Almost Engel linear groups

Abstract

A group G is almost Engel if for every $g\in G$ there is a finite set ${\mathcal {E}}(g)$ such that for every $x\in G$ all sufficiently long commutators $[x,{}_n\, g]$ belong to ${\mathcal {E}}(g)$, that is, for every $x\in G$ there is a positive integer n(x, g) such that $[x,{}_n\, g]\in {\mathcal {E}}(g)$ whenever $n(x,g)\le n$. A group G is almost nil if it is almost Engel and for every $g\in G$ there is a positive integer n depending on g such that $[x,{}_s g]\in {\mathcal {E}}(g)$ for every $x\in G$ and every $s\ge n$. We prove that if a linear group G is almost Engel, then G is finite-by-hypercentral. If G is almost nil, then G is finite-by-nilpotent.

1 Introduction

By a linear group we understand here a subgroup of GL(m, F) for some field F and a positive integer m. An element g of a group G is called a (left) Engel element if for any $x\in G$ there exists $n=n(x,g)\ge 1$ such that $[x,{}_n\, g]=1$. As usual, the commutator $[x,{}_n\, g]$ is defined recursively by the rule

$$\begin{aligned}{}[x,{}_n\, g]=[[x,{}_{n-1}\, g],g] \end{aligned}$$

assuming $[x,{}_0\, g]=x$. If n can be chosen independently of x, then g is a (left) n-Engel element. A group G is called Engel if all elements of G are Engel. It is called n-Engel if all its elements are n-Engel. A group is said to be locally nilpotent if every finite subset generates a nilpotent subgroup. Clearly, any locally nilpotent group is an Engel group. It is a long-standing problem whether any n-Engel group is locally nilpotent. Engel linear groups are known to be locally nilpotent (cf. [2, 3]).

We say that a group G is almost Engel if for every $g\in G$ there is a finite set ${\mathcal {E}}(g)$ such that for every $x\in G$ all sufficiently long commutators $[x,{}_n\, g]$ belong to ${\mathcal {E}}(g)$, that is, for every $x\in G$ there is a positive integer n(x, g) such that $[x,{}_n\, g]\in {\mathcal {E}}(g)$ whenever $n(x,g)\le n$. (Thus, Engel groups are precisely the almost Engel groups for which we can choose ${\mathcal {E}}(g)=\{ 1\}$ for all $g\in G$.) We say that a group G is nil if for every $g\in G$ there is a positive integer n depending on g such that g is n-Engel. The group G will be called almost nil if it is almost Engel and for every $g\in G$ there is a positive integer n depending on g such that $[x,{}_s\,g]\in {\mathcal {E}}(g)$ for every $x\in G$ and every $s\ge n$.

Almost Engel groups were introduced in [6] where it was proved that an almost Engel compact group is necessarily finite-by-(locally nilpotent). The purpose of the present article is to prove the following related result.

Theorem 1.1

Let G be a linear group.

• 1. If G is almost Engel, then G is finite-by-hypercentral.

• 2. If G is almost nil, then G is finite-by-nilpotent.

Recall that the union of all terms of the (transfinite) upper central series of G is called the hypercenter. The group G is hypercentral if it coincides with its hypercenter. The hypercentral groups are known to be locally nilpotent (see [10, P. 365]). By well-known results obtained in [2, 3], if under the hypotheses of Theorem 1.1 the group G is Engel or nil, then G is hypercentral or nilpotent, respectively.

As a warning to the reader, we mention that in many articles (including some of the author) the expression “the group G is almost an X-group” for a property X means “G has an X-subgroup of finite index”. In the present paper, however, the meaning of the term “almost Engel” is different. It is hoped that this discrepancy does not lead to a confusion.

