Proof. Let

\begin{equation*}\{1\}=N_0\leqslant N_1\leqslant \cdots\leqslant N_t=N\end{equation*}

be a G-invariant series with cyclic factors. Then $G/C_G\big(N_{i+1}/N_i\big)$ is abelian for each $0\leqslant i \lt t$ (see Remark 2.3), so $G'\leqslant C_G\big(N_{i+1}/N_i\big)$, and hence

\begin{equation*}\{1\}=G'\cap N_0\leqslant G'\cap N_1\leqslant \cdots\leqslant G'\cap N_t=G'\cap N\end{equation*}

is a finite central series for $G'\cap N$. It follows that $G'\cap N$ is nilpotent and we are done.

Proof. Let

\begin{equation*}\{1\}=N_0\leqslant N_1\leqslant \cdots\leqslant N_m=N\end{equation*}

be a G-invariant series with paracentral factors. Set

\begin{equation*}C=\bigcap_{0\leqslant i \lt m}C_G\big(N_{i+1}/N_i\big).\end{equation*}

Then $G/C$ is abelian and C centralizes a finite series of N. Thus, it follows from a well-known result of Philip Hall (see [Reference Hall20] and [Reference Newell and Trombetti23]) that $C/C_G(N)$ is soluble, so also  $G/C_G(N)$ is soluble. Finally, if N is polycyclic, we have that $G/C_G(N)$ is polycyclic by Theorem 4.27 of [Reference Robinson27].

Proof. Of course, we may assume A is a p-group for some prime p. Let pⁿ be the exponent of A. If n = 1, then the statement is obvious. Suppose n > 1, and consider the map

\begin{equation*}\varphi:\, a\in A\mapsto a^{p^{n-1}}\in A,\end{equation*}

which is a G-homomorphism with non-trivial image $U=\operatorname{Im}(\varphi)\leqslant \operatorname{Frat}(A)$. Since $\operatorname{Ker}(\varphi)\geqslant \operatorname{Frat}(A)$, then U is G-paranilpotent. Moreover, the exponent of $A/U$ is $p^{n-1}$, so induction yields that $A/U$ is G-paranilpotent. Therefore, A is G-paranilpotent and the statement is proved.

Proof. It follows from Lemma 2.6 that $P/P'$ is G-paranilpotent. Since G-paranilpotent G-modules form a tensorial class, it follows from Theorem 2.26 of [Reference Robinson27] that P is G-paranilpotent.

Proof. We denote by g_i, $i\in\mathbb{N}$, elements of order pⁱ in P such that $g_{j+1}^p=g_j$ for all $j\in\mathbb{N}$. In order to show that $P\leqslant Z(G)$, we only need to show that $[N,g_i]=1$ for all $i\in\mathbb{N}$.

Assume first that $\pi(N)$ does not contain any odd prime, and let  $x\in N$ and $i\in\mathbb{N}$. Then $H=\langle x,g_{i+1}\rangle$ is supersoluble, so $Y=\langle g_{i+1}\rangle^H$ has an H-invariant series ${\mathcal{S}}$ with cyclic factors that are either isomorphic with $\mathbb{Z}$ or with $\mathbb{Z}_2$. Thus,  $g_{i+1}^2$ centralizes each factor of this series. If p > 2, we get that $g_{i+1}$ centralizes every factor of ${\mathcal{S}}$, so an easy induction (on the length of ${\mathcal{S}}$) yields that $g_{i+1}$ centralizes Y. If p = 2, then Y is torsion-free, and a further easy induction (on the length of ${\mathcal{S}}$) shows that $g_{i+1}^2=g_i$ centralizes Y. In any case, the arbitrariness of i gives $[N,P]=\{1\}$.

Next, let Q be a normal Sylow q-subgroup of N for some prime q. Let pⁿ be the maximal power of p dividing q − 1. If  $x\in Q$ and $i\in\mathbb{N}$, then $H=\langle x,g_{n+i}\rangle$ is supersoluble and consequently $g^{p^n}_{n+i}=g_i$ centralizes a finite (central) series of  $\langle x\rangle^H$, so $\langle x,g_i\rangle$ is nilpotent and hence $\big[x,g_i\big]=1$. The arbitrariness of x and i shows that  $[Q,P]=\{1\}$.

