Periodic linear groups in which permutability is a transitive relation

Abstract

A PT-group is a group in which the relation of being a permutable subgroup is transitive. The main aim of this paper is to show that a (homomorphic image of a) periodic linear group is a soluble PT-group if and only if each subgroup of a Sylow subgroup is permutable in the corresponding Sylow normalizer (see Theorem 4.7); for a fixed prime p, the latter condition is denoted by $\mathfrak {X}_p$. In order to prove our main theorem, we need (i) to characterize (homomorphic images of) periodic linear groups that are PT-groups (see Sect. 2), (ii) to develop a fusion theory for locally finite groups (see Sect. 3), (iii) to carefully study (homomorphic images of) periodic linear groups with the property $\mathfrak {X}_p$ for a fixed prime p (see for instance Theorem 4.6). As a by-product we obtain (among other results) a characterization of (homomorphic images of) periodic linear $\mathfrak {X}_p$-groups in terms of pronormality (see Theorem 4.11) that will allow us to show that, on some occasions, the property $\mathfrak {X}_p$ is inherited by subgroups.

1 Introduction

Let G be a group. A subgroup X of G is permutable (or quasinormal) in G if $XH=HX$ for every $H\le G$. The consideration of the alternating group of degree 4 shows that permutability, as normality, is not a transitive relation in general. It is therefore interesting to understand the structure of the so-called PT-groups, i.e. groups G in which if K is a permutable subgroup of G, and H is a permutable subgroup of K, then H is permutable in G. In the finite case, the structure of soluble PT-groups was determined by Zacher [30], while Menegazzo [18, 19] dealt with the infinite case; results on the subject can also be found in [4,5,6,7,8].

It is well known that, for finite groups, being a permutable subgroup is equivalent to being a subnormal subgroup that is also a modular element of the subgroup lattice. This means that a finite PT-group is just a finite group whose subnormal subgroups are permutable, so in particular this class contains the class of finite T-groups, i.e. finite groups in which normality is a transitive relation. It therefore makes sense to understand how far the class of (finite) PT-groups is from that of (finite) T-groups. This has been made, among many other things, in [1].

In the general case, permutable subgroups are ascendant and need not be subnormal (see [24, Theorem 6.2.10]). However, the situation is better if we restrict our attention to the universe ${\mathcal {L}}$ of homomorphic images of periodic linear groups. Here, in fact, permutable subgroups are again subnormal (see [10]) and so again, in this universe, PT-groups are precisely the groups in which every subnormal subgroup is permutable (see Lemma 2.1); in particular, this class contains that of (homomorphic images of) periodic linear T-groups. Again, it makes sense to understand how far PT-groups in ${\mathfrak {L}}$ are from T-groups. It turns out that the difference in these types of groups occurs in the elementary abelian sections (see Theorem 2.8), and that a soluble PT-group is just a subdirect product of a locally nilpotent PT-group and soluble T-group (see Theorem 2.9).

Although it may seem that (soluble) PT-groups (in ${\mathcal {L}}$) should really behave like the finite ones, this is not true. The reason for this being the fact that the class of finite soluble PT-groups is subgroup closed (this is an easy corollary of Zacher’s theorem), while that of soluble PT-groups in ${\mathcal {L}}$ is not (see for example the structure outlined in Lemma 2.3). As we will see, this bad behaviour is due to the prime 2, and in fact if this prime is not involved in the PT-group, then we obtain the desired subgroup inheritance (see Corollary 2.6).

In order to “locally” study the class of finite soluble PT-groups, the authors of [1] introduced, for any prime p, the class ${\mathfrak {X}}_p$ of all finite groups G in which the subgroups of a Sylow p-subgroup P of G are permutable in the Sylow normalizer $N_G(P)$. Their main result is that a finite group is a soluble PT-group if and only if it satisfies ${\mathfrak {X}}_p$ for every prime p. Moreover, they carry out a really nice study of the properties ${\mathfrak {X}}_p$; for example they show that the class ${\mathfrak {X}}_p$ is subgroup closed, and that groups satisfying ${\mathfrak {X}}_2$ are soluble.

Moving from finite groups to groups in ${\mathfrak {L}}$, we see that their results are false in general. For example, the property ${\mathfrak {X}}_2$ is not inherited by subgroups now, and there exist infinite simple ${\mathcal {L}}$-groups satisfying ${\mathfrak {X}}_2$ (see Example 4.2). However, it turns out that these bad examples look all very similar to each other (see Theorem 4.3), and this enables us to show that, on many occasions, an ${\mathcal {L}}$-group G satisfying ${\mathfrak {X}}_p$, where p is the smallest prime in $\pi (G)$, must be p-nilpotent (that is, $G/O_{p'}(G)$ is a p-group, where $O_{p'}(G)$ is the unique maximal $p'$-subgroup of G). In turns, the latter fact makes it possible to prove our main result: a homomorphic image of a periodic linear group is a soluble PT-group if and only if it satisfies ${\mathfrak {X}}_p$ for every prime p (see Theorem 4.7); this is a full generalization of the main result of [1] to the universe ${\mathcal {L}}$.

As a by-product of our main result, we have some interesting characterizations of the classes ${\mathfrak {X}}_p$ (see Theorems 4.8 and 4.11 ). In particular, Theorem 4.11 characterizes the class ${\mathfrak {X}}_p$ in terms of pronormal subgroups and enables us to show that, apart from some bad cases, the class ${\mathfrak {X}}_p$ is subgroup closed.

The layout of the paper is the following one. In Sect. 2, we characterize (homomorphic images of) periodic linear groups that are PT-groups; as consequences, we describe how far PT-groups in ${\mathcal {L}}$ are from being T-groups, and we briefly deal with those PT-groups G in ${\mathcal {L}}$ whose graph $\Gamma (G)$ is connected (see later for the definition, or [2]). In Sect. 3, we develop a fusion theory for locally finite groups; in particular, we prove an analogue of the focal subgroup theorem for finite groups (see Theorem 3.2), an analogue of Frobenius’ p-nilpotency criterion, and we describe some of the properties of nil-focal and hypo-focal subgroups in the class ${\mathcal {L}}$. In the final Sect. 4, we prove our main theorem and we deal with the property ${\mathfrak {X}}_p$.

Our notation is standard and can for instance be found in [22,23,24, 29]. In particular, we refer to [29] for results and properties of (homomorphic images of periodic) linear groups, to [23] for the structure of (soluble) T-groups, and to (Theorem 2.4.14 of) [24] for the structure of periodic, locally nilpotent groups in which all subgroups are permutable (note that this is precisely the class of all periodic, modular, locally nilpotent groups). Moreover, as usual, if G is a group, then:

• $\pi (G)$ denotes the set of all prime numbers p such that G has some non-trivial p-element;

• if $\pi$ is a set of primes, then $O_\pi (G)$ is the unique maximal normal $\pi$-subgroup of G;

• if p is a prime, then ${\text {Syl}}_p(G)$ denotes the set of all Sylow p-subgroups of the periodic group G;

• $\zeta _n(G)$ denotes the n-th centre of G (note that $\zeta _1(G)$ is also denoted by $\zeta (G)$), while ${\overline{\zeta }}(G)$ denotes the last term of the upper central series of G;

• $\varrho _{{\textbf{L}}{\mathfrak {N}}}(G)$ is the unique maximal locally nilpotent subgroup of G (this is also called the locally nilpotent radical of G, or the Hirsch-Plotkin radical of G);

• $\varrho ^{{\textbf{L}}{\mathfrak {N}}}(G)$ denotes the locally nilpotent residual of G, that is the intersection of all normal subgroups N of G such that G/N is locally nilpotent (if G is locally finite, then $G/\varrho ^{{\textbf{L}}{\mathfrak {N}}}(G)$ is locally nilpotent).

2 Periodic linear PT-groups and related properties

The main aim of this section is to characterize those homomorphic images of periodic linear groups that are PT-groups (see Theorem 2.5). A fundamental step in doing this is the analysis of Černikov primary PT-groups; the structure of these groups can actually be deduced from combining Theorem C2 of [18] and Lemma 4.2.1 of [23], but since the former paper is in Italian and we do not need the full force of these results, we give here direct proofs of those results we need; actually, we should also say that it seems that case (2f) of our Lemma 2.3 does not to appear in the statement of Theorem C2 of [18]. Before getting to the hearth of these results (see Lemmas 2.3 and 2.4 ), we prove two useful and easy lemmas.

Lemma 1.1

Let G be a homomorphic image of a periodic linear group. Then G is a PT-group if and only if every subnormal subgroup of G is permutable.

Proof

As we have already remarked, every permutable subgroup of a periodic linear group is subnormal (see [10]), so this also holds for every homomorphic image of a periodic linear group.

Assume every subnormal subgroup of G is permutable. If K is a permutable subgroup of G, and H is a permutable subgroup of K, then H is subnormal in G. Therefore H is permutable in G, and consequently G is a PT-group. Since normal subgroups are permutable, the converse is clear. $\square$

Lemma 1.2

Let $\pi$ be a set of primes and let G be a homomorphic image of a periodic linear group. If N is a normal maximal $\pi$-subgroup of G such that

• (1) G/N is a PT-group, and

• (2) every subnormal subgroup of N is normal in G,

then G is a PT-group.

Proof

By Lemma 2.1 we need to prove that every subnormal subgroup of G is permutable in G. Let H be a subnormal subgroup of G. By (2), $H\cap N$ is a normal subgroup of G and clearly $G/(H\cap N)$ satisfies the hypotheses of the statement. Thus, in order to prove that H is permutable in G, we may certainly assume $H\cap N=\{1\}$. Let $\pi =\pi (N)$. Application of Theorem 9.24 of [29] (also note the remark below it) shows that G has a maximal $\pi '$-subgroup X such that $G=X\ltimes N$, and that all the maximal $\pi '$-subgroups are conjugate. But H is subnormal in G and $H\cap N=\{1\}$, so $H^G$ is a $\pi '$-subgroup and consequently $H\le H^G\le X$ and $[H,N]\le [H^G,N]=\{1\}$.

Let Y be any subgroup of G of prime power order $p^m$. If $p\in \pi$, then $YH=HY$ since $[H,N]=\{1\}$. If $p\in \pi '$, then Y is contained in some conjugate $X^g$ of X. But, $H\le H^G\le X^g$ too, so again $YH=HY$ because $X\simeq G/N$ is a PT-group. Therefore H is permutable in G, and the statement is proved. $\square$

Lemma 1.3

Let G be a non-abelian infinite Černikov 2-group. Then G is a PT-group if and only if $G=\langle x\rangle (D\times F)$, where

• (1) D is divisible and $h^x=h^{-1}$ for all $h\in D$;

• (2) $F=\langle g\rangle \times A$, where A is elementary abelian, and one of the following cases holds:

  • (2a) $o(g)=4=o(x)$, $h^x=h^{-1}$ for all $h\in F$, and $x^2=g^2$.