2 Preliminaries

Let G be a group and $g\in G$ an almost Engel element, so that there is a finite set ${\mathcal {E}}(g)$ such that for every $x\in G$ there is a positive integer n(x, g) with the property that $[x,{}_n\, g]$ belongs to ${\mathcal {E}}(g)$ whenever $n(x,g)\le n$. If ${\mathcal {E}}'(g)$ is another finite set with the same property for possibly different numbers $n'(x,g)$, then ${\mathcal {E}}(g)\cap {\mathcal {E}}'(g)$ also satisfies the same condition with the numbers $n''(x,g)=\max \{n(x,g),n'(x,g)\}$. Hence there is a minimal set with the above property. The minimal set will again be denoted by ${\mathcal {E}}(g)$ and, following [6], called the Engel sink for g, or simply g -sink for short. From now on we will always use the notation ${\mathcal {E}}(g)$ to denote the (minimal) Engel sinks. In particular, it follows that for each $x\in {\mathcal {E}}(g)$ there exists $y\in {\mathcal {E}}(g)$ such that $x=[y,g]$. More generally, given a subset $K\subseteq G$ and an almost Engel element $g\in G$, we write ${\mathcal {E}}(g,K)$ to denote the minimal subset of G with the property that for every $x\in K$ there is a positive integer n(x, g) such that $[x,{}_n\, g]$ belongs to ${\mathcal {E}}(g,K)$ whenever $n(x,g)\le n$. Throughout the article we use the symbols $\langle X\rangle$ and $\langle X^G\rangle$ to denote the subgroup generated by a set X and the minimal normal subgroup of G containing X, respectively.

A group is said to have a property virtually if some subgroup of finite index has the property. The following lemma can be found in [8, Ch. 12, Lemma 1.2] or in [5, Lemma 21.1.4].

Lemma 2.1

A virtually abelian group contains a characteristic abelian subgroup of finite index.

As usual, we write $Z_i(G)$ for the ith term of the upper central series of G and $\gamma _i(G)$ for the ith term of the lower central series. A well-known theorem of Schur states that if G is central-by-finite, then the commutator subgroup $G'$ is finite (see [10, 10.1.4]). Baer proved that if, for a positive integer k, the quotient $G/Z_k(G)$ is finite, then so is $\gamma _{k+1}(G)$ (see [10, 14.5.1]). Recently, the following related result was obtained in [1] (see also [7]).

Theorem 2.2

Let G be a group and let H be the hypercenter of G. If G / H is finite, then G has a finite normal subgroup N such that G / N is hypercentral.

We will also require the Dicman Lemma (see [10, 14.5.7]).

Lemma 2.3

In any group a normal finite subset consisting of elements of finite order generates a finite subgroup.

In [9] Plotkin proved that if a group G has an ascending series whose quotients locally satisfy the maximal condition, then the Engel elements of G form a locally nilpotent subgroup. In particular we have the following lemma.

Lemma 2.4

Let G be a group having an ascending series whose quotients locally satisfy the maximal condition and let $a\in G$ be an Engel element. Then $\langle a^G\rangle$ is locally nilpotent.

Linear groups are naturally equipped with the Zariski topology. If G is a linear group, the connected component of G containing 1 is denoted by $G^0$. We will use (sometimes implicitly) the following facts on linear groups. All these facts are well-known and are provided here just for the reader’s convenience.

• If G is a linear group and N a normal subgroup which is closed in the Zarissky topology, then G / N is linear (see [12, Theorem 6.4]).

• Since finite subsets of G are closed in the Zariski topology, it follows that any finite subgroup of a linear group is closed. Hence G / N is linear for any finite normal subgroup N.

• If G is a linear group, the connected component $G^0$ has finite index in G (see [12, Lemma 5.3]).

• Each finite conjugacy class in a linear group centralizes $G^0$ (see [12, Lemma 5.5]).

• In a linear group any descending chain of centralizers is finite.

• A linear group generated by normal nilpotent subgroups is nilpotent (see Gruenberg [3]).

• Tits alternative: A finitely generated linear group either is virtually soluble or contains a subgroup isomorphic to a nonabelian free group (see [11]).

• The Burnside–Schur theorem: A periodic linear group is locally finite (see [12, 9.1]).

• Zassenhaus theorem: A locally soluble linear group is soluble. Every linear group contains a unique maximal soluble normal subgroup (see [12, Corollary 3.8]).

