|, we
obtain
Corollary 1.3. x|G:H| ∈ H if and only if |H⟨x⟩ : H| divides |G : H|.
If H,K ⩽f G, then |HK : H| divides |G : H| if and only if |HK : K| divides
|G : K|. If G is finite, both are equivalent to |HK| dividing |G|. In Section 3, we prove
the following two results:
Proposition 1.4. Let H ◁◁G. Then |HK : K| divides |G : K| for every K ⩽f G.
Theorem 1.5. Let H ⩽f G. Then H ◁ ◁ G if and only if |HK : H| divides |G : H| for
every K ⩽ G.
The converse of Proposition 1.4 is not true in general (see Example 5.11). In particular,
some attention is needed with subgroups of infinite index. During the preparation of this
manuscript, the author found out that the finite version of Theorem 1.5 already appeared in
[5, Theorem 2]. In Section 4, we study the following class of subgroups
Definition 1.6. A subgroup H ⩽f G is exponential if x|G:H| ∈ H for every x ∈ G.
This is a generalization of subnormality, and we prove that it is equivalent to normal-
ity in some cases, namely for the Hall subgroups of a finite group and for the maximal
subgroups of a solvable group. From the dual point of view, in Section 5 we study the set
S(G) := {x ∈ G : x|G:H| ∈ H for every H ⩽f G}.
At first glance S(G) is quite elusive, and indeed working directly with the definition is
not easy. Using the results of Section 3, we give an elementary proof of the next theorem.
Given N ◁ G, let FN (G) be the preimage of F (G/N), where F (G) denotes the Fitting
subgroup of G.
Theorem 1.7. If G is any group, then S(G) = ∩N◁fGFN (G).
In particular, S(G) = F (G) when G is finite (Proposition 5.1). Of course, Theorem 1.7
implies that S(G) is closed under multiplication, a fact which is not immediately clear from
the definition.
2 Preliminaries
We start with the proof of the key Lemma 1.2.
Proof of Lemma 1.2. Let ordH(x) := min{n ≥ 1 : xn ∈ H}. We first notice that
oH(x) = ordH(x). Indeed, from the definitions we have oH(x) = oH∩⟨x⟩(x) and
ordH(x) = ordH∩⟨x⟩(x). The fact that oH∩⟨x⟩(x) = ordH∩⟨x⟩(x) is a simple exercise.
Now, the “if” part of the statement is trivial. On the other hand, if xn ∈ H for some n ≥ 1,
then clearly ordH(x) < ∞. Let n = q · ordH(x) + r with r, q ≥ 0 and r < ordH(x).
Since H is a subgroup, the fact that xn = xq·ordH(x)xr ∈ H implies that xr ∈ H , which
in turn means r = 0.
L. Sabatini: Products of subgroups, subnormality, and relative orders of elements 3
The bulk of this paper is about finite groups. We summarize here the basic tools and
notation that are used with regard to general non-finite groups. Let G be an arbitrary group
and H,K ⩽ G. If H and K have finite index, then so has H ∩ K, and |G : H ∩ K| =
|G : H||H : H ∩ K|. As we have said in the introduction, we write |HK : H| for the
cardinality of the set of all cosets of H which are intersected by K. This is not accidental,
because the product set HK = {hk : h ∈ H, k ∈ K} is a union of cosets of H . It is not
relevant to distinguish between left-cosets and right-cosets, since k ∈ Hx if and only if
k−1 ∈ x−1H . We also observe that |HK : H| = |K : H ∩K| = |KH : H|.
The finite residual R(G) is the intersection of the subgroups of G of finite index. If R(G) =
1, then G is said to be residually finite. It is easy to check that G/R(G) is always residually
finite. Finally, the Fitting subgroup F (G) is defined as the subgroup generated by the
nilpotent normal subgroups, and coincides with the set of the elements x ∈ G such that the
normal closure ⟨x⟩G is nilpotent [1]. In general, this is a stronger condition than ⟨x⟩ being
subnormal in G. If G is finite, then F (G) itself is nilpotent, i.e. it is the largest nilpotent
normal subgroup.
3 Products of subgroups
The proof of Proposition 1.4 follows immediately from the following
Lemma 3.1. Let H ◁M ⩽ G, and let K ⩽f G. Then |HK : K| divides |MK : K|.