  • (2b) $g=x^2$, $\langle [x,a]\rangle \le \Omega \big (\langle x\rangle \big )$ for all $a\in A$, and $o(x)\ge 8$.

  • (2c) $g=1$, $\langle x\rangle \cap (D\times A)=\{1\}$, $[x,a]=1$ for all $a\in A$, and $o(x)\le 4$.

  • (2d) $o(g)=2$, $x^2=gd$ with $d\in D$ and $o(d)=4$, $g^x=gd^2$, $[x,a]=1$ for all $a\in A$.

  • (2e) $g=1$, $\langle x\rangle \cap (D\times A)=\langle x\rangle \cap D=\Omega \big (\langle x\rangle \big )$ and $[x,a]=\{1\}$ for all $a\in A$, and ${\text {o}}(x)=4$.

  • (2f) $g=1$, $\langle x\rangle \cap (D\times A)=\langle x\rangle \cap D=\Omega \big (\langle x\rangle \big )\ge \langle [x,a]\rangle$ for all $a\in A$, and ${\text {o}}(x)\ge 8$.

In particular, $C_G(D)=\langle x^2\rangle (D\times F)$ has index 2 in G.

Proof

Assume first G is a PT-group. Let D be the finite residual of G and let F be a finite subgroup of G such that $G=FD$. Then $N=F\cap D$ is finite and is normal in G. Let’s temporary assume $N=\{1\}$, so in particular $G=F\ltimes D$. Let $H\le D$. Since H is subnormal in G, we have that H is permutable in G (see Lemma 2.1), so

$$\begin{aligned} H^F=H^F\cap HF=H \end{aligned}$$

and hence H is normal in G. It follows that $|G/C_G(D)|\le 2$. Let’s get back to the general case and put $C/N=C_{G/N}\big (D/N\big )$, so $|G/C|\le 2$. Now, C is nilpotent, so modular, being a PT-group. Therefore Theorem 2.4.14 of [24] shows that C is abelian, and we can write $C=D\times E$, for some finite abelian subgroup E of G. In particular, $|G/C|=2$ and $C=C_G(D)$. Let $x\in G$ with $G=\langle x\rangle C$, and note that $\langle x^2\rangle =\langle x\rangle \cap C$. A previous argument entails that x acts as the inversion on $C/\langle x^2\rangle$, so, if $g\in D$, then $g^x=g^{-1}v_g$ for some $v_g\in \langle x^2\rangle \le \zeta (G)$. Suppose there is $g\in D$ with $v_g\ne 1$, then the divisibility of D entails the existence of $g_1\in D$ with $g_1^{\text {o}(x)}=g$. Consequently $o\big (v_{g_1}\big )>{\text {o}}(x)$, contradicting the fact that $v_{g_1}$ belongs to $\langle x^2\rangle$. This means that x acts as the inversion on D.

In order to understand the action of x on E, we note that G/D is nilpotent and modular. Suppose first G/D contains a subgroup that is isomorphic to $Q_8$, the quaternion group of order 8. Then xD has order 4 and there is $g\in E$ of order 4 such that $g^x=g^{-1}d$ for some $d\in D$. This yields that

$$\begin{aligned} g=g^{x^2}=\big (g^x\big )^x=\big (g^{-1}d\big )^x=gd^{-2}, \end{aligned}$$

so d has order at most 2. Moreover, write $x^2=uv$, where $u\in D$ and $v\in E$; then

$$\begin{aligned} uv=x^2=(uv)^x=u^{-1}v^x, \end{aligned}$$

so $v^xD=vD$, which means that v has order at most 2 and consequently u has order at most 4. It follows that x has order at most 8.

Assume $o(x)=4$ and by contradiction that $d\ne 1$. Since $\langle g\rangle \langle x\rangle =\langle g\rangle \langle x\rangle$, we have that $dg^3=g^nx^m$, for some $n,m\in \{0,1,2,3\}$. If $n=3$, then $m=2$, so $x^2\in D$, a contradiction. If $n=0,2$, then $\langle gD\rangle =\langle xD\rangle$, a contradiction. Thus $n=1$ and $dg^2=x^2$. Replace x by xg, noting that

$$\begin{aligned} (xg)^2=xgxg=x^2g^{-1}dg=g^2d^2=g^2. \end{aligned}$$

Repeating the previous argument, we have that $dg^2=g^2$, a contradiction. Consequently, $d=1$. Let h be any element of E of order 2. If $h^x\ne h$, then gh is an element of E of order 4 such that $(gh)^x\ne (gh)^{-1}$, contradicting what we have already proved. Therefore x acts as the inversion on C and we are in case (2a).

Now, if $o(x)=8$, then $G/\langle u^2\rangle$ satisfies the previous conditions. Thus we get that $x^2\langle u^2\rangle =g^2\langle u^2\rangle$, which implies the contradiction $o(x)=4$.

Suppose G/D does not contain subgroups that are isomorphic to $Q_8$. Put ${\widetilde{G}}=G/\langle x^2\rangle$, so ${\widetilde{G}}=\langle {\widetilde{x}}\rangle \ltimes {\widetilde{C}}$ since $\langle x^2\rangle =\langle x\rangle \cap C$. In this case, the argument we employed in the first paragraph of the proof shows that ${\widetilde{x}}$ acts as (a power automorphism and so as) the inversion on ${\widetilde{C}}$. Write ${\widetilde{C}}={\widetilde{D}}\times {\widetilde{S}}$ for some finite abelian group ${\widetilde{S}}$. Since ${\widetilde{G}}/{\widetilde{D}}$ does not have any section that is isomorphic to $Q_8$ (see [24, Theorem 2.3.8]), it follows from [24, Theorems 2.4.14 and 2.3.15], that ${\widetilde{S}}$ has exponent $\le 2$. Then it is not difficult to see that $E=\langle g\rangle \times A$, where A is elementary abelian, and that we may even assume $x^2\in D\times \langle g\rangle$. There are only two possible cases.

First, $x^2\equiv g^2$ mod D. It follows from [24, Theorem 2.3.15], that $g^x\equiv g^{1+2^{m}}$ mod D for some $m\ge 2$. Now,

$$\begin{aligned} \big (xg^{-1}\big )^2=xg^{-1}xg^{-1}=x^2\big (g^{-1}\big )^xg^{-1}=x^2g^{-2}g^{-2^m}d=g^{-2^{m}}d_1 \end{aligned}$$

for some $d,d_1\in D$, and this is only possible if $g^2=1$. Replacing g with 1, we are in the situation $x^2\in D$. Since x acts as the inversion on D, we have $x^2=1$. If there is $a\in A$ such that $[a,x]=x^2$, then xa has order 2 and we are in case (2c); otherwise we are in case (2e).

Second, $x^2=gd$ for some $d\in D$. Clearly $[x,a]\le \Omega \big (\langle x\rangle \big )$ for any $a\in A$.

Assume $o(d)\le o(g)$; in this case a suitable replacement of g allows us to assume $x^2=g$ too, so if $o(x)\ge 8$, then we are in case (2b). If $o(x)=2$, then we are in case (2c). If $o(x)=4$ and $[x,a]\ne 1$ for some $a\in A$, then $o(xa)=2$, so we are again in case (2c). The only remaining possibility is still case (2c).

Assume $o(d)>o(g)$. Since x acts as the inversion on D, we have $o(d)=2\cdot o(g)$. Thus, in ${\widetilde{G}}=G/\Omega \big (\langle x\rangle \big )$ we have $o\big ({\widetilde{d}}\big )=o\big ({\widetilde{g}}\big )$, so the previous case applies and we get that ${\widetilde{G}}$ satisfies either (2b) or (2c). Of course, we may also assume $g\ne 1$, otherwise the argument of the final part of the second-to-last paragraph yields that we are in case (2c) or (2e). Thus $o(g)\ge 2$, $o(d)\ge 4$ and $o(x)\ge 8$. It follows that the socles of the subgroups generated by the elements in $G\setminus C$ coincide. If ${\widetilde{G}}$ satisfies (2b), then G satisfies (2f). Assume ${\widetilde{G}}$ satisfies (2c). In particular, $o(x)=8$, so $o(g)=2$ and $o(d)=4$. Moreover, $gd=(gd)^x=g^xd^{-1}$, so $g^x=gd^2=gx^4$. If there is $a\in A$ with $[a,x]=x^4$, then we replace it with ga. This is case (2d).

Conversely, assume G satisfies the conditions in the statement, and let H be a subnormal subgroup of G. If $H\not \le C=C_G(D)$, then H contains an element acting as the inversion on D, so it also contains D; since G/D is Dedekind, we have that H is permutable in G. Thus, without loss of generality, we have $H\le C$; of course, we may further assume $H\not \le D$ and that $H=\langle h\rangle$ is cyclic. By symmetry, we only need to show that $\langle h\rangle \langle x\rangle =\langle x\rangle \langle h\rangle$, and this can be really easily checked in all cases of the statement. $\square$

Lemma 1.4

Let p be an odd prime and let G be an infinite Černikov p-group. If G is a PT-group, then G is abelian.

Proof

Let D be the finite residual of G and let F be a finite subgroup of G such that $G=FD$. Put $N=F\cap D$. Then N is normal in G and $G/N=F/N\ltimes D/N$. For simplicity sake, we now temporary assume $N=\{1\}$, so $G=F\ltimes D$. If $g\in D$, then

$$\begin{aligned} \langle g\rangle ^F=\langle g\rangle ^F\cap \langle g\rangle F=\langle g\rangle \big (\langle g\rangle ^F\cap F\big )=\langle g\rangle , \end{aligned}$$

so every subgroup of D is normal in G. It follows that D is central in G, so G is nilpotent in this case.

Let us come back to the general case. Now, G/N is nilpotent. On the other hand, N is finite and G is locally nilpotent, which means that $N\le \zeta _n(G)$ for some non-negative integer n. Therefore G is nilpotent and the statement follows from [24, Theorem 2.4.14]. $\square$

Theorem 1.5

Let G be a homomorphic image of a periodic linear group. Then G is a soluble PT-group if and only if the following conditions hold:

• (1) $\varrho _{{\textbf{L}}{\mathfrak {N}}}(G)=\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)\times {\overline{\zeta }}(G)$;

• (2) $\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)$ is abelian and is also a maximal $\pi$-subgroup of G for some set of odd primes $\pi$;

• (3) every subgroup of $\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)$ is normal in G;

• (4) $O_{2'}\big (G/\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)\big )$ is nilpotent and modular;

• (5) If $O_{2}\big (G/\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)\big )$ is non-modular, then it satisfies the conditions of Lemma 2.3.