• Since the closure in the Zariski topology of a soluble subgroup is again soluble (see [12, Lemma 5.11]), it follows that the unique maximal soluble normal subgroup of a linear group is closed. In particular, if G is linear and R is the unique maximal soluble normal subgroup of G, then G / R is linear and has no nontrivial normal soluble subgroups.

• A locally nilpotent linear group is hypercentral (see [2] or [3]).

• Gruenberg theorem: The set of Engel elements in a linear group G coincides with the Hirsch–Plotkin radical of G. The set of right Engel elements coincides with the hypercenter of G (see [3]).

Here, as usual, the Hirsch–Plotkin radical of a group is the maximal normal locally nilpotent subgroup. An element $g\in G$ is a right Engel element if for each $x\in G$ there exists a positive integer n such that $[g,{}_n\, x]=1$.

3 Almost Engel elements in virtually soluble groups

In the present section we give certain criteria for a group containing almost Engel elements to be finite-by-nilpotent or finite-by-hypercentral. In particular, we prove that a virtually soluble group generated by finitely many almost Engel elements is finite-by-nilpotent (Theorem 3.3).

Lemma 3.1

Let $G=H\langle a_1,\dots ,a_s\rangle$, where H is a normal subgroup and $a_i$ are almost Engel elements. Assume that G / H is nilpotent. If $N\le H$ is a finite normal subgroup of H, then $\langle N^G\rangle$ is finite.

Proof

Suppose first that $s=1$ and write a in place of $a_1$. Let M be the subgroup generated by all commutators of the form $[x,{}_j\, a]$, where $x\in N$ and j is a nonnegative integer. Since both N and ${\mathcal {E}}(a)$ are finite, it follows that there exists an integer k such that M is contained in the product $\prod _{i=0}^kN^{a^i}$. It is clear that the product $\prod _{i=0}^kN^{a^i}$ is normal in H and a normalizes M. Therefore $\langle M^H\rangle$ is normal in G and is contained in $\prod _{i=0}^kN^{a^i}$. Moreover, $\langle N^G\rangle =\langle M^H\rangle$ so in the case where $s=1$ the lemma follows.

Therefore we will assume that $s\ge 2$ and use induction on s. Assume additionally that G / H is abelian. Set $H_0=H$ and $H_i=H_{i-1}\langle a_i\rangle$ for $i=1,\dots ,s$. The subgroups $H_i$ are normal in G and $H_s=G$. By induction, $K=\langle N^{H_{s-1}}\rangle$ is finite. Since $G=H_{s-1}\langle a_s\rangle$, the above paragraph shows that $\langle K^G\rangle$ is finite. Obviously, $\langle K^G\rangle =\langle N^G\rangle$ and so in the case where G / H is abelian the lemma follows.

We will now allow G / H to be nonabelian, say of nilpotency class c. We will use induction on c. Set $B=\langle a_s^G\rangle$ and $G_1=HB$. Since G / H is a finitely generated nilpotent group, it follows that each subgroup of G / H is finitely generated and so B has finitely many conjugates of $a_s$, say $a_s^{g_1}\dots ,a_s^{g_r}$ such that $G_1=H\langle a_s^{g_1}\dots ,a_s^{g_r}\rangle$. Since $G_1/H$ has nilpotency class at most $c-1$, by induction $\langle N^{G_1}\rangle$ is finite. We now note that $G=G_1\langle a_1,\dots ,a_{s-1}\rangle$ so the induction on s completes the proof. $\square$

Lemma 3.2

Let $G=H\langle a\rangle$, where H is a virtually abelian normal subgroup and a is an almost Engel element. Then $\langle a^G\rangle$ is finite-by-(locally nilpotent).

