Proof. Let E be any finitely generated subgroup of G. Then

$$ \begin{align*}[E/E\cap Q]\simeq[\langle E,Q\rangle/Q]\end{align*} $$

is a distributive lattice, and so, in particular, the finitely generated group $E/(E\cap Q)^E$ is cyclic. Moreover, the index $|(E\cap Q)^E:E\cap Q|$ is finite because $E\cap Q$ is permutable in G (see [Reference Schmidt13, Lemma 6.2.8]). Thus, the lattice $[E/E\cap Q]$ satisfies the maximal condition so that there exists an element x such that $E=\langle x,E\cap Q\rangle $ (see [Reference De Falco, de Giovanni and Musella3, Corollary 2.4]) and hence E is quasihamiltonian. Since the property of being quasihamiltonian is local, it follows that G is also a quasihamiltonian group.

Proof. Assume for a contradiction that N is not contained in $\zeta (G)$ so that $N\cap \zeta (G)$ is a proper subgroup of $N\cap \zeta _2(G)$ . Consider an element a of $\bigl (N\cap \zeta _2(G)\bigr )\setminus \zeta (G)$ and let k be the order of a. Since the map $g\mapsto [a,g]$ is an epimorphism from G onto $[a,G]$ with kernel $C_G(a)$ , the nontrivial subgroup $[a,G]$ is a homomorphic image of $G/C_G(N)$ , which is impossible because $[a,G]^k=[a^k,G]=\{1\}$ .

Proof. By Schlette’s result, we may suppose that Q is not contained in $\zeta (G)$ so that there is a $g\in G$ such that $[Q,g]\neq \{1\}$ . Since the subgroup $\langle g,Q\rangle $ is quasihamiltonian, it follows from Lemma 2.1 that Q has torsion-free rank $1$ . Let E be any finitely generated subgroup of G and put $X=\langle Q,E\rangle $ and $Y=Q^X$ . Then $X/Y$ is finitely generated and $\mathfrak {L}(X/Y)$ is a Černikov lattice so that $X/Y$ is finite. Moreover, the index $|Y:Q|$ is finite (see [Reference Schmidt13, Lemma 6.2.8]) and hence Q has finite index in X. It follows now from [Reference De Falco, de Giovanni and Musella2, Theorem 3.5] that X contains a finite normal subgroup L such that $X/L$ is quasihamiltonian. Thus, $X'$ is locally finite so the arbitrary choice of E yields that $G'$ is locally finite and hence the elements of finite order of G form a locally finite subgroup T containing $G'$ . If x is any element of T, the nonperiodic subgroup $\langle x,Q\rangle $ is quasihamiltonian so that $\langle x,Q\cap T\rangle $ is abelian by Lemma 2.1 and hence $Q\cap T\leq \zeta (T)$ .

Let a be any element of infinite order of G. Then $\langle a,Q\rangle $ is quasihamiltonian so that a normalises all subgroups of $Q\cap T$ and centralises each element of Q which has order either a prime or $4$ (see [Reference Schmidt13, Theorem 2.4.11]). Since G is generated by its elements of infinite order, the proof is complete.

Proof of Theorem 1.1

Since $G/Q$ is Černikov, the subgroup J has finite index in G and so it is enough to prove that J is quasihamiltonian. Thus without loss of generality, we may assume that $G/Q$ is a direct product of $r\geq 1$ Prüfer subgroups. Suppose first $r=1$ so that $G/Q$ is a group of type $q^\infty $ for some prime number q and hence the subgroup lattice of $G/Q$ is distributive. In this case, the group G is quasihamiltonian by Lemma 3.1. Assume now $r>1$ . Let p be any prime in the set $\pi =\pi (G/Q)$ and $P/Q$ any subgroup of type $p^\infty $ of $G/Q$ . Thus,

$$ \begin{align*}G/Q=P/Q\times V/Q,\end{align*} $$

where $V/Q$ is the direct product of $r-1$ Prüfer subgroups so that P is quasihamiltonian by the first part of the proof, while V is quasihamiltonian by induction on r. In particular, P and V are hypercentral and hence $G=PV$ is also a hypercentral group (see [Reference Robinson10, Part 1, page 51]).