Proof

We start dealing with the necessity of (1)–(5). It follows from Theorem D of [18] that G has a subgroup N such that $2\not \in \pi (N)$, $\pi (N)\cap \pi (G/N)=\emptyset$, G/N is the direct product of primary PT-groups, and every subgroup of N is normal in G; in particular, N is abelian and G/N is locally nilpotent. Let $L=\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)$ and $H=\varrho _{{{\textbf{L}}{\mathfrak {N}}}}(G)$, so $L\le N\le H$. If P is any Sylow p-subgroup of N whose socle is not central in G, we have that $P\le L$; if the socle of P is central, then $P\cap L=\{1\}$ (and P is contained in the hypercentre of G). Thus $\pi (L)\cap \pi (G/L)=\emptyset$ and $H=L\times Z$, where Z is contained in the hypercentre of G. But the hypercentre of G cannot be larger than Z, and consequently $Z={\overline{\zeta }}(G)$. This proves (1)–(3).

Obviously, the Sylow subgroups of G/L are PT-groups, so (5) is proved by Lemma 2.3 (and [29], 2.6). In order to prove (4), we need to write $G/L=K/M$ for some periodic linear group K over a field of characteristic r and some normal subgroup M of K. For each prime p, the Sylow p-subgroup of K/M is the image of a Sylow p-subgroup $K_p$ of K (see [29, Theorem 9.15]). Now, 2.6 of [29] yields that $K_r$ is nilpotent, while $K_p$ is Černikov if $p\ne r$; so, if $r\ne p\ne 2$, Lemma 2.4 implies that $K_pM/M$ is nilpotent. Using Theorem 3.8 of [11], we see that $O_{2'}\big (G/\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)\big )$ is nilpotent.

We turn now to the sufficiency of the conditions. Put again $L=\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)$. Since G is locally finite, we have that G/L is locally nilpotent, so

$$\begin{aligned} G/L=O_{2'}(G/L)\times O_2(G/L). \end{aligned}$$

It follows from (4), (5) and Lemma 2.3 that G/L is a PT-group. Since every subgroup of L is normal in G, and L is maximal $\pi$-subgroup of G for some set of primes $\pi$, we have that G is a PT-group by Lemma 2.2. $\square$

As a consequence of the above characterization, we get that the class of all soluble PT-groups which are homomorphic images of periodic linear groups and do not contain subgroups of even order is subgroup closed. This is included in the following result.

Corollary 1.6

Let G be a soluble homomorphic image of a periodic linear group. If G is a PT-group, then the following hold:

• (1) G is metabelian;

• (2) If $G'\cap \zeta (G)=\{1\}$, then G is a T-group;

• (3) If $H\le G$ and $2\notin \pi (G)$, then H is a PT-group.

Proof

Let $L=\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)$ and $H=\varrho _{{{\textbf{L}}{\mathfrak {N}}}}(G)$. We use the structure described in Theorem 2.5 without any further notice.

(1)   Since G induces power automorphisms on L, we have that $G'\cap L$ is contained in the centre of $G'$. But G/L is metabelian, so $G'/(G'\cap L)$ is abelian and hence $G'\le H$ is abelian.

(2)   Certainly, $L\le G'$. Moreover, by (1), $G'$ is abelian, so $G'\le H=L\times {\overline{\zeta }}(G)$. Since $G'\cap \zeta (G)=\{1\}$, we get that $G'=L$, so G is a T-group by Lemma 5.2.2 of [23].

(3)   If H is any subgroup of G, then $H\cap L$ is a normal subgroup of H that is a maximal $\pi$-subgroup of G for some set of primes $\pi$. Moreover, all subgroups of $H\cap L$ are normal in H, and $H/(H\cap L)\simeq HL/L\le G/L$ is a PT-group since $2\not \in \pi (G/L)$. It follows from Lemma 2.2 that H is a PT-group. $\square$

Let G be a group. We define $\Gamma (G)$ as the graph whose vertices are the conjugacy classes of cyclic subgroups of G, and such that two conjugacy classes ${\mathcal {A}},{\mathcal {B}}$ are joined by an edge if there is $A\in {\mathcal {A}}$ and $B\in {\mathcal {B}}$ such that $AB=BA$. It has been proved in [2] that a finite group G is a soluble PT-group if and only if $\Gamma (G)$ is complete. Moving to (homomorphic images of) linear groups, we can at least prove one implication.

Corollary 1.7

Let G be a soluble homomorphic image of a periodic linear group. If G is a PT-group, then $\Gamma (G)$ is complete.

Proof

We start noting that if G is a 2-group, then Lemma  2.3 (and a simple computation) gives that $\Gamma (G)$ is complete.

Now to the general case. Write $G=X\ltimes N$, where $N=\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(G)$ and X is a $\pi '$-group. If Y and Z are cyclic subgroups of G, we can write $Y=Y_0Y_1$ and $Z=Z_0Z_1$, where $Y_0$ and $Z_0$ are subgroups of N and $Y_1,Z_1$ are $\pi '$-groups. Replacing Y and Z by conjugates (if necessary), we have that $Y_1$ and $Z_1$ are contained in X. By (4), (5) of Theorem 2.5, and our previous remark, we get that $\Gamma (X)$ is complete, so a further (possible) replacing of Y and Z by suitable conjugates (through elements of X) yields that $Y_1Z_1=Z_1Y_1$. It follows now from (3) of Theorem 2.5 that

$$\begin{aligned} YZ=Y_0Y_1Z_0Z_1=Z_0Z_1Y_0Y_1=ZY \end{aligned}$$

and the statement is proved. $\square$

It follows from Lemma 2.1 that every T-group is a PT-group in the universe ${\mathcal {L}}$. Our final two results describe how these two types of groups are related.

Theorem 1.8

Let G be a homomorphic image of a periodic linear group. Assume G is a PT-group. Then G is a T-group if and only if for each elementary section H/K of order $p^2$, with p a prime and $H\trianglelefteq \trianglelefteq G$, the factor group $N_G(H/K)/C_G(H/K)$ is a $p'$-group.

Proof

Assume first G is a T-group and let H/K be an elementary abelian section of order $p^2$, p a prime, with H subnormal in G. Then every subgroup of H/K is normal in G, so every p-element of G acts trivially on the elementary abelian p-group H/K, and we are done.

Assume conversely that the condition of the statement holds in G, and that G is not a T-group. Then we can find a subnormal non-normal subgroup H of G, and we can clearly assume $H_G=\{1\}$. Now, Theorem 6.2.14 of [24] shows that H is locally nilpotent (see also Lemma 6.2.2 of [24]). This means we may assume H is a p-group for some prime p. Moreover, H is a PT-group, so the structure of Theorem 2.5 applies, and, since $H_G=\{1\}$, every non-trivial subnormal subgroup of H cannot be normal in G. Therefore we may replace H by a subnormal subgroup of prime order p.

Let F be any finite subgroup of G containing H as a non-normal subgroup.Then $H_F=\{1\}$, so the Maier–Schmidt theorem (see Theorem 5.2.3 of [24]) entails that H is contained in the hypercentre of F, which means that H is centralized by all $p'$-elements of F. The arbitrariness of F shows that H is centralized by all $p'$-elements of G.

Let Q be the subgroup generated by all $p'$-elements of G. Then G/Q is a p-group. But G/Q is a PT-group, so it follows from Theorem 2.5 that G/Q either is modular or satisfies the conditions of Lemma 2.3. If G/Q is modular, then a combination of our hypothesis with Theorem 2.4.14 of [24] shows that G/Q is a Dedekind. If G/Q satisfies the conditions of Lemma 2.3, then our hypothesis (and some simple computations) implies that G/Q is a T-group. In both cases, HQ/Q is normal (and so even central) in G/Q.

Finally, $H\le \zeta (HQ)$, so $L=H^G$ is a finite elementary abelian p-subgroup of $\zeta (HQ)$. Let x be any p-element of G that does not normalize H; in particular, x does not centralize L. Thus, L contains an elementary abelian section of order $p^2$ that is not centralized by x, contradicting the hypotheses of the statement. $\square$

Theorem 1.9

Let G be a homomorphic image of a periodic linear group. Then G is a soluble PT-group if and only if there are a locally nilpotent PT-group M and a soluble T-group W such that:

• (1) $\pi (M)\cap \pi \big (\varrho ^{{{{\textbf {L}}}{\mathfrak {N}}}}(W)\big )=\emptyset$;

• (2) there is a monomorphism $\alpha :\, G\longrightarrow M\times W$ with $\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(W)\le \alpha (G)$, and such that the natural projections $G\longrightarrow M$ and $G\longrightarrow W$ are surjective;

• (3) if $p\in \pi (M)\cap \pi (W)$, then the Sylow p-subgroups of G are isomorphic to the Sylow p-subgroups of M.

Proof

Suppose first G is a soluble PT-group, so the description of Theorem 2.5 applies. Let L be the locally nilpotent residual of G, and let Z be the hypercentre of G; in particular, $H=L\times Z$ is the Hirsch–Plotkin radical of G. Since G/Z has trivial centre, we get that G/Z is a T-group by Corollary 2.6. Clearly, G/L is locally nilpotent.

Put $M=G/L$ and $W=G/Z$. Now,

$$\begin{aligned} \varrho ^{{\textbf{L}}{\mathfrak {N}}}(W)=LZ/Z\simeq L, \end{aligned}$$

so $\pi \big (\varrho ^{{{\textbf {L}}}{\mathfrak {N}}}(W)\big )$ consists of odd primes and is disjoint from $\pi (G/L)=\pi (M)$. This proves (1). Moreover, $\alpha :\, g\mapsto (gL,gZ)$ is the map required in (2), since

$$\begin{aligned} \varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(W)=\big \{(L,gZ):\, g\in L\big \}=\big \{(gL,gZ):\, g\in L\big \}\le \alpha (G), \end{aligned}$$

so (2) holds. Finally, if p is a prime in $\pi (M)\cap \pi (W)$, then p does not belong to $\pi (L)$ and (3) is established.

Assume (1), (2) and (3) hold. It follows from (2) that both M and W are homomorphic images of a periodic linear group. Thus, since M is locally soluble, it is also soluble (see [29, Corollary 9.21]), so $G\le M\times W$ is soluble. Let $L=\varrho ^{{{\textbf{L}}{\mathfrak {N}}}}(W)$. Therefore $\pi (L)\cap \pi (W/L)=\emptyset$ (see [23, Lemma 2.2.1 and Theorem 4.2.2]). Since L is abelian (see [23, Lemma 2.4.1]), W (and hence G) induces power automorphisms in L. Moreover, (1) yields that $\pi (G/L)\cap \pi (L)=\emptyset$.