Proof

Assume that G is a counter-example with $|{\mathcal {E}}(a)|$ as small as possible. In view of Lemma 2.1 we can choose a maximal characteristic abelian subgroup V in H. Since V is abelian, we have $[v_1,a][v_2,a]=[v_1v_2,a]$ for any $v_1,v_2\in V$. In other words, a product of two commutators of the form [v, a], where $v\in V$, again has the same form. Therefore ${\mathcal {E}}(a,V)$ is a finite subgroup. Obviously, the normalizer in G of ${\mathcal {E}}(a,V)$ has finite index. It follows that ${\mathcal {E}}(a,V)$ is contained in a finite normal subgroup N. If ${\mathcal {E}}(a,V)\ne 1$, we pass to the quotient G / N and use induction on $|{\mathcal {E}}(a)|$. Therefore without loss of generality we will assume that ${\mathcal {E}}(a,V)=1$, that is, a is Engel in $V\langle a\rangle$. Since ${\mathcal {E}}(a)$ consists of commutators of the form [x, a] with $x\in {\mathcal {E}}(a)$, it follows that ${\mathcal {E}}(a)\cap V=\{1\}$. Let $C_0=1$ and

$$\begin{aligned} C_i=\{v\in V\ \vert \ [v,a]\in C_{i-1}\} \end{aligned}$$

for $i=1,2,\dots$. Since a is Engel in V, we have $V=\cup _i C_i$.

Let $T=\langle {\mathcal {E}}(a),a\rangle$ and $U=V\cap T$. We observe that U is a finitely generated abelian subgroup. In view of the fact that V is the union of the $C_i$ we deduce that there exists a positive integer n such that $U=C_n\cap U$.

For $i=0,\dots ,n$ set $U_i=C_i\cap U$. Thus, $U=U_n$. Observe that $U_1$ centralizes a and therefore $U_1$ normalizes the set ${\mathcal {E}}(a)$. Denote by $W_{1}$ the intersection $U_1\cap C_G({\mathcal {E}}(a))$. Since ${\mathcal {E}}(a)$ is finite, it follows that $W_1$ has finite index in $U_1$. Further, it is clear that $W_1$ is contained in the center Z(T).

The finiteness of the index $[U_1:W_1]$ implies that $U_2$ contains a normal in T subgroup $W_2$ such that the index $[U_2:W_2]$ is finite, and $[W_2,T]\le W_1$. Thus, $W_2$ is contained in $Z_2(T)$, the second term of the upper central series of T.

Next, in a similar way we conclude that $U_3\cap Z_3(T)$ has finite index in $U_3$ and so on. Eventually, we deduce that $U\cap Z_n(T)$ has finite index in U. Thus, $T/Z_n(T)$ is finite-by-cyclic and therefore there exists a positive integer k such that $a^k\in Z_{n+1}(T)$. Hence, $T/Z_{n+1}(T)$ is finite and so, in view of Baer’s theorem, we deduce that T is finite-by-nilpotent. In particular, for some positive integer r the subgroup $\gamma _r(T)$ is finite. The observation that for each $x\in {\mathcal {E}}(a)$ there exists $y\in {\mathcal {E}}(a)$ such that $x=[y,g]$ guarantees that $\mathcal E(a)$ is contained in $\gamma _r(T)$. In particular, we proved that the subgroup $\langle {\mathcal {E}}(a)\rangle$ is finite. Because V is abelian, it is obvious that V normalizes $V\cap \langle {\mathcal {E}}(a)\rangle$. Thus, $V\cap \langle \mathcal E(a)\rangle$ is a finite subgroup with normalizer of finite index. It follows that $V\cap \langle \mathcal E(a)\rangle$ is contained in a finite normal subgroup of G. We can factor out the latter and without loss of generality assume that $V\cap \langle {\mathcal {E}}(a)\rangle =1$.