Proof. We have to prove that the ratio
|MK : K|
|HK : K|
=
|M : M ∩K|
|H : H ∩K|
is an integer. Now H ◁ M implies that H(M ∩ K) is a subgroup of M , and so we can
write
|M : M ∩K| = |M : H(M ∩K)||H(M ∩K) : M ∩K|
= |M : H(M ∩K)||H : H ∩K|.
In particular, the original ratio equals |M : H(M ∩K)|.
We continue with the easiest direction of Theorem 1.5.
Lemma 3.2. Let H ⩽f M ⩽f G, and let K ⩽ G. Then |HK:H||MK:M | = |M ∩K : H ∩K|.
Proof. We have
|HK : H|
|MK : M |
=
|K : H ∩K|
|K : M ∩K|
=
|K : M ∩K||M ∩K : H ∩K|
|K : M ∩K|
= |M ∩K : H ∩K|.
We prove the claim of Theorem 1.5 by induction on the subnormal defect of H , so let
H ◁f M ◁◁f G, and K ⩽ G. Using Lemma 3.2, we have
|G : H|
|HK : H|
=
|G : M ||M : H|
|MK : M ||M ∩K : H ∩K|
.
4 Ars Math. Contemp. 24 (2024) #P1.09
By induction, it is sufficient to prove that |M :H||M∩K:H∩K| is an integer. Now H ◁M implies
that H(M ∩K) is a subgroup of M , and so we can write
|M : H| = |M : H(M ∩K)||H(M ∩K) : H|
= |M : H(M ∩K)||M ∩K : H ∩K|.
This concludes the proof of the “only if” part.
3.1 The Kegel-Wielandt-Kleidman theorem, revisited
Definition 3.3. Let G be a finite group, H ⩽ G, and let p be a prime. Then H is p-
subnormal in G if H ∩ P is a p-Sylow of H for every p-Sylow P of G.
We characterize p-subnormality with the following
Lemma 3.4. A subgroup H is p-subnormal if and only if |HP | divides |G| for every p-
Sylow P ⩽ G.
Proof. We have that H ∩ P is a p-Sylow of H if and only if |H : H ∩ P | = |HP : P | is
not divisible by p. Since |H : H ∩ P | is a divisor of |G|, the last condition is equivalent to
|HP : P | dividing |G : P |, i.e. |HP | | |G|.
The famous Kegel-Wielandt conjecture [3, 7], proved by Kleidman [4] using the clas-
sification of the finite simple groups, says that H ◁ ◁ G whenever H is p-subnormal for
every p.
Theorem 3.5 (Kegel-Wielandt conjecture). If |HP | divides |G| for every Sylow subgroup
P ⩽ G, then H ◁◁G.
See [2] for some consequences of p-subnormality for a single p. The “if” part of Theo-
rem 1.5 follows easily. Let H ⩽f G, and assume that |HK : H| divides |G : H| for every
K ⩽ G. Let N ◁f G be the normal core of H , and let N ⩽ K ⩽ G be any intermediate
subgroup. Working with G/N and K/N , Theorem 3.5 gives H/N◁◁G/N , i.e. H◁◁G.
We point out that Kegel [3] did not use the classification to prove Theorem 3.5 when H
is solvable. We give a very short proof in the case where H is nilpotent, which is enough
for the characterization of S(G) we will present in Section 5.
Lemma 3.6 (Kegel-Wielandt for nilpotent subgroups). Let H ⩽ G be a nilpotent subgroup
of the finite group G. If |HP | divides |G| for every Sylow subgroup P ⩽ G, then H◁◁G.
Proof. Suppose that H is not subnormal, and in particular H ⩽̸ F (G). So there exists a
p-element x such that x ∈ H \ F (G). Since x /∈ Op(G), there exists a p-Sylow P of G
such that x /∈ P . By hypothesis H ∩P is a p-Sylow of H and, since H is nilpotent, H ∩P
contains all the p-elements of H . This contradicts the fact that x /∈ P .
Levy [5] proves the same result when H is a p-subgroup of G. Another consequence
of Theorem 1.5 is that p-subnormality for every p implies that |HK| divides |G| for every
K ⩽ G. We provide an elementary proof of this fact.