If $p\in \pi (M)\cap \pi (W)$, then the Sylow p-subgroups of G are PT-groups by (3).If $p\not \in \pi (M)\cap \pi (W)$, then a Sylow p-subgroup P of G is isomorphic either to a Sylow p-subgroup of M or of W; in both cases, P is a PT-group. Therefore G/L is a PT-group and Lemma 2.2 shows that also G is a PT-group. The statement is proved. $\square$

3 Fusion systems

In this section we show how to develop a theory of fusion systems for (possibly infinite) locally finite groups, extending in this way certain well-known statements of the corresponding theory for finite groups. Some of these results will be employed in the next section to prove our main theorem, while others are of an independent interest. The core idea can be found in the proof of Theorem 3.2, and in fact all other results depend on similar arguments.

We start with an easy lemma, but first we need some definitions. Let $\pi$ be a set of primes. A locally finite group G is $\pi$ -integrated if the Sylow $\pi$-subgroups of any subgroup H of G are conjugate in H; if $\pi =\{p\}$, we also speak of p-integrated groups. Results concerning $\pi$-integrated groups and related concepts can be found in [9].

Let A and B be subgroups of a group G. Suppose further $A=B^g$ for some $g\in G$. In this case we write $\varphi _{g}^{A,B}:\, a\in A\mapsto a^g\in B$ for the conjugation map determined by g, A and B (note that we usually omit A and B if they are obvious from the context, so we usually only write $\varphi _g$). Now, if $H\le K$ are subgroups of G, we say that K controls G -fusion in H if, whenever two subgroups of H are conjugate in G via $\varphi _g$, then there is some $k\in K$ such that $\varphi _g=\varphi _k$ (of course, this is stronger than simply requiring any two subgroups of H conjugate in G to be conjugate in K).

Lemma 1.10

Let p be a prime and let P be a Sylow p-subgroup of the p-integrated, locally finite group G.

• (1) If P is abelian, then $N_G(P)$ controls G-fusion in P.

• (2) If X, Y are normal subsets of P which are conjugate in G, then they are also conjugate in $N_G(P)$.

Proof

(1)   Let X and Y be subgroups of P such that $X^g=Y$ for some $g\in G$. We need to find $g_1\in N_G(P)$ such that $\varphi _{g_1}=\varphi _{g}$. Of course,

$$\begin{aligned} P^g\le C_G(X)^g=C_G(X^g)=C_G(Y). \end{aligned}$$

Now, P and $P^g$ are Sylow p-subgroups of $C_G(Y)$ and p-integrability entails the existence of $h\in C_G(Y)$ such that $P^{gh}=P$. Then gh belongs to $N_G(P)$ and

$$\begin{aligned} x^{gh}=(x^g)^h=x^g \end{aligned}$$

for each $x\in X$, so $\varphi _{gh}=\varphi _g$ and (1) is proved.

(2)   There is $g\in G$ with $X^g=Y$. Since X is normal in P, then Y is normal in both P and $P^g$. It follows that P and $P^g$ are Sylow p-subgroups of $N_G(Y)$ so there is $h\in N_G(Y)$ such that $\big (P^g\big )^h=P$. Then $gh\in N_G(P)$ and hence $X^{gh}=Y^h=Y$. $\square$

Let X be a subgroup of a group G. We recursively define the focal commutator series of X in G as follows (see also [15]): set $[[X,_0G]]:=X$ and

$$\begin{aligned}{}[[X,_1G]]=[[X,G]]:=\big \langle xy^{-1}:\, x,y\in X,\, \exists g\in G,\, x=y^g\big \rangle ; \end{aligned}$$

if $\alpha$ is an ordinal, then

$$\begin{aligned}{}[[X,_{\alpha +1}G]]:=[[Y,G]], \end{aligned}$$

where $Y=[[X,_{\alpha }G]]$; if $\lambda$ is a limit ordinal, we put

$$\begin{aligned}{}[[X,_\lambda G]]=\bigcap _{\alpha <\lambda }[[X,_\alpha G]]. \end{aligned}$$

We say that X is hypo-focal in G if $[[X,_\alpha G]]=\{1\}$ for some ordinal $\alpha$; if $[[X,_kG]]=\{1\}$, for some $k\in {\mathbb {N}}$, we also say that X is nil-focal; and in special case $k=1$, we say X is a focal subgroup of G. In each of the previous cases, the smallest ordinal $\alpha$ such that $[[X,_\alpha G]]=\{1\}$ is the focal length of X.

Some remarks are in order. Let X be a subgroup of a group G.

• (1) If X is a normal nil-focal subgroup of G, we have that $X\le \zeta _k(G)$ for some non-negative integer k. Conversely, if k is a non-negative integer, every subgroup of $\zeta _k(G)$ is nil-focal in G of length at most k. Thus, every nilpotent group is nil-focal in itself; similarly, every hypocentral group is hypo-focal in itself: this also shows that the hypo-focal length can essentially be arbitrary (even in the locally finite case).

• (2) If X is hypo-focal (resp. nil-focal), then X is hypocentral (resp. nilpotent). Thus, if X is locally finite, then X is locally nilpotent.

• (3) If X is hypo-focal (resp. nil-focal) in G, and $Y\le X\le H\le G$, then Y is a hypo-focal (resp. nil-focal) in H.

• (4) If X is nil-focal in G and N is a normal subgroup of G contained in X, then X/N is nil-focal in G.

Our next result generalizes the well-known Focal Subgroup Theorem for finite groups (see [15, Theorem 3.6]).

Theorem 1.11

Let $\pi$ be a set of primes and let G be a $\pi$-integrated, locally finite group. If $k\in {\mathbb {N}}$ and S is a Sylow $\pi$-subgroup of G, then $S\cap \gamma _{k+1}(G)=[[S,_kG]]$.

Proof

Let $S^*=[[S,_kG]]$. It is clear that $S^*$ is contained in $S\cap \gamma _{k+1}(G)$. Assume by contradiction that $S^*<S\cap \gamma _{k+1}(G)$. Since the Sylow $\pi$-subgroups of G are conjugate, it follows that $R^*<R\cap \gamma _{k+1}(G)$ for every Sylow $\pi$-subgroup R of G.

First, we prove the statement in case G is countable. In this case, there is a sequence of finite subgroups

$$\begin{aligned} E_1\le E_2\le \cdots \le E_n\le E_{n+1}\le \cdots \end{aligned}$$

whose union is G. Then there is a Sylow $\pi$-subgroup S of G such that $S\cap E_i$ is a Sylow $\pi$-subgroup of $E_i$ for every i (start with a Sylow $\pi$-subgroup of $E_1$, extend it to a Sylow $\pi$-subgroup of $E_2$ and so on). Let $h\in \big (S\cap \gamma _{k+1}(G)\big )\setminus S^*$. Then there are finitely many elements $x_1,\ldots ,x_m$ of G such that $h\in \gamma _{k+1}\big (\langle x_1,\ldots ,x_m\rangle \big )$. Let $\ell$ be such that $x_1,\ldots ,x_m\in E_\ell$. Since $S\cap E_\ell$ is a Sylow p-subgroup of the finite group $E_\ell$, we have that

$$\begin{aligned} h\in S\cap \gamma _{k+1}(E_\ell )=[[S\cap E_\ell ,_kE_\ell ]]\le [[S,_kG]]=S^*, \end{aligned}$$

a contradiction. This proves the statement in the countable case.

Let’s move to the general case. Again, let $h\in \big (S\cap \gamma _{k+1}(G)\big ){\setminus } S^*$ and let $x_1,\ldots ,x_m$ be elements of G with $h\in \gamma _{k+1}\big (\langle x_1,\ldots ,x_m\rangle \big )$. Let $E_1=\langle x_1,\ldots ,x_m\rangle$. Assume $E_j$ is defined for some positive integer j. Now, $S\cap E_j$ is contained in a Sylow $\pi$-subgroup $S_j$ of $E_j$. For each $g\in S_j{\setminus } (S\cap E_j)$, there is a finite subset ${L}_g^j$ of S such that $\langle g,{L}_g^j\rangle$ is not a $\pi$-group. Let

$$\begin{aligned} E_{j+1}=\langle E_j,{L}_g^j:\, g\in S_j\setminus (S\cap E_j)\rangle . \end{aligned}$$

We have thus defined the finite subgroups $E_i$ of G for every positive integer i. Let

$$\begin{aligned} E=\bigcup _{i\ge 1}E_i. \end{aligned}$$

Then E is a countable subgroup of G. Moreover, $S\cap E$ is a Sylow $\pi$-subgroup of E. In fact, if this is not the case, then there is a Sylow $\pi$-subgroup Q of E strictly containing $S\cap E$. Let $x\in Q\setminus (S\cap E)$. Since $E=\bigcup _{i\ge 1}E_i$, there is a positive integer u such that x belongs to $E_u$. But then $\langle x,{\mathcal {L}}_x^u\rangle$ is contained in Q and is not a $\pi$-subgroup. Therefore $S\cap E$ is a Sylow $\pi$-subgroup of E. Finally, since $h\in (E\cap S)\cap \gamma _{k+1}(E)$, we get $[[E\cap S,_kE]]<(E\cap S)\cap \gamma _{k+1}(E)$, and this contradicts the previous paragraph. The statement is proved. $\square$

The argument we employed in the above theorem makes it possible to generalize many other results of [15] to locally finite groups. For example, we can prove the following generalization of Corollary 4.4 of [15].

Lemma 1.12

Let $\pi$ be a set of primes and let G be a locally finite group. If S is an abelian Sylow $\pi$-subgroup of G, then S is hypo-focal in G if and only if S is focal in G.

Others can be obtained combining Theorem 3.2 with some known results. For example, it has been showed in [11, Theorem 3.8], that the lower central series of (a homomorphic image of) a periodic linear group stops after finitely many steps. Combining this fact with our Theorem 3.2, we get the following results.

Corollary 1.13

Let $\pi$ be a set of primes and let G be a homomorphic image of a periodic linear group. If S is a hypo-focal Sylow $\pi$-subgroup of G, then S is nil-focal.

Corollary 1.14

(see also [15, Corollary 4.7]) Let G be a homomorphic image of a periodic linear group. Then G is nilpotent if and only if for each prime p there is a Sylow p-subgroup of G which is nil-focal.

Corollary 1.15

(see also [15, Corollary 4.8]) Let G be a homomorphic image of a periodic linear group, and let S be a subgroup of G. Then S is hypo-focal in G if and only if for each prime p there exists a Sylow p-subgroup of S which is hypo-focal in G.