Recall that $C_1=C_V(a)$. Therefore $C_1$ normalizes $\langle {\mathcal {E}}(a)\rangle$ and in view of the fact that $V\cap \langle {\mathcal {E}}(a)\rangle =1$ we conclude that $C_1$ centralizes $\langle {\mathcal {E}}(a)\rangle$. So $C_1\le Z(VT)$. Same argument shows that $C_2/C_1\le Z(VT/C_1)$ and, more generally, $C_{i+1}/C_i\le Z(VT/C_i)$ for $i=0,1,2\dots$. Thus, $V\le Z_\infty (VT)$ where $Z_\infty (VT)$ stands for the hypercenter of T. Of course, it follows that there exists a positive integer k such that $a^k\in Z_\infty (VT)$. We deduce that $Z_\infty (VT)$ has finite index in VT. Theorem 2.2 now tells us that VT has a finite normal subgroup N such that the quotient group (VT) / N is hypercentral. The hypercentral groups are locally nilpotent and so VT is finite-by-(locally nilpotent). The observation that for each $x\in {\mathcal {E}}(a)$ there exists $y\in {\mathcal {E}}(a)$ such that $x=[y,g]$ guarantees that ${\mathcal {E}}(a)$ is contained in N.

Since VT has finite index in G, Dicman’s lemma tells us that G contains a finite normal subgroup R such that ${\mathcal {E}}(a)\subseteq N\le R$. The image of a in G / R is Engel and the required result follows from Lemma 2.4. $\square$

Theorem 3.3

A virtually soluble group generated by finitely many almost Engel elements is finite-by-nilpotent.

Proof

Let G be a virtually soluble group generated by finitely many almost Engel elements $a_1,\dots ,a_s$ and let S be a normal soluble subgroup of finite index in G. We assume that $S\ne 1$ and let V be the last nontrivial term of the derived series of S. By induction on the derived length of S we assume that G / V is finite-by-nilpotent. Therefore G contains a normal subgroup H such that V has finite index in H and the quotient G / H is nilpotent. For $i=1,\dots ,s$ set $G_i=H\langle a_i\rangle$. By Lemma 3.2 each subgroup $\langle a_i^{G_i}\rangle$ has a finite normal subgroup $N_i$ such that $\langle a_i^{G_i}\rangle /N_i$ is locally nilpotent. Since $G_i/H$ are abelian, it is clear that all quotients $G_i/H\cap N_i$ are locally nilpotent and so, replacing if necessary $N_i$ by $H\cap N_i$, without loss of generality we can assume that all subgroups $N_i$ are normal subgroups of H. Therefore the product of the subgroups $N_i$ is finite. By Lemma 3.1 the product of $N_1\cdots N_s$ is contained in a finite subgroup N which is normal in G. Obviously the images in G / N of the generators $a_1,\dots ,a_s$ are Engel. Thus, G / N is a virtually soluble group generated by finitely many Engel elements. It follows from Lemma 2.4 that G / N is nilpotent. The proof is complete. $\square$

The next lemma is well-known. For the reader’s convenience we provide the proof.

Lemma 3.4

Let $G=H\langle a\rangle$, where H is a nilpotent normal subgroup and a is a nil element. Then G is nilpotent.

Proof

Suppose that a is n-Engel. Let $K=Z(H)$ and set $K_0=K$ and $K_{i+1}=[K_i,a]$ for $i=0,1,\dots$. Then $K_{n-1}\le K\cap C_K(a)$ and so $K_{n-1}\le Z(G)$. Moreover we observe that $[K_{i-1},G]\le K_i$ and it follows that $K_{n-i}\le Z_{i}(G)$ for $i=1,2,\dots ,n$. Therefore $K\le Z_{n}(G)$. Passing to the quotient $G/Z_{n}(G)$ and using induction on the nilpotency class of H we deduce that if H is nilpotent with class c, then G is nilpotent with class at most cn. $\square$

Lemma 3.5

Let $G=H\langle a\rangle$, where H is a hypercentral normal subgroup and a is an Engel element. Then G is hypercentral.

Proof

It is sufficient to show that $Z(G)\ne 1$. Let $Z=Z(H)$. Since a is an Engel element, $C_Z(a)\ne 1$. Obviously, $C_Z(a)\le Z(G)$. The proof is complete. $\square$

Lemma 3.6

Let a be an almost Engel element in a group G and assume that ${\mathcal {E}}(a)$ is contained in a locally nilpotent subgroup. Then the subgroup $\langle {\mathcal {E}}(a)\rangle$ is finite.