Lemma 3.7. Let G be a finite group and H ⩽ G. If |HP | divides |G| for every Sylow
P ⩽ G, then |HK| divides |G| for every K ⩽ G.
L. Sabatini: Products of subgroups, subnormality, and relative orders of elements 5
Proof. Let K ⩽ G. We have to show that |HK : K| = |H : H ∩K| divides |G : K|. Let
pα be a prime power that divides |H : H ∩K|. Since pα is arbitrary, it is sufficient to prove
that pα | |G : K|. Let P0 ⩽ K be a p-Sylow of K, and let P ⩽ G be a p-Sylow of G such
that P ∩K = P0. Of course, pα | |H : H ∩ P0|. By hypothesis |H : H ∩ P | = |HP : P |
divides |G : P |, and so is not divisible by p. Therefore, pα | |H ∩ P : H ∩ P0|. Now
|H∩P : H∩P0| = |(H∩P )P0 : P0|, and this divides |P : P0| because P is a p-group. So
pα | |P : P0|, and then of course pα | |G : P0|. Since p ∤ |K : P0|, we obtain pα | |G : K|
as desired.
4 Exponential subgroups
We write H ⩽exp G if x|G:H| ∈ H for all x ∈ G. We observe immediately that exponen-
tiality is preserved by quotients.
Lemma 4.1. Let N ◁ G, and N ⩽ H ⩽ G. Then H ⩽exp G if and only if H/N ⩽exp
G/N .
Proof. Let x ∈ G and H ⩽exp G. Then (Nx)|G/N :H/N | = Nx|G:H| ∈ H/N and so
H/N ⩽exp G/N . If H/N ⩽exp G/N , then Nx|G:H| = (Nx)|G/N :H/N | ∈ H , and so
x|G:H| ∈ H .
Since exponential subgroups have finite index, we can apply Lemma 4.1 with the nor-
mal core, and work with a finite group. Let G be a finite group and H ⩽ G. From
Corollary 1.3 and Theorem 1.5, we have
• H ◁◁G if and only if |HK| divides |G| for every K ⩽ G;
• H ⩽exp G if and only if |HC| divides |G| for every cyclic C ⩽ G.
We stress that H ⩽exp G whenever |G : H| is a multiple of the exponent exp(G).
Remark 4.2. Every finite group of order other than a prime has a non-trivial exponential
subgroup: if exp(G) < |G|, then it is sufficient to take any subgroup whose order divides
|G|/ exp(G). Otherwise, all the Sylow subgroups of G are cyclic, and it is well known
that G is solvable. In particular, G has a non-trivial normal subgroup, which is certainly
exponential.
We notice a difference with the stronger condition that HK is a subgroup for every K i.e.
H is a permutable subgroup. Indeed, it is easy to prove that if HC is a subgroup for every
cyclic C ⩽ G, then HK is a subgroup for every K ⩽ G.
For every n ≥ 1, let Gn := ⟨{xn : x ∈ G}⟩. The exponential subgroups of G of index
n are in correspondence with the subgroups of G/Gn of index n. Since Gn is characteristic,
the property of being exponential is preserved by automorphisms. Moreover, we have the
following
Lemma 4.3. Let H ⩽ G have a trivial characteristic core. Then H ⩽exp G if and only if
|G : H| is a multiple of the exponent of G.
Proof. Let n = |G : H|. By the exponentiality of H we have Gn ⩽ H . Since Gn is a
characteristic subgroup of G contained in H , we obtain Gn = 1. But this means exactly
that n is a multiple of exp(G). The converse is trivial.
6 Ars Math. Contemp. 24 (2024) #P1.09
In general, there exist non-subnormal exponential subgroups whose index is not a mul-
tiple of the exponent. A simple example is G = C4 × Sym(3) and H ∼= C2 × C2. The
following corollaries of Lemma 4.3 are obtained with the same strategy.
Corollary 4.4. Let H ⩽ G be a Hall subgroup. If H ⩽exp G, then H ◁G.
Proof. Suppose that H is not normal, and let N ◁G be the normal core of H . Since H/N
is a Hall subgroup of G/N , by induction and Lemma 4.1, we can assume that H is core-
free. Now exp(G) captures every prime dividing |G|, and so the contradiction is given by
Lemma 4.3.