Proof

Clearly, every hypo-focal subgroup of G has the above mentioned property. Assume S is a subgroup of G which, for each prime p, has a Sylow p-subgroup which is hypo-focal in G. Then the same is true in S, hence S is nilpotent by Corollary 3.5. Thus S has only one Sylow p-subgroup for each prime p, and is the direct product of these. Obviously, a direct product of coprime hypo-focal subgroups is hypo-focal. $\square$

As a final application of the argument in the proof of Theorem 3.2 we prove a generalization of Frobenius’ Normal p-Complement Theorem, and a small extension of [3, Theorem 1]. Recall that a periodic group G has a normal p-complement if $G=P\ltimes N$ for some p-subgroup P and some $p'$-subgroup N.

Theorem 1.16

Let p be a prime and let P be a Sylow p-subgroup of the p-integrated, locally finite group G. The following statements are equivalent:

• (1) G has a normal p-complement.

• (2) P controls G-fusion in P.

• (3) If $X\le P$, then $N_G(X)$ has a normal p-complement.

• (4) $N_G(X)/C_G(X)$ is a p-group for every non-trivial subgroup X of P.

• (5) $N_G(X)/C_G(X)$ is a p-group for every non-trivial finite subgroup X of P.

Proof

The implications (2) $\implies$ (4) $\implies$ (5) and (3) $\implies$ (4) are obvious.

(1) $\implies$ (2), (3)   Write $G=P\ltimes N$ for some normal $p'$-subgroup N. Let A, B be subgroups of P with $A^g=B$ for some $g\in G$. If $x\in P$ and $h\in N$, then $x^h=xh'$, with $h'\in H$, lies in P if and only if $x^h=x$. Now, write $g=xh$ for some $x\in P$ and $h\in N$. Then $B=A^{xh}=\big (A^x\big )^h$, so $A^x$ is centralized by h and $\varphi _g^{A,B}=\varphi _x^{A,B}$. This proves (2).

Now, assume g belongs to $N_G(A)$, so $A=B$. In this case, $A^{h^{-1}}\le P$, so h centralizes A and belongs to $C_G(A)$. This means that x belongs to $N_G(A)$ and consequently that $N_G(A)$ has a normal p-complement, proving (3).

(5) $\implies$ (1)   Suppose first G is countable, so there is a sequence of finite subgroups

$$\begin{aligned} E_1\le E_2\le \cdots \le E_n\le E_{n+1}\le \cdots \end{aligned}$$

such that $\bigcup _{i>0}E_i=G$ and $P\cap E_i\in {\text {Syl}}_p(E_i)$ for every i. Thus, for each i, $E_i$ satisfies the hypotheses of the finite version of Frobenius’ theorem; consequently, the set of all $p'$-elements of $E_i$ is a subgroup, say $D_i$. Therefore the set of all $p'$-elements of G is a subgroup D, because it coincides with the union of all the $D_i$’s, and clearly $G=P\ltimes D$.

Assume now G is arbitrary and let ${\mathcal {C}}$ be the set of all countable subgroups of G. For each $X\in {\mathcal {C}}$, we can find a countable subgroup $C_X$ of G containing X and such that $P\cap C_X$ is a Sylow p-subgroup of $C_X$ (this can be done using the argument of the final part of the proof of Theorem 3.2). Then ${\mathcal {S}}=\{C_X\,:\, X\in {\mathcal {C}}\}$ is a local system for G, and by the first paragraph $C_X=(P\cap C_X)\ltimes N_X$ for some $p'$-subgroup $N_X$. It follows that

$$\begin{aligned} N=\bigcup _{X\in {\mathcal {C}}}N_X \end{aligned}$$

is a $p'$-subgroup such that $G=P\ltimes N$. $\square$

Note that if we wish to replace “normal p-complement” by “p-nilpotent” in the above statement, we need to get rid of (2); otherwise, we can restrict our attention the universe of (homomorphic images of) periodic linear groups, since here “normal p-complement” is equivalent to “p-nilpotent”.

Theorem 1.17

Let p be a prime and let G be a p-integrated, locally finite group having an abelian Sylow p-subgroup P of finite rank whose finite residual R is normal in G. Then G is p-nilpotent if and only if $N_G(P)$ is p-nilpotent.

Proof

Only the sufficiency of the condition is in doubt. Write $P=RF$ for some finite subgroup F of G. Suppose first G is countable. In this case we can find a chain of finite subgroups

$$\begin{aligned} F=F_0\le F_1\le \cdots \le F_n\le F_{n+1}\le \cdots \end{aligned}$$

whose union is G and such that $P\cap F_i$ is a Sylow p-subgroup of $F_i$. If g belongs to $N_i:=N_{F_i}(P\cap F_i)$, then $[g,E]\subseteq P\cap F_i$, so g normalizes $P=RE$, and hence $N_i$ is p-nilpotent. It follows from [3, Theorem 1], that $F_i$ is p-nilpotent. Consequently, G is p-nilpotent.

In the general case, we just need to observe that every countable subgroup containing P (which is countable) is p-nilpotent by the previous paragraph, so G is (locally) p-nilpotent. $\square$

4 PT-groups and the property ${\mathfrak {X}}_p$

In this final section we prove our main result (Theorem 4.7) and we deal with some nice features of the property ${\mathfrak {X}}_p$. As we will see, the proof of the main result involves the description of Černikov 2-groups that are PT-groups, and Theorem 3.2. Note also that here we make a heavy use of the structure of periodic linear groups: this can be found in [29], or in the introduction to Section 3 of [14].

Let p be a prime. We define ${\mathfrak {X}}_p$ as the class of all periodic groups G such that, for every Sylow p-subgroup P of G, $X\trianglelefteq \trianglelefteq P$ (i.e. X subnormal in P) implies that X is permutable in $N_G(P)$. The consideration of any of the examples described in the statement of Lemma 2.3 shows that the class ${\mathfrak {X}}_2$ is not closed with respect to forming subgroups; however, we will see that these are the only bad possibilities, and in fact, for example, ${\mathfrak {X}}_p$ is subgroup closed when $p>2$ (see Corollary 4.12). Concerning the inheritance by quotients, we can certainly say that if $G\in {\mathfrak {X}}_p$ and N is a normal p-subgroup of G, then $G/N\in {\mathfrak {X}}_p$; our first result shows that this is still true if G is a homomorphic image of a periodic linear group and N is a normal $p'$-subgroup of G.

Lemma 1.18

Let G be a homomorphic image of a periodic linear group. If $X\le G$ and $N\trianglelefteq G$ are such that $\pi (X)\cap \pi (N)=\emptyset$, then $N_G(X)N=N_G(XN)$.

Proof

Let $g\in N_G(XN)$. Then g normalizes XN, so X and $X^g$ are complements to N in XN. It follows from Theorem 9.24 of [29] that there is $h\in XN$ such that $X^h=X^g$. Therefore $gh^{-1}=x\in N_G(X)$, so $g=xh\in N_G(X)N$ and hence $N_G(XN)$ is contained in $N_G(X)N$. Since the inclusion $N_G(X)N\le N_G(XN)$ is obvious, we are done. $\square$

It turns out that a finite ${\mathfrak {X}}_2$-group is soluble (see [1, Theorem C]). The following example shows that this is not the case for infinite, locally finite groups.

Example 4.2

There is an infinite periodic linear simple group satisfying ${\mathfrak {X}}_2$.

Proof

Let p be an odd prime which is $\equiv _8\pm 1$. Put $q_1=p$. If $q_n$ is defined, put $q_{n+1}=q_n^2$. This defines a sequence of integers $\{q_n\}_{n\in {\mathbb {N}}}$. Now, for each $m\in {\mathbb {N}}$, let $G_m={\text {PSL}}\big (2,q_m\big )$. Naturally,

$$\begin{aligned} G_1\le G_2\le \cdots \le G_i\le G_{i+1}\le \cdots \end{aligned}$$

Let G be the union of all the $G_i$’s. Of course, G is an infinite periodic linear simple group. Let’s prove that G satisfies ${\mathfrak {X}}_2$.

For each i, the Sylow 2-subgroups of $G_i$ are dihedral 2-groups of order $2^{i+1}$ or greater, so the Sylow 2-subgroups of G are infinite locally dihedral 2-groups. Let P be a Sylow 2-subgroup of G such that $P\cap G_i$ is a Sylow 2-subgroup of $G_i$ (start by taking a Sylow 2-subgroup of $G_1$ and then extend it to one of $G_2$ and so on). Since the Sylow 2-subgroups of the $G_i$ are self-centralizing, it follows that also P is self-centralizing. On the other hand, ${\text {Aut}}(P)\simeq P$, so $P=N_G(P)$ and G satisfies ${\mathfrak {X}}_2$. $\square$

Our next two results show that the “minimal” counterexamples to an analogue of [1, Theorem C], look like Example 4.2.

Theorem 1.20

Let p be a prime and let G be an infinite locally finite simple group. If the Sylow p-subgroups of G are soluble PT-groups, then G is linear.

Moreover, if G is linear of characteristic p, then G is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}})$ for some infinite locally finite field ${\mathfrak {K}}$ of characteristic p; in particular, G does not satisfy ${\mathfrak {X}}_p$.

Proof

It follows from Theorems C1 and C2 of [18] that the Sylow p-subgroups of G are abelian-by-cyclic (see also Lemma 4.2.1 of [23], and Theorem 2.4.14 of [24]). Let ${\mathcal {K}}=\big \{(H_i,N_i):\, i\in I\big \}$ be a Kegel cover of G (see [16]). By definition, (1) the factors $H_i/N_i$ of ${\mathcal {K}}$ are simple groups, (2) $H_i$ is finite for all $i\in I$, and (3) for each finite subgroup F of G, there is $i\in I$ such that $F\le H_i$ and $F\cap N_i=\{1\}$. If the factors of ${\mathcal {K}}$ involve arbitrarily large alternating groups, we see that the Sylow p-subgroups of G cannot be abelian-by-cyclic. Therefore the factors of ${\mathcal {K}}$ may be assume to be finite simple groups of Lie type of bounded Lie ranks, otherwise their Weyl groups will contain alternating groups of arbitrarily large degrees. Since every finite subgroup of G is isomorphic to a subgroup of a factor of ${\mathcal {K}}$, which is now linear of bounded degree, it follows that G is linear (see [29, Theorem 2.7]).