Suppose that G is periodic and write $P=P_1\times P_2$ , where $P_1$ is a p-subgroup and $P_2$ is a $p'$ -subgroup. Since $P_1$ has infinite exponent, it is abelian (see [Reference Schmidt13, Theorem 2.4.14]) and so $P_1\leq \zeta (P)$ . Thus, $Q\leq C_G(P_1)$ and hence $G/C_G(P_1)$ is divisible. Application of Lemma 3.2 yields that $P_1$ is contained in $\zeta (G)$ . It follows that the largest $\pi $ -subgroup $O_\pi (G)$ of G lies in $\zeta (G)$ . However, the largest $\pi '$ -subgroup $O_{\pi '}(G)$ of G is contained in Q and hence $G=O_\pi (G)\times O_{\pi '}(G)$ is quasihamiltonian.

Assume finally that G is not periodic so that Q is not periodic. It follows from Lemma 3.3 that $G'$ is locally finite, whence the elements of finite order of G form a locally finite subgroup T and $G'\leq Q\cap T\leq \zeta (T)$ . Let x be any element of T and put $C=C_G(x)$ . Then $G'\leq C$ and so C is normal in G. Since $\langle x,Q\rangle $ is a nonperiodic quasihamiltonian group, the subgroup $\langle x\rangle $ is normalised by Q so that $N_G(\langle x\rangle )$ is normal in G and $G/N_G(\langle x\rangle )$ is a Černikov group. It follows that $G/C$ is also Černikov. If $R/C$ is the largest divisible subgroup of $G/C$ , Lemma 3.2 yields that R centralises x and so x has only finitely many conjugates in G. Thus, $G/C$ is finite so that $G=QC$ and $\langle x\rangle $ is normal in G. Moreover, by Lemma 2.1, $[x,Q]=\{1\}$ whenever the order of x is a prime or $4$ and so $[x,G]=\{1\}$ under the same assumption. Therefore the group G is quasihamiltonian (see [Reference Schmidt13, Theorem 2.4.11]) and the proof is complete.

Proof. The map $a\longmapsto a^e$ defines an epimorphism of A onto $A^e$ whose kernel contains B so that $A^e$ is divisible and hence $A=A^e\times C$ for a suitable subgroup C. Obviously, C has finite exponent and so $A^e$ is the largest divisible subgroup of A. If $A/B$ is a group of type $p^\infty $ for some prime p, then $A^e$ is also of type $p^\infty $ since it is a nontrivial homomorphic image of $A/B$ .

Proof. Of course, it can be assumed that $[Q,g]\neq \{1\}$ for a suitable element g of G. Since the subgroup $\langle Q,g\rangle $ is quasihamiltonian, it follows from [Reference Schmidt13, Theorem 2.4.14] that Q has finite exponent e. Let x be an element of Q of order p and let y be an arbitrary element of G. Then $\langle y\rangle $ is a permutable subgroup of $\langle Q,y\rangle $ and so, in particular, $\langle y\rangle ^x=\langle y\rangle $ . Thus, x normalises all subgroups of G and hence it belongs to $\zeta _2(G)$ (see for instance [Reference Schenkman11]). Therefore, $Q\zeta _2(G)/\zeta _2(G)$ has exponent strictly smaller than e and so by induction Q is contained in some term of the upper central series of G with finite ordinal type.

Proof. Each primary subgroup of Q is permutably embedded in G and so by Lemma 4.2, it is contained in some term with finite ordinal type of the upper central series of G. Therefore, $Q\leq \zeta _\omega (G)$ .

Proof. Let $p_1<p_2<\cdots <p_n<p_{n+1}<\cdots $ be the sequence of all prime numbers. For each positive integer n, let $P_n$ be the unique Sylow $p_n$ -subgroup of G and put $G_n=\mathop {\textrm {Dr}}_{k\geq n}P_k$ . Then,

$$ \begin{align*}G=G_1=XG_1\geq XG_2\geq\cdots\geq XG_n\geq XG_{n+1}\geq\cdots\end{align*} $$

is a descending sequence of subgroups of G containing X and so there exists m such that $XG_n=XG_{n+1}$ for all $n\geq m$ . If $X_n$ is the unique Sylow $p_n$ -subgroup of X, we have $[X_n,G_{n+1}]=\{1\}$ so that $X_n$ is normal in $XG_{n+1}$ and $XG_{n+1}/X_n$ has no elements of order $p_n$ . Thus, $P_n=X_n\leq X$ for all $n\geq m$ and hence $G_m$ is contained in X.