Corollary 4.5. Let M ⩽ G be a maximal subgroup of the solvable group G. If M ⩽exp G,
then M ◁G.
Proof. Suppose that M is not normal, and let N◁G be the normal core of M . Since M/N
is a maximal subgroup of G/N , by induction and Lemma 4.1, we can assume that M is
core-free. Now |G : M | = qα for some prime power qα. If G is a q-group we are done.
Otherwise, the contradiction is given by Lemma 4.3.
We cannot drop the hypothesis of solvability in Corollary 4.5: the alternating group
G = Alt(10) has a conjugacy class of maximal subgroups M of size 720. Since exp(G) =
2520 = |G : M |, it appears that M is an exponential maximal subgroup which is not
normal.
We conclude this section with the hereditary properties of exponential subgroups.
Lemma 4.6. The following are true:
• If H ⩽exp M ⩽exp G, then H ⩽exp G;
• The intersection of exponential subgroups is exponential.
Proof. Let x ∈ G. Since M ⩽exp G, we have m = x|G:M | ∈ M . Then x|G:H| =
m|M :H| ∈ H . To prove the second statement, it is sufficient to notice that |G : H ∩K| is
a multiple of both |G : H| and |G : K|.
Other important properties of the lattice of the subnormal subgroups are not true for
exponential subgroups, and the dihedral group G = D12 is a good source of counterexam-
ples. Every subgroup of G whose order is 2 is exponential in G, since exp(G) = 6. Let H
be any non-central subgroup of order 2. Now
• The subgroup H1 = ⟨H,Z(G)⟩ ∼= C2 × C2 provides a counterexample to the state-
ment that two exponential subgroups generate an exponential subgroup: choosing
any involution x ∈ G \H1 we get x|G:H1| = x /∈ H1.
• The subgroup H2 which satisfies H < H2 ∼= Sym(3) provides a counterexample to
the statement that the intersection of an exponential subgroup of G with any subgroup
of G is exponential in that subgroup: choosing any involution x ∈ H2 \ H , we get
that H is not exponential in H2 although it is exponential in G.
L. Sabatini: Products of subgroups, subnormality, and relative orders of elements 7
5 The set S(G)
Let us recall the definition of S(G) given in the introduction:
S(G) := {x ∈ G : x|G:H| ∈ H for every H ⩽f G}.
From Corollary 1.3, we have
S(G) = {x ∈ G : |H⟨x⟩ : H| divides |G : H| for every H ⩽f G}.
The results of Section 3 allow to settle the finite case easily:
Proposition 5.1. If G is finite, then S(G) = F (G).
Proof. Let x ∈ G. Then x ∈ S(G) if and only if |H⟨x⟩| divides |G| for every H ⩽ G.
From Proposition 1.4 and Lemma 3.6, this is equivalent to ⟨x⟩◁◁G, i.e. x ∈ F (G).
5.1 A top-down approach
Let G be an arbitrary group and let R(G) = ∩H⩽fGH be its finite residual. The condition
in the definition of S(G) is empty on R(G), and so R(G) ⊆ S(G). In fact, S(G) is the
preimage of S(G/R(G)) under the projection G ↠ G/R(G).
Lemma 5.2. Let N ◁G. Then S(G/N) = {Nx : x|G:H| ∈ H for every N ⩽ H ⩽f G}.
In particular, S(G/R(G)) = S(G)/R(G).
Proof. Let x ∈ G and N ⩽ H ⩽f G. The equality (Nx)|G:H| = Nx|G:H| implies that
Nx ∈ H/N if and only if x|G:H| ∈ H , and the first part follows because H is arbitrary.
The second part follows because R(G) contains all the subgroups of G of finite index.
As a consequence of Lemma 5.2, we can assume that G is residually finite. Given
N ◁G, let FN (G) be the preimage of F (G/N).
Proof of Theorem 1.7. We have to prove that S(G) = ∩N◁fGFN (G). Let x ∈ S(G) and
N ◁f G. From Lemma 5.2 and Proposition 5.1 we have Nx ∈ S(G/N) = F (G/N), i.e.
x ∈ FN (G).
On the other hand, let x ∈ ∩N◁fGFN (G) and H ⩽f G. If N ◁f G is the normal core of
H , then in particular x ∈ FN (G). From Proposition 5.1 we have
Nx ∈ FN (G)
N
= F (G/N) = S(G/N),
and so Lemma 5.2 provides x|G:H| ∈ H . The proof follows because H is arbitrary.
