Products of subgroups, subnormality, and relative orders of elements

Let $G$ be a group. We give an explicit description of the set of elements $x \in G$ such that $x^{|G:H|} \in H$ for every subgroup of finite index $H \leqslant G$. This is related to the following problem: given two subgroups $H$ and $K$, with $H$ of finite index, when does $|HK:H|$ divide $|G:H|$?


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.Let x ∈ G and H G. The relative order of x with respect to H is The following result is proved in Section 2. The converse of Proposition 1.3 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 1.4 already appeared in [5, Theorem 2].In Section 4, we study the following class of subgroups 2020 Mathematics Subject Classification.Primary 20D40, 20D25, 20F99.Key words and phrases.Relative order; product of subgroups; subnormal subgroup.This is a generalization of subnormality, and we prove that it is equivalent to normality 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 F N (G) be the preimage of F (G/N ), where F (G) denotes the Fitting subgroup of G.
In particular, S(G) = F (G) when G is finite (Proposition 5.1).Of course, Theorem 1.5 implies that S(G) is closed under multiplication, a fact which is not immediately clear from the definition.

Preliminaries
We start with the proof of the key Lemma 1.1.
Proof of Lemma 1.1.Let ord H (x) := min{n ≥ 1 : x n ∈ H}.We first notice that o H (x) = ord H (x). Indeed, from the definitions we have o H (x) = o H∩ x (x) and ord H (x) = ord H∩ x (x).The fact that o H∩ x (x) = ord H∩ x (x) is a simple exercise.Now, the "if" part of the statement is trivial.On the other hand, if x n ∈ H for some n ≥ 1, then clearly ord H (x) < ∞.Let n = q • ord H (x) + r with r, q ≥ 0 and r < ord H (x). Since H is a subgroup, the fact that x n = x q•ord H (x) x r ∈ H implies that x r ∈ H, which in turn means r = 0.
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 −1 H.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.

Products of subgroups
The proof of Proposition 1.3 follows immediately from the following Proof.We have to prove that the ratio , and so we can write In particular, the original ratio equals |M : We continue with the easiest direction of Theorem 1.4.
We prove the claim of Theorem 1.4 by induction on the subnormal defect of H, so let H ⊳ f M ⊳ ⊳ f G, and K G. Using Lemma 3.2, we have 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 This concludes the proof of the "only if" part.The famous Kegel-Wielandt conjecture [3,7], proved by Kleidman [4] using the classification of the finite simple groups, says that H ⊳ ⊳ G whenever H is p-subnormal for every p.We point out that Kegel [3] did not use the classification to prove Theorem 3.4 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. 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 / ∈ O p (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 1.4 is that p-subnormality for every p implies that |HK| divides |G| for every K G.We provide an elementary proof of this fact.

Exponential subgroups
We write H exp G if x |G:H| ∈ H for all x ∈ G.We observe immediately that exponentiality is preserved by quotients.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 G n := {x n : x ∈ G} .The exponential subgroups of G of index n are in correspondence with the subgroups of G/G n of index n.Since G n is characteristic, the property of being exponential is preserved by automorphisms.Moreover, we have the following Proof.Let n = |G : H|.By the exponentiality of H we have G n H. Since G n is a characteristic subgroup of G contained in H, we obtain G n = 1.But this means exactly that n is a multiple of exp(G).The converse is trivial.
In general, there exist non-subnormal exponential subgroups whose index is not a multiple of the exponent.A simple example is G = C 4 × Sym(3) and H ∼ = C 2 × C 2 .The following corollaries of Lemma 4.3 are obtained with the same strategy.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.
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 | = q α for some prime power q α .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.Other important properties of the lattice of the subnormal subgroups are not true for exponential subgroups, and the dihedral group G = D 12 is a good source of counterexamples.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  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 1.3.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. 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.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.

Lemma 1 . 1 .
Let n ≥ 1.Then x n ∈ H if and only if o H (x) is finite and divides n.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 o H (x) = |H x : H|, we obtain Corollary 1.2.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 1.3.Let H ⊳ ⊳ G. Then |HK : K| divides |G : K| for every K f G. Theorem 1.4.Let H f G. Then H ⊳ ⊳ G if and only if |HK : H| divides |G : H| for every K G.

Theorem 3 . 4 (
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 Theorem 1.4 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 3.4 gives H/N ⊳ ⊳ G/N , i.e.H ⊳ ⊳ G.

Lemma 3 . 5 (
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.

Lemma 3 . 6 .
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. Proof.Let K G.We have to show that |HK : K| = |H : H ∩ K| divides |G : K|.Let p α be a prime power that divides |H : H ∩ K|.Since p α is arbitrary, it is sufficient to prove that p α | |G : K|.Let P 0 K be a p-Sylow of K, and let P G be a p-Sylow of G such that P ∩ K = P 0 .Of course, p α | |H : H ∩ P 0 |.By hypothesis |H : H ∩ P | = |HP : P | divides |G : P |, and so is not divisible by p.Therefore, p α | |H ∩ P : H ∩ P 0 |.Now |H ∩ P : H ∩ P 0 | = |(H ∩ P )P 0 : P 0 |, and this divides |P : P 0 | because P is a p-group.So p α | |P : P 0 |, and then of course p α | |G : P 0 |.Since p ∤ |K : P 0 |, we obtain p α | |G : K| as desired.

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 (N x) |G/N :H/N | = N x |G:H| ∈ H/N and so H/N exp G/N .If H/N exp G/N , then N x |G:H| = (N x) |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 normal core, and work with a finite group.Let G be a finite group and H G. From Corollary 1.2 and Theorem 1.4, 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.

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.

Corollary 4 . 4 .
Let H G be a Hall subgroup.If H exp G, then H ⊳ G.

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|.

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 1.2, we have S(G) = {x ∈ G : |H x : H| divides |G : H| for every H f G}.

Example 5 .
11 (No converse to Proposition 1.3).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 1.4 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.
a counterexample to the statement that two exponential subgroups generate an exponential subgroup: choosing any involutionx ∈ G \ H 1 we get x |G:H 1 | = x / ∈ H 1 .•Thesubgroup H 2 which satisfies H < H 2 ∼ = 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 ∈ H 2 \ H, we get that H is not exponential in H 2 although it is exponential in G.