Proof. By Corollary 4.3, the subgroup Q is contained in $Z=\zeta _\omega (G)$ . Thus, the subgroup lattice of $G/Z$ is Černikov and so $G/Z$ is a Černikov group. Let $J/Z$ be the largest divisible subgroup of $G/Z$ . Then $G/J$ is finite and J is hypercentral so that it follows from Lemma 4.4 that there exists a finite set $\pi $ of prime numbers such that $\pi (G/J)\subseteq \pi $ and Q contains the $\pi '$ -component $J_{\pi '}$ of J. Then $Q_{\pi '}=J_{\pi '}$ is normal in G and $G/Q_{\pi '}$ is a $\pi $ -group. Application of Lemma 4.2 to each primary component of $Q/Q_{\pi '}$ yields that $Q/Q_{\pi '}\leq \zeta _k(G/Q_{\pi '})$ for a suitable nonnegative integer k, and hence $Q/Q_G$ lies in $\zeta _k(G/Q_G)$ .

Proof of Theorem 1.2

The subgroup Q is contained in $\zeta _\omega (G)$ by Corollary 4.3 so that $G/\zeta _\omega (G)$ is a Černikov group and hence there exists a Černikov normal subgroup W of G such that $G/W$ is hypercentral (see [Reference Dixon, Kurdachenko and Otal5, Corollary B4]). Since the hypotheses are inherited by homomorphic images, we may replace G by $G/W$ and assume that G is hypercentral. In particular, G is the direct product of its Sylow subgroups. Put $\pi =\pi (G/Q)$ . Then the largest $\pi '$ -subgroup $G_{\pi '}$ of G is contained in Q and hence it is quasihamiltonian. Clearly, the set $\pi $ is finite and $G=G_\pi \times G_{\pi '}$ so that it is enough to prove that the statement is true for every Sylow subgroup of G. Thus, we may suppose that G is a p-group for some prime number p.

Since Q is quasihamiltonian, it is abelian-by-finite (see [Reference Schmidt13, Theorem 2.4.14]) and so it contains an abelian characteristic subgroup A of finite index (see for instance [Reference Kargapolov and Merzljakov7, Lemma 21.1.4]). Then A is a permutably embedded normal subgroup of G and $G/A$ is Černikov so that Q can be replaced by A and we may assume that Q is abelian. Moreover, by the results of Černikov and Schlette, we may also suppose that Q is not contained in $\zeta (G)$ . Thus there exists $g\in G$ such that $\langle g,Q\rangle $ is a nonabelian quasihamiltonian group, whence Q must have finite exponent. It follows from Lemma 4.2 that $Q\leq \zeta _k(G)$ for some nonnegative integer k so that the factor group $G/\zeta _k(G)$ is Černikov. Therefore, $\gamma _{k+1}(G)$ is also a Černikov group (see [Reference Robinson10, Part 1, Corollary 2 to Theorem 4.21]) and hence it is enough to prove the statement for the factor group $G/\gamma _{k+1}(G)$ . Thus, we may assume without loss of generality that G is nilpotent.

Let $J/Q$ be the largest divisible subgroup of the Černikov group $G/Q$ . Then Q is contained in $\zeta (J)$ by Lemma 3.2 so that $J'$ is Černikov, and again we may replace G by $G/J'$ and assume that J is abelian. Write

$$ \begin{align*}J/Q=J_1/Q\times\cdots\times J_t/Q,\end{align*} $$

where each $J_i/Q$ is a group of type $p^\infty $ . It follows from Lemma 4.1 that $J_i=P_i\times B_i$ , where $P_i$ is a group of type $p^\infty $ and $B_i$ has finite exponent i ( $i=1,\ldots ,t$ ). Since G is nilpotent, $J/Q$ lies in $\zeta (G/Q)$ so that each $J_i$ is normal in G and hence also $P_i$ is normal in G for every $i=1,\ldots ,t$ . It follows that $P=\langle P_1,\ldots ,P_t\rangle $ is a Černikov normal subgroup of G. Moreover, $J/P$ has finite exponent and so $J=PQ$ . Therefore, $J/P$ is a permutably embedded subgroup of finite index of $G/P$ and hence $G/P$ contains a finite normal subgroup $N/P$ such that $G/N$ is quasihamiltonian (see [Reference De Falco, de Giovanni and Musella2, Theorem 3.5]). Since N is Černikov, the proof is complete.