Proof

Set $D=\langle {\mathcal {E}}(a)\rangle$. Without loss of generality we can assume that $G=D\langle a\rangle$. Since ${\mathcal {E}}(a)$ is finite, D is nilpotent and we can use induction on the nilpotency class of D. Thus, by induction assume that the quotient of D over its center is finite. By Schur’s theorem the derived group $D'$ is finite as well. Factoring out $D'$ we can assume that D is abelian. So now D is abelian and $D=[D,a]$. By [6, Lemma 2.3], $D={\mathcal {E}}(a)$ and hence D is finite. $\square$

Lemma 3.7

Let $G=H\langle a\rangle$, where H is a hypercentral normal subgroup.

• 1. If a is almost Engel, then G is finite-by-hypercentral.

• 2. If H is nilpotent and a is almost nil, then G is finite-by-nilpotent.

Proof

We will prove Claim 1 first. Assume that a is almost Engel. Let N be the product of all normal subgroups of G whose intersection with ${\mathcal {E}}(a)$ is $\{1\}$. It is easy to see that $N\cap \mathcal E(a)=\{1\}$ and N is the unique maximal normal subgroup with that property. Therefore $K\cap {\mathcal {E}}(a)\ne \{1\}$ whenever K is a normal subgroup containing N as a proper subgroup. Since ${\mathcal {E}}(a)$ is finite, the group G contains a minimal normal subgroup M such that $N<M$. Taking into account that H is hypercentral, we observe that M / N is central in H / N.

Let $D=\langle {\mathcal {E}}(a)\rangle \cap M$. It follows that $M=ND$. Suppose that D is not normal in M and set $L=N_M(N_M(D))$. Since M is hypercentral, it satisfies the normalizer condition and so $L\ne N_M(D)$. Obviously a normalizes both L and $N_M(D)$. Since a acts on $L/N_M(D)$ as an Engel element, the centralizer of a in $L/N_M(D)$ is nontrivial. Thus, L has a subgroup C such that $N_M(D)<C$ and C normalizes $N_M(D)\langle a\rangle$. Of course, D is normal in $N_M(D)\langle a\rangle$. By Lemma 3.5 the quotient of $N_M(D)\langle a\rangle$ by D is hypercentral. It is easy to see that D is a unique minimal normal subgroup of $N_M(D)\langle a\rangle$ whose quotient is hypercentral. Therefore D is characteristic in $N_M(D)\langle a\rangle$ and so C normalizes D. This is a contradiction since $N_M(D)<C$.

Hence, D is normal in M. Again, it is easy to see that D is a unique minimal normal subgroup of $M\langle a\rangle$ whose quotient is hypercentral. Therefore D is characteristic in M and so it is normal in G. We pass to the quotient G / D and Claim 1 now follows by straightforward induction on $|{\mathcal {E}}(a)|$.

We now assume that H is nilpotent and a is almost nil. We already know that G is finite-by-hypercentral. Factoring out a finite normal subgroup we can assume that G is hypercentral. In that case a is actually nil and so by Lemma 3.4 G is nilpotent. The proof of the lemma is complete. $\square$

4 Linear groups

Lemma 4.1

A virtually soluble almost Engel linear group is finite-by-hypercentral.