Assume G is linear of characteristic p. Corollary 9.32 of [29] shows that G is locally finite-simple; let ${\mathcal {S}}=\big \{S_i\big \}_{i\in I}$ be a local system of G made by finite simple groups. As above, we see that ${\mathcal {S}}$ can be assumed to be made by finite simple groups of Lie type with bounded Lie ranks; let $\ell$ be a bound to these Lie ranks. We prove that each $S_i$, $i\in I$, is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}}_i)$ for some finite field ${\mathfrak {K}}_i$. Let $q_i=p_i^{n_i}$ be the order of the field corresponding to the Lie type of $S_i$. Let $q_i^{m_i}$ be the order of a Sylow $p_i$-subgroup of $S_i$. If ${\mathcal {S}}$ contains a sub-local system in which every $p_i\ne p$, then the ranks of the abelian p-subgroups are bounded in terms of $\ell$ and consequently the Sylow p-subgroups of G are finite. But then Corollary 9.7 of [29] gives a contradiction. Therefore ${\mathcal {S}}$ does not contain such sub-local systems, which means that it contains a sub-local systems such that $p_i=p$ for all i; there is no loss of generality in assuming that ${\mathcal {S}}$ coincides with this sub-local system.

Suppose now there is a sub-local system of ${\mathcal {S}}$ such that $q_i\le n$ for all $i\in I$. Since the Sylow p-subgroups are not finite (see [29, Corollary 9.7]), we have that the ranks of the Lie groups in ${\mathcal {S}}$ must increase indefinitely. However, as above, this means that their Weyl groups will contain alternating groups of arbitrarily large degrees and this contradicts the fact that the Sylow subgroups are abelian-by-cyclic. Therefore, if n is a fixed positive integer, then ${\mathcal {S}}$ can be assumed satisfying the condition $q_i\ge n$ for all i.

Now, a maximal abelian p-subgroup of $S_i$ has order $\ge q_i^{m_i-1}$ (for $q_i$ large enough, at least), because it is abelian-by-cyclic and it has finite exponent. Using [28, Table 4], and [17, Theorem], we see that this is only possible if $S_i$ is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}}_i)$ for some finite field ${\mathfrak {K}}_i$ (see also Theorem 3.7 of [26] and its proof, and [27, Theorem 3.1]). Finally, Theorem 4.18 of [16] shows that G is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}})$ for some infinite, locally finite field ${\mathfrak {K}}$ of characteristic p. $\square$

Corollary 1.21

Let G be an infinite locally finite simple group whose Sylow p-subgroups are soluble PT-groups for every prime p. Then G is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}})$ for some infinite locally finite field ${\mathfrak {K}}$.

In connection with the above corollary, we recall that also an infinite, locally finite simple group whose Sylow subgroups are either Černikov or abelian is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}})$ for some infinite locally finite field ${\mathfrak {K}}$ of positive characteristic (see for instance [13, Lemma 6.1]).

Corollary 1.22

Let G be an infinite periodic linear simple group whose Sylow p-subgroups are PT-groups for every prime p. Then G is isomorphic to ${\text {PSL}}(2,{\mathfrak {K}})$ for some infinite locally finite field ${\mathfrak {K}}$.

Proof

The Sylow p-subgroups of G are either Černikov or nilpotent (see [29, 2.6]), so we may apply Corollary 4.4. $\square$

The following result makes use of the description we gave of Černikov 2-groups and of the fusion theory described in Sect. 3.

Theorem 1.23

Let G be a homomorphic image of a periodic linear group satisfying one of the following conditions:

• (1) the Sylow 2-subgroups of G are nilpotent;

• (2) G is soluble-by-finite.

If p is the smallest prime in $\pi (G)$ and $G\in {\mathfrak {X}}_p$, then G is p-nilpotent.

Proof

Let P be a Sylow p-subgroup of G. Then $N_G(P)=X\ltimes P$ for some $p'$-subgroup X of G (see [29, Corollary 9.25]). If H is a cyclic subnormal subgroup of P, then

$$\begin{aligned} H^X=H^X\cap XH=\big (H^X\cap X\big )H=H, \end{aligned}$$

and hence H is normalized by X; but p is the smallest prime number in $\pi (G)$, and this means that H is actually centralized by X. Thus, if R is the Fitting subgroup of P, we get that R is centralized by X. Now, P is either nilpotent or satisfies the conditions of Lemma 2.3, so P/R is cyclic of order 2 at most. It follows that $X\le C_G(P)=N_G(P)$ and in particular $N_G(P)$ is p-nilpotent.

Now, case (1) is divided in three cases (see our Lemma 2.3, [29, 2.6], and [24, Theorem 2.4.14]): (1$_a$) $p=2$ and P is finite; (1$_b$) $p=2$ and P is abelian; (1$_c$) $p=2$ and P is nilpotent of finite exponent.

(1$_a$)   Every finite subgroup containing P is p-nilpotent by Theorem C of [1], and hence G is p-nilpotent itself.

(1$_b$)   In this case, Lemma 3.1 shows that $N_G(P)$ controls G-fusion in P, and this means that $[[P,G]]=\{1\}$. Theorem 3.2 yields that $P\cap G'=\{1\}$, so G is certainly p-nilpotent in this case.

Before proving (1$_c$) and (2), let’s fix some notation. Write $G=H/N$, where H is a periodic linear group of characteristic r and $N\trianglelefteq H$. The structure of a periodic linear group shows that G has normal subgroups $U\le A\le S\le B$ such that U is a nilpotent r-group of finite exponent, A/U is an abelian $r'$-group whose Sylow subgroups are Černikov, S/A is finite, B/S is the direct product of finitely many infinite linear simple groups, and G/B is finite.

(2)   Suppose G is soluble-by-finite, $p=2$ and P is infinite non-nilpotent; in particular, $p\ne r$ and P satisfies the conditions of Lemma 2.3. By Lemma 4.1 we may assume that $U=\{1\}$ and that A is the finite residual of P. Let $C=C_G(A)$; in particular, $|PC/C|=2$. Now, G/C is isomorphic to an automorphism group of A, so PC/C is contained in the centre of G/C and in particular $PC\trianglelefteq G$. Suppose we can prove PC is 2-nilpotent; by Lemma 4.1, $G/W\in {\mathfrak {X}}_2$, where $W=O_{2'}(PC)$, so G/W is 2-nilpotent since $PW/W=PC/W\trianglelefteq G/W$ (see also the first paragraph of the proof); thus G is 2-nilpotent and we are done. Assume therefore $G=PC$. Let $P_1=P\cap C$ and $L_1=N_C(P_1)$. Of course, $L_1$ is P-invariant, and if we can prove $PL_1$ is 2-nilpotent, then also C is 2-nilpotent by Theorem 3.8 (recall that $P_1$ is an abelian Sylow 2-subgroup of C) and hence G is 2-nilpotent. We may thus assume $C=N_C(P_1)$ and $G=PN_C(P_1)$, so $P_1$ is normal in G. Now, we use induction on the index |G : P|; this immediately yields that $P_1=C_C(P_1)$.

Let J be the socle of $P'$; in particular, $J=J_1\times \langle a\rangle$, where $J_1$ is the socle of A and $o(a)\le 2$. Note that J is normal in G since $J_1\trianglelefteq G$ and, if $a\ne 1$, then JA/A is the only cyclic subgroup of $P_1/A$ made by elements of maximal p-height, so (the socle J of) AJ is characteristic in $P_1$ and hence normal in G

Suppose G/J is 2-nilpotent. Then $G=PQ$, where Q is a $2'$-subgroup of G such that $QJ\trianglelefteq G$. Now $A\trianglelefteq G$, so AJ/J is centralized by Q. Since AJ/J is divisible (and non-trivial) we have that A is centralized by Q. But $|AJ:A|\le 2$, and hence Q centralizes AJ. Then Q is normalized by P and G is 2-nilpotent. Therefore, recalling the structure of P, it is possible to assume that the socle S of $P_1$ is contained in the centre of P. In particular, $S\trianglelefteq G$.

Let $Z=C_G(S)$. Then $P\le Z$ and $Z\cap C=P_1$, so $Z=P$ is normal in G. But the normalizer of P is 2-nilpotent, and hence G is 2-nilpotent as required.

Suppose finally G is soluble-by-finite and either $p>2$, or $p=2$ and P is nilpotent. If $p=r$, then $D:=U$ is a G-invariant subgroup of finite index of P; while, if $p\ne r$, we can factor $O_{p'}(G)$ out (see Lemma 4.1), so now $D:=A$ is a G-invariant subgroup of finite index of P.

Write $P=DF$ for some finite subgroup F. Assume finally G is a counterexample to the statement of smallest possibly cardinality. By Theorem C of [1], $|G|\ge \aleph _0$. Now, G can be written as a union of an ascending chain of subgroups

$$\begin{aligned} F=E_0\le E_1\le E_2\le \cdots E_\alpha \le E_{\alpha +1}\le \cdots \bigcup _{\beta <\lambda }E_\beta =G \end{aligned}$$

of strictly smaller cardinality for some ordinal number $\lambda$. We construct a Sylow p-subgroup as follows: put $Q_0=\{1\}$; if $Q_\alpha$ is defined for some ordinal $\alpha <\lambda$ and is contained in $E_\alpha$, let $Q_{\alpha +1}$ be a Sylow p-subgroup of $E_{\alpha +1}$ extending $Q_\alpha$; if $\mu$ is a limit ordinal $<\lambda$, let $Q_\mu$ be the union of all $Q_\alpha$’s with $\alpha <\mu$. Then

$$\begin{aligned} Q=\bigcup _{\alpha <\lambda }Q_\alpha \end{aligned}$$

is a Sylow p-subgroup of G. Since the p-subgroups of G are conjugate, we may actually assume P has the property that, for each ordinal $\alpha <\lambda$, $P\cap E_\alpha$ is a Sylow p-subgroup of $E_\alpha$. Set $\beta <\lambda$. If x is any element of $E_\beta$ normalizing $P\cap E_\beta$, then $F^x\le P$ and consequently $P^x=(DF)^x=P$, so $x\in N_G(P)$. But $N_G(P)$ is p-nilpotent, and hence the minimality assumption entails that $E_\beta$ is p-nilpotent. Therefore G is locally p-nilpotent and hence p-nilpotent. The statement is proved.

(1$_c$)   In this case P is modular. If $r\ne 2$, then P is finite and we are in case (1$_a$). Thus, $r=p=2$ and the linear simple groups in B/S have characteristic 2. It follows from Theorem 4.3 that these groups are isomorphic to ${\text {PSL}}(2,{\mathfrak {K}})$ for some infinite locally finite fields ${\mathfrak {K}}$. We show that $B=S$, so the statement is proved by case (2). In order to do this, we may assume by Lemma 4.1 that $S=\{1\}$. Let F be a finite subgroup of G such that $P=F(P\cap B)$; note that, since B is normal, we have that $P\cap B$ is a Sylow 2-subgroup of B. Let L be a normal simple subgroup of B; in particular, $L\cap P$ is a Sylow 2-subgroup of L. If F does not fix L by conjugation, we find a section of P that is dihedral of order 8, contradicting [24, Theorem 2.4.14]. Therefore F normalizes L and $P\cap L$; in particular, $F(P\cap L)$ is a Sylow 2-subgroup of FL. Now, every element of FL normalizing $F(P\cap L)$ is easily seen to normalize $P=F(P\cap B)$, so $FL\in {\mathfrak {X}}_2$. Finally, the normalizer of $F(P\cap L)$ in FL is soluble, so it must be 2-nilpotent by (2). On the other hand, the normalizer in L of $P\cap L$ is a $2'$-group that does not centralize $P\cap L$. This contradiction shows that $B=S$, so G is soluble-by-finite and (2) complete the proof. $\square$

It is now possible to prove our main theorem.