The following observation deletes a bunch of terms from ∩N◁fGFN (G).
Lemma 5.3. Let G be a finite group and N ◁G. Then F (G) ⩽ FN (G).
Proof. We have that NF (G)/N ∼= F (G)/(N ∩ F (G)) is a nilpotent normal subgroup of
G/N . Then NF (G)/N ⩽ F (G/N) = FN (G)/N , and so NF (G) ⩽ FN (G).
Corollary 5.4. If N,K ◁f G and K ⩽ N , then FK(G) ⩽ FN (G).
As a particular case of Theorem 1.7, we have
8 Ars Math. Contemp. 24 (2024) #P1.09
Proposition 5.5. Let G be a group. The following are equivalent:
(A) G = S(G);
(B) every subgroup of finite index of G is exponential;
(C) every finite quotient of G is nilpotent;
(D) every subgroup of finite index of G is subnormal.
Proof. This follows easily from Theorem 1.7.
We say that a group G is S-free if S(G) = 1.
Lemma 5.6. Let G be a group which is residually S-free. Then S(G) = 1.
Proof. Let 1 ̸= x ∈ G. By definition, there exists N ◁G such that x /∈ N and S(G/N) =
1. In particular Nx /∈ S(G/N), and so from Lemma 5.2 we obtain x /∈ S(G). Since x is
arbitrary, it follows that S(G) = 1.
Corollary 5.7. If F is a finitely generated free group, then S(F ) = 1.
5.2 Baer groups and S-groups
Following a different approach, now we study S(G) starting from the subgroups of G. This
will provide a counterexample to the converse of Proposition 1.4.
Let B(G) := {x ∈ G : ⟨x⟩ ◁ ◁ G} be the Baer radical of G. It is clear that B(G) is
a characteristic subgroup. Moreover, B(G) coincides with F (G) if G is finite, but it can
be much larger in general (see [1, Example 85]). A group which equals its Baer radical
is called a Baer group. The same argument in the proof of Proposition 5.1 shows that
B(G) ⊆ S(G). We say that a group is an S-group if it satisfies the equivalent conditions of
Proposition 5.5. It is easy to see that the class of S-groups is closed by subgroups of finite
index and quotients. Of course, every Baer group is an S-group.
Proposition 5.8 (Theorem 73 in [1]). A group is a Baer group if and only if every its finitely
generated subgroup is subnormal and nilpotent. In particular, every finitely generated Baer
group is nilpotent.
By Propositions 5.5 and 5.8, every finitely generated non-nilpotent p-group is an S-
group which is not Baer. The next theorem of Wilson [8] provides many groups with trivial
Baer radical. We recall that an infinite group is just-infinite if every its proper quotient is
finite.
Theorem 5.9 (Theorem 2 in [8]). Let G be a just-infinite group. If B(G) ̸= 1, then B(G)
is a free abelian group of finite rank, which coincides with its own centralizer in G.
Lemma 5.10. Let G be a just-infinite p-group. Then S(G) = G, but B(G) = 1.
Proof. The fact that G = S(G) follows from Proposition 5.5 and the fact that finite p-
groups are nilpotent. If B(G) ̸= 1, then B(G) is a free abelian group by Theorem 5.9,
which contraddicts that G is a p-group.
L. Sabatini: Products of subgroups, subnormality, and relative orders of elements 9
Example 5.11 (No converse to Proposition 1.4). Let G be a just-infinite p-group, and let
K ⩽ G be any nilpotent subgroup. Since every subgroup of finite index of G is subnormal,
from Theorem 1.5 we have that |HK : H| divides |G : H| for every H ⩽f G. On the
other hand, K is not subnormal in G, because B(G) = 1.
Finally, it is worth to mention the following theorem of Robinson [6]. Given a group
property P , a group is hyper-P if every its non-trivial homomorphic image has some non-
trivial normal subgroup with the property P .
Theorem 5.12 (Theorem 1 in [6]). Let G be a finitely generated hyperabelian or hyperfinite
group. If G is an S-group, then G is nilpotent.
ORCID iDs
Luca Sabatini https://orcid.org/0000-0002-4781-5579
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