Proof

Suppose that G is a virtually soluble almost Engel linear group. Let S be a normal soluble subgroup of finite index in G. By induction on the derived length of S we assume that $S'$ is finite-by-hypercentral. Passing to the quotient over a normal finite subgroup without loss of generality we can assume that $S'$ is hypercentral. By Lemma 3.7 the subgroup $\langle S',x\rangle$ is finite-by-hypercentral for each $x\in G$. Thus, for each $x\in G$ there exists a finite characteristic subgroup $R_x\le \langle S',x\rangle$ such that $\langle S',x\rangle /R_x$ is hypercentral. Since $\langle S',x\rangle$ is normal in S, it follows that each element in $R_x$ has centralizer of finite index in S, hence centralizer of finite index in G. Therefore $G^0$ centralizes $R_x$ and it follows that $\langle S',x\rangle$ is hypercentral for each $x\in G^0$. The subgroup $\prod \langle S',x\rangle$, where x ranges over $S\cap G^0$, is locally nilpotent and therefore hypercentral. In particular $N=S\cap G^0$ is hypercentral and so G is virtually hypercentral. By Lemma 3.7 the subgroup $\langle N,x\rangle$ is finite-by-hypercentral for each $x\in G$. In other words, for each $x\in G$ there exists a finite characteristic subgroup $Q_x\le \langle N,x\rangle$ such that the quotient $\langle N,x\rangle /Q_x$ is hypercentral. Since N has finite index in G, it follows that G contains only finitely many subgroups of the form $\langle N,x\rangle$. Set $N_0=\prod _{x\in G}Q_x$. We see that $N_0$ is a finite normal subgroup. Pass to the quotient $G/N_0$. Now the subgroup $\langle N,x\rangle$ is hypercentral for each $x\in G$. It follows that N consists of right Engel elements and so, by the result of Gruenberg, N is contained in the hypercenter of G. It follows from Theorem 2.2 that G is finite-by-hypercentral, as required. $\square$

We are now ready to prove Theorem 1.1 in its full generality. For the reader’s convenience we restate it here.

Theorem 4.2

Let G be a linear group. If G is almost Engel, then G is finite-by-hypercentral. If G is almost nil, then G is finite-by-nilpotent.

Proof

Assume that G is almost Engel. In view of Lemma 4.1 it is sufficient to show that G is virtually soluble. By the Zassenhaus theorem a linear group is soluble if and only if it is locally soluble. Therefore it is sufficient to show that G is virtually locally soluble. It is clear that G does not contain a subgroup isomorphic to a nonabelian free group. Hence, by Tits alternative, any finitely generated subgroup of G is virtually soluble. Therefore, by Theorem 3.3, any finitely generated subgroup of G is finite-by-nilpotent. It becomes obvious that elements of finite order in G generate a periodic subgroup. Moreover, the quotient of G over the subgroup generated by all elements of finite order is locally nilpotent. Hence, G is virtually locally soluble if and only if so is the subgroup generated by elements of finite order. Therefore without loss of generality we can assume that G is an infinite periodic (and locally finite) group.

Let R be the soluble radical of G. We can pass to the quotient and without loss of generality assume that $R=1$. So in particular G has no nontrivial Engel elements. By the theorem of Hall–Kulatilaka G contains an infinite abelian subgroup [4]. We conclude that some centralizers in G are infinite. Since G satisfies the minimal condition on centralizers, it follows that G has a subgroup $D\ne 1$ such that the centralizer $C=C_G(D)$ is infinite while $C_G(\langle D,x\rangle )$ is finite for each $x\in G\setminus D$. Using that C is infinite we deduce from the Hall–Kulatilaka theorem that C contains an infinite abelian subgroup A. Obviously $A\le C_G(\langle D,A\rangle )$ and it follows that $A\le D$. Thus, $A\le Z(C)$.

Now choose $1\ne a\in A$. The centralizer C normalizes the finite set $\mathcal E(a)$ because $a\in Z(C)$. Hence, C contains a subgroup of finite index which centralizes ${\mathcal {E}}(a)$. It follows that $C_G(\langle D,{\mathcal {E}}(a)\rangle )$ is infinite and we conclude that $\mathcal E(a)$ is contained in D and C centralizes ${\mathcal {E}}(a)$. In particular, a centralizes ${\mathcal {E}}(a)$ and so ${\mathcal {E}}(a)=\{1\}$. Thus, a is an Engel element, a contradiction. This completes the proof of Claim 1.

Suppose now that G is almost nil. We already know that G is finite-by-hypercentral. Passing to a quotient over a finite normal subgroup we can assume that G is hypercentral. Then obviously G, being both hypercentral and almost nil, must be nil. By the result of Gruenberg, G is nilpotent. $\square$