Theorem 1.24

Let G be a homomorphic image of a periodic linear group. Then G is a soluble PT-group if and only if it satisfies ${\mathfrak {X}}_p$ for every $p\in \pi (G)$.

Proof

Suppose G is a soluble PT-group. Then it satisfies the conditions of Theorem 2.5. Let $N\!=\,\varrho _{{{\textbf {L}}}{\mathfrak {N}}}(G)$. Then N is abelian and is a maximal $\pi$-subgroup of G for some set of odd primes $\pi$; in particular, there is a $\pi '$-subgroup X of G such that $G=X\ltimes N$. Moreover, G acts on N as power automorphisms. Now, let p be any prime and let P be a Sylow p-subgroup of G. If $p\in \pi$, then all subgroups of P are normal in $G=N_G(P)$, so $G\in {\mathfrak {X}}_p$. If $p\not \in \pi$, we may assume $P\le X$, so we can find a $p'$-subgroup Y of X such that $G=P\ltimes (YN)$. It is now clear that $C_G(P)=N_G(P)$, so again G satisfies ${\mathfrak {X}}_p$.

Suppose conversely that G satisfies ${\mathfrak {X}}_p$ for every $p\in \pi (G)$. Write $G=H/N$, where H is a periodic linear group of characteristic r and $N\trianglelefteq H$. The structure of a periodic linear group shows that G has normal subgroups $U\le A\le S\le B$ such that U is a nilpotent r-group of finite exponent, A/U is an abelian $r'$-group whose Sylow subgroups are Černikov, S/A is finite-soluble, B/S is the direct product of finitely many infinite linear simple groups, and G/B is finite.

Suppose $B/S\ne \{1\}$. Since the Sylow subgroups of G are PT-groups, we also get that the Sylow subgroups of every homomorphic image of G are PT-groups: this follows from (Theorem 9.15 of [29] and) the fact that the Sylow subgroups of odd order are modular, while the Sylow 2-subgroups satisfy Lemma 2.3. Therefore Corollary 4.4 yields that B/S is the direct product of groups $B_i/S$ isomorphic to ${\text {PSL}}(2,{\mathfrak {K}}_i)$ for certain infinite locally finite fields ${\mathfrak {K}}_i$ of characteristic $p_i$.

Fix i, and note that $p_i=2$ implies that the Sylow 2-subgroups of G are nilpotent of finite exponent, so G is soluble by Theorem 4.6, a contradiction. Thus $p_i\ne 2$. Repeated applications of Lemma 4.1 yield that G/S satisfies ${\mathfrak {X}}_{p_i}$. Let P/S be a Sylow $p_i$-subgroup of G/S. Then there is a finite subgroup F/S of P/S such that $P=F(P\cap B)$; note here that $(P\cap B)/S$ is a Sylow $p_i$-subgroup of B/S, so $(P\cap B_i)/S$ is a Sylow $p_i$-subgroup of $B_i/S$. Clearly, all elements of $P\cap B$ normalize $B_i/S$. Moreover, if there is an element of F that does not fix (by conjugation) $B_i/S$, then we get a section of P isomorphic to $C_p\wr C_p^m$, a contradiction because the latter is not modular. Thus F/S normalizes $B_i/S$ and consequently a Sylow $p_i$-subgroup of $FB_i/S$ is $F(P\cap B_i)/S$. But any element of $FB_i/S$ normalizing $F(P\cap B_i)/S$ must also normalize $P/S=F(P\cap B)/S$, so $FB_i/S\in {\mathfrak {X}}_{p_i}$. Let T/S be the normalizer of $(P\cap B_i)/S$ in $FB_i/S$. Then

$$\begin{aligned} T/(P\cap B_i)=F(P\cap B_i)/(P\cap B_i)\ltimes J/(P\cap B_i), \end{aligned}$$

where $J/(P\cap B_i)$ is an abelian $p_i'$-group with locally cyclic Sylow subgroups. Of course, $T/(P\cap B_i)\in {\mathfrak {X}}_{p_i}$ is soluble. Let $K/(P\cap B_i)$ be either an infinite Sylow q-subgroup of $J/(P\cap B_i)$ centralizing $F(P\cap B_i)/(P\cap B_i)$, or the subgroup generated by all Sylow q-subgroups of $J/(P\cap B_i)$ such that $q>p_i$; write $K=Q(P\cap B_i)$ for some $p_i'$-subgroup Q. In the former case we have that $[K,F]\le P\cap B_i$, so Q normalizes $F(P\cap B_i)$, and hence Q acts as power automorphisms on $P\cap B_i$; since $Q\simeq K/(P\cap B_i)$ is an infinite locally cyclic q-group, and $q\ne p_i$, we get that K is abelian, an obvious contradiction (recall that $B_i\simeq {\text {PSL}}(2,{\mathfrak {K}}_i)$). In the latter case, $F(P\cap B_i)/S$ is a Sylow $p_i$-subgroup of the ${\mathfrak {X}}_{p_i}$-group FK/S. Then Theorem 4.6 shows that FK/S is $p_i$-nilpotent, so $[F,K]\le S$ and we obtain a contradiction similarly to the previous case.

Thus G is soluble-by-finite. Let $p=p_0$ be the smallest prime in $\pi (G)$. Then Theorem 4.6 yields that G is p-nilpotent. Set $N=O_{p'}(G)$. Then $N\in {\mathfrak {X}}_q$ for every $q\in \pi (N)$, and again N is $p_1'$-nilpotent, where $p_1'$ is the smallest prime in $\pi (N)$. Continuing in this way, we can define a (possible finite) strictly increasing series of primes

$$\begin{aligned} p_0=p<p_1<p_2<\cdots<p_\ell<p_{\ell +1}<\cdots \end{aligned}$$

and a (possible finite) strictly descending series of normal subgroups of G

$$\begin{aligned} G=N_0>N=N_1>N_2>\cdots>N_\ell>N_{\ell +1}>\cdots \end{aligned}$$

such that

$$\begin{aligned} \bigcap _{i}N_i=\{1\} \end{aligned}$$

and $N_i/N_{i+1}$ is a $p_i$-group for every i.

Since $G/N_1$ is a $p_0$-group and $N_1$ is $p_0'$-subgroup, it follows that $G/N_1$ is a PT-group. Assume $G/N_i$ is a PT-group for some i. Then $N_i/N_{i+1}$ is a normal Sylow $p_i$-subgroup of $G/N_{i+1}\in {\mathfrak {X}}_{p_i}$, so Lemma 2.2 implies that $G/N_{i+1}$ is a PT-group. This means that $G/N_i$ is a PT-group for every i.

Let R be the locally nilpotent residual of G. Then $RN_i/N_i$ is the locally nilpotent residual of $G/N_i$, so Theorem 2.5 yields that R is an abelian Sylow $\pi$-subgroup of G for some set of primes $\pi$. Since G/R is locally nilpotent, we have that G/R is a PT-group, so a final application of Lemma 2.2 completes the proof. $\square$

Note that the argument in fourth paragraph of the proof proves the following statement: Let p be a prime and let G be a homomorphic image of a periodic linear group of characteristic p . If $G\in {\mathfrak {X}}_p$ , then G is soluble-by-finite.

In the remaining of the section we give some more information about the property ${\mathfrak {X}}_p$. We start characterizing the case in which p is odd.

Theorem 1.25

Let p be an odd prime, G a homomorphic image of a periodic linear group, and $P\in {\text {Syl}}_p(G)$. Then $G\in {\mathfrak {X}}_p$ if and only if one of the following alternatives holds:

• (1) G is p-nilpotent;

• (2) P is abelian and all subgroups of P are normal in $N_G(P)$.

Proof

Only the necessity of the conditions is in doubt. It is certainly possible to assume P is non-abelian (see the argument at the beginning of the proof of Theorem 4.6). Write $G=H/N$, where H is a linear group of characteristic r, and $N\trianglelefteq H$. If $p\ne r$, then P is finite by Lemma 2.4. Then it easily follows from Theorem B of [1] that G is locally p-nilpotent, so G is p-nilpotent and we are done. Assume $p=r$, so P has finite exponent. As we have already seen several times, G has a normal soluble subgroup S and a normal subgroup B such that B/S is the direct product of infinite simple subgroups and G/B is finite. If PS/S is not finite, there must be some normal subgroup L/S of B/S that is an infinite linear simple group of characteristic p, and contains an infinite Sylow p-subgroup. Then $L/S\simeq {\text {PSL}}(2,{\mathfrak {K}})$ for some infinite locally finite field ${\mathfrak {K}}$ (see Corollary 4 of [25], Theorem 3.7 of [26], Corollary 9.32 of [29], and Theorem 4.18 of [16]). Now, the argument employed in the fourth paragraph of the proof of Theorem 4.7 gives a contradiction (see the above remark).

This means that $P=FD$ for some normal subgroup D of G and some finite subgroup F. The argument employed in the final paragraph of Theorem 4.6 yields that G is (locally) p-nilpotent (here we need also Theorem B of [1]). The statement is proved. $\square$

Now, we characterize property ${\mathfrak {X}}_p$ in terms of pronormal subgroups (see Theorem 4.11). Recall that if G is a group, a subgroup H is pronormal in G if H and $H^g$ are conjugate in $\langle H,H^g\rangle$ for any $g\in G$. Clearly, normal subgroups, maximal subgroups and Sylow p-subgroups of (homomorphic images of) linear groups are pronormal subgroups, so this property encompasses many natural embedding properties. In time, pronormal subgroups have proved to be a very useful tool in group theory. For example, Peng [20] showed that a finite soluble T-group is just a finite group in which all subgroups are pronormal; see also [12, 21],...and their introductions for many other results on pronormal subgroups.

One nice feature of pronormal subgroups is that a subgroup of an arbitrary group is normal if and only it is pronormal and subnormal. It turns out that in case of (homomorphic images of) periodic linear groups, one can replace subnormality with seriality. Recall that a subgroup X of a group G is serial if there is a chain of subgroups ${\mathcal {S}}$ running from X to G such that $H\trianglelefteq K$ when $H,K\in {\mathcal {S}}$ and there is no other subgroup of ${\mathcal {S}}$ between H and K; serial subgroups usually play a very relevant role in the theory of embedding properties of locally finite groups (see for instance [14]).

Lemma 1.26

Let G be a homomorphic image of a periodic linear group, and let X be a subgroup of G. Then X is normal in G if and only if X is serial and pronormal in G.

Proof

Assume X is serial and pronormal in G. Theorem 2.14 of [11] entails that X is ascendant in G. Then there is an ascending series

$$\begin{aligned} X=X_0\trianglelefteq X_1\trianglelefteq \ldots \, X_\alpha \trianglelefteq X_{\alpha +1}\trianglelefteq \ldots \, X_\mu =G \end{aligned}$$

running from X to G (here $\alpha <\mu$ are ordinal numbers). Since a pronormal subnormal subgroup is normal, an easy induction on $\mu$ yields that X is normal in G. $\square$

In order to prove our characterization we need an extension of a result of Rose that characterizes pronormal primary subgroups of a finite group.

Theorem 1.27

Let p be a prime and let G be a homomorphic image of a periodic linear group. If P is a p-subgroup of G, then P is pronormal in G if and only if P is normal in the normalizer of any Sylow p-subgroup containing it.

Proof

Suppose first P is pronormal in G, and let $P_1$ be any Sylow p-subgroup of G containing P. Then $P_1$ is locally nilpotent, so P is serial in $P_1$ (see for instance [22, 12.4.4]) and consequently is also serial in $N_G(P_1)$. By Lemma 4.9 we get that P is normal in $N_G(P_1)$.

Suppose conversely that P is normal in the normalizer of any Sylow p-subgroup containing it. Let $g\in G$ and put $X=\langle P,P^g\rangle$. If $P_1$ is a Sylow p-subgroup of X containing P, then there is $x\in X$ such that $P^g\le P_1^x$. Now, let $P_2$ be a Sylow p-subgroup of G containing $P_1$, so in particular $P\le P_2$. Thus there is $h\in \langle P_2,P_2^{xg^{-1}}\rangle$ such that $P_2^{xg^{-1}}=P_2^{h}$, so $z=hgx^{-1}\in N_G(P_2)$ and, by hypothesis, $P^z=P$. Now,

$$\begin{aligned} P^{gx^{-1}}\le P_1\le P_2 \end{aligned}$$

so $P\le P_2^{xg^{-1}}$. By hypothesis P is normal in $P_2$ and in $P_2^{xg^{-1}}$, and hence $h\in N_G(P)$. Therefore

$$\begin{aligned} P=P^z=P^{hgx^{-1}}=P^{gx^{-1}} \end{aligned}$$

and consequently $P^x=P^g$. The statement is proved. $\square$

Theorem 1.28

Let p be a prime and let G be a homomorphic image of a periodic linear group. Let P be a Sylow p-subgroup of G and let D be its finite residual. Then $G\in {\mathfrak {X}}_p$ if and only if the following conditions hold:

• (1) P is a PT-group;

• (2) if P is nilpotent, then all normal subgroups of P are pronormal in G;

• (3) if P is non-nilpotent, then:

  • (3a) $D\le X\trianglelefteq P$ and $|X/D|={\text {exp}}(P/D)$ imply that X is pronormal in G;

  • (3b) $X\le D$ and $X=X^p$ imply that X is pronormal in G.

Proof

Suppose first $G\in {\mathfrak {X}}_p$. Then P is certainly a PT-group and the usual argument entails that every subnormal subgroup of P is normalized by any $p'$-element of $N_G(P)$, so in particular every normal subgroup of P is normal in $N_G(P)$. If P is abelian, Theorem 4.10 proves (2). Assume P is non-abelian but nilpotent. If $p=2$, then G is 2-nilpotent by Theorem 4.6; if $p>2$, then G is p-nilpotent by Theorem 4.8. Thus, in any case, G is p-nilpotent. Let $N=O_{p'}(G)$; in particular, $G=PN$. If X is any normal subgroup of P and $g\in G$, then $XN=X^gN$. Since X and $X^g$ are Sylow p-subgroups of $\langle X,X^g\rangle \le XN$, we have that X is conjugate to $X^g$ in $\langle X,X^g\rangle$, so X is pronormal in G.

Assume finally, P is non-nilpotent, so $p=2$ and P satisfies the conditions of Lemma 2.3. Of course, if X is any divisible subgroup of D, then X is contained in the finite residual of any Sylow subgroup containing it, so it is a normal subgroup of (the normalizer of) that Sylow subgroup, and consequently Theorem 4.10 shows that it is pronormal in G.

If P/D is non-Dedekind, then P satisfies condition (2b) of Lemma 2.3. Let X/D be any (normal) cyclic subgroup of P/D with $|X/D|={\text {exp}}(P/D)$. If $P_1$ is any Sylow p-subgroup containing X, then the finite residual of $P_1$ is D and X/D is normal in $P_1/D$ since $|X/D|={\text {exp}}(P_1/D)$. Thus X is pronormal in G by Theorem 4.10.

If P/D is Dedekind, X is any normal subgroup of P containing D, and $P_1$ is any Sylow p-subgroup of G containing X, then the finite residual of $P_1$ is D, so again X is normal in $P_1$ and pronormal in G by Theorem 4.10. The sufficiency is thus proved.

Conversely, suppose P is a Sylow p-subgroup of G satisfying the requirements in the statement. Assume first P is nilpotent. Thus, if $X\trianglelefteq P$, then X is pronormal in G, so X is subnormal and pronormal in $N_G(P)$, and hence even normal in $N_G(P)$. Therefore, if P is abelian, the result follows.

Assume P is non-abelian (but still nilpotent). Then $P=\langle x\rangle A$, where A is abelian and every subgroup of A is normal in P. Let g be a $p'$-element in $N_G(P)$. Then g induces power automorphisms in $P/P'$ and in A. If $P=P'[P,g]$, then

$$\begin{aligned}{}[P,A]\le \big [[x,g],A\big ]=\{1\} \end{aligned}$$

since power automorphisms are central; it follows that P is abelian, a contradiction. Thus $P\ne P'[P,g]$. Since g is a $p'$-element, we have that g centralizes $P/P'$, so g centralizes P. This proves the result in this case.

Assume P is non-nilpotent, so P satisfies the conditions of Lemma 2.3. Let D be the finite residual of P, and choose a $2'$-element $g\in N_G(P)$. Since the locally cyclic subgroups of D are normal in P, we see that g centralizes D. Now, every cyclic subgroup C/D of G/D of maximal order is normal in G/D by Lemma 2.3, so C is normalized (and actually centralized) by g. Since $P/P^2$ is generated by elements $aP^2$, where $o(aD)={\text {exp}}(P/D)$, we get that $P/P^2$ is centralized by g, so g centralizes $P/P^2$ and hence even P. The statement is proved. $\square$

As one can easily deduce from Lemma 2.3, the property ${\mathfrak {X}}_2$ is not inherited by subgroups. However, the following easy consequence of Theorem 4.11 shows that only the cases described by Lemma 2.3 have such a bad behaviour.

Corollary 1.29

Let p be a prime and let G be a homomorphic image of a periodic linear group. Set $H\le G\in {\mathfrak {X}}_p$. If either $p>2$, or $p=2$ and the Sylow 2-subgroups of G are nilpotent, then $H\in {\mathfrak {X}}_p$.

Proof

Let P be a Sylow p-subgroup of G. If P is non-abelian, it follows from Theorems 4.6 and 4.8 that G is p-nilpotent. In this case it is almost obvious that any subgroup of G satisfies ${\mathfrak {X}}_p$.

Assume P is abelian, and let $P_1$ be a Sylow p-subgroup of H. If X is any subgroup of $P_1$, then X is normal in any Sylow p-subgroup of G containing it, so it is pronormal in G by Theorem 4.11. This means that X is pronormal in H, so a further application of Theorem 4.11 shows that H satisfies ${\mathfrak {X}}_p$. $\square$

A further consequence of Theorem 4.11 is that the class ${\mathfrak {X}}_p$ is local when $p>2$ and we restrict our attention to (homomorphic images of) periodic linear groups. Recall that, given a group class ${\mathfrak {X}}$, a group G is locally ${\mathfrak {X}}$ if every finite subset of G is contained in an ${\mathfrak {X}}$-subgroup.

Corollary 1.30

Let p be a prime and let G be a homomorphic image of a periodic linear group. If G is locally ${\mathfrak {X}}_p$, then $G\in {\mathfrak {X}}_p$ provided that the Sylow p-subgroups of G are nilpotent.

Proof

Let P be a Sylow p-subgroup of G, and assume every finite subset E of G is contained in an ${\mathfrak {X}}_p$-subgroup $X_E$. It follows from Corollary 4.12 that we may take $X_E=\langle E\rangle$. Now, we need to prove that every element g of $N_G(P)$ normalizes every cyclic subgroup $\langle x\rangle$ of P. Clearly, G can be written as a union of an ascending chain of subgroups $\{F_\alpha \}_{\alpha <\lambda }$ of strictly smaller cardinality for some ordinal number $\lambda$. We may also assume $F_\alpha \cap P$ is a Sylow p-subgroup of $F_\alpha$. Let $\beta <\lambda$ with $g,x\in F_\beta$. Then g normalizes $F_\beta \cap P$, and hence by induction $g\in N_F\big (\langle x\rangle \big )$. $\square$

The previous result should be seen in comparison with the fact that the class of PT-groups is local, that is, it contains every group in which the finite subsets are contained in PT-subgroups.

Lemma 1.31

The following classes of groups are local:

• (1) the class of PT-groups;

• (2) the class of groups in which every subnormal subgroup is permutable.

Proof

(1)   Let G be a group in which every finite subset E is contained in a PT-subgroup $X_E$, and let $H\le K$ be subgroups of G such that H is permutable in K, and K is permutable in G. Let F be any finite subset of G, and let $g\in X_F\cap K$. Then $H\langle g\rangle$ is a subgroup, so also

$$\begin{aligned} \langle g\rangle H\cap X_F=\langle g\rangle (X_F\cap H) \end{aligned}$$

is a subgroup, and consequently $X_F\cap H$ is permutable in $X_F\cap K$. Similarly, $X_F\cap K$ is permutable in $X_F=X_F\cap G$. Therefore $X_F\cap H$ is permutable in $X_F$. It easily follows that H is permutable in G and we are done.

(2)   This can be proved similarly to (1). $\square$

Note finally that if we require all countable subsets of a homomorphic image G of a periodic linear group to be contained in ${\mathfrak {X}}_p$-subgroups, then G itself satisfies ${\mathfrak {X}}_p$: this is because Černikov groups are countable.