What we have just proved implies that, if $\pi(N)$ is finite, then N has a finite G-invariant series whose factors are centralized by P and contain no elements of order p. By induction on the length of such a series, we obtain $[N,P]=\{1\}$.

Assume finally that $\pi(N)$ is infinite. In this case, if q is any odd prime in $\pi(N)$, then the statement holds for  $G/N_{ \gt q}$; in particular, $[P,N]\leqslant N_{ \gt q}$. The arbitrariness of q then shows that $[P,N]=\{1\}$ and completes the proof.

Proof. Since G is locally finite and contains p-elements of arbitrarily large order, it follows that G contains an infinite Sylow p-subgroup. Thus, $G=P \lt imes N$ and we may apply Lemma 2.10.

Proof. The structure we outlined above shows that G has normal subgroups  $U\leqslant V$ such that: $|G:V|$ is finite; $G/U$ is linear of characteristic p; U is a p-group and $V/U$ is a pʹ-group in case p > 0; and $V/U$ is abelian of finite rank. In particular, $G/U$ is hypercyclic by [Reference Wehrfritz34, Theorem 11.21]. Also, if p = 0, then G is abelian-by-finite and even (locally nilpotent)-by-(finite abelian) since the derived subgroup of a locally supersoluble group is locally nilpotent (recall that the product of two locally nilpotent normal subgroups is locally nilpotent).

Assume p > 0. Let $P=O_p(G)$ and $C=C_G(P)$. Since G is locally supersoluble, we have that $G_{ \gt p}\leqslant C_G(P)$; thus, since $V/U$ is a pʹ-group and $U\leqslant P$, we have that the index $|G:CP|$ is finite (see Corollary 2.11).

We now claim that P is G-paranilpotent (this will yield in particular that G is always hypercyclic). To this aim, since P is nilpotent of finite exponent (see [Reference Wehrfritz34, 2.6 and Lemma 9.13]), it is enough by Corollary 2.7 to show that $P/\operatorname{Frat}(P)$ is G-paranilpotent. Therefore, replacing G by $G/\operatorname{Frat}(P)$, we may assume that P is an elementary abelian p-group; in particular, C = CP has finite index in G. Let F be a finite subgroup of G such that $G/C=FC/C$. Then S = FP is a locally supersoluble linear group of characteristic p (see [Reference Wehrfritz34, Theorem 2.2 and Lemma 2.3]), so P is S-hypercyclic by [Reference Wehrfritz34, Theorem 11.21] and S-paranilpotent by [Reference Wehrfritz34, Corollary 11.4]. Consequently, P is G-paranilpotent and the claim is proved.

It remains to show that G is nilpotent-by-finite and (locally nilpotent)-by-(finite abelian). To this aim, let $H/P$ be an abelian pʹ-subgroup of finite index of $CP/P$. Then H is nilpotent and has finite index in G. Finally, since Gʹ is locally nilpotent, we get that G is (locally nilpotent)-by-(finite abelian).

Proof. Let F be the Fitting subgroup of G. By Theorem 2.12, F is nilpotent-by-finite. By Fitting’s theorem, every element of F is contained in a nilpotent normal subgroup of G, so F is the product of finitely many nilpotent normal subgroups and is consequently nilpotent.

Proof. Let G be a homomorphic image of a periodic linear group H of characteristic p. For each prime q, the unique Sylow q-subgroup of G is the image of a suitable Sylow q-subgroup H_q of H (see [Reference Wehrfritz34, Theorem 9.15]). It follows from 2.6 of [Reference Wehrfritz34] that H_p is nilpotent if p > 0, and that H_q is Černikov (and so hypercentral) for each prime q ≠ p. Thus, G is hypercentral. The fact that G is nilpotent-by-finite follows at once from Theorem 2.12.

Proof. Replacing G by $G/G^{{\mathbf{P}}{\mathcal{N}}}$, we may assume G is residually paranilpotent; in particular, since G is locally finite, we also have that G is locally supersoluble.

Now, if G is a homomorphic image of a periodic linear group of characteristic p, then Theorem 2.12 shows that G is hypercyclic and that, if p > 0, then $O_p(G)$ is G-paranilpotent. Put $P=\{1\}$ if p = 0, and $P=O_p(G)$ if p > 0. By the structure of soluble periodic linear groups, there is a finite-index normal abelian subgroup  $A/P$ of  $G/P$ of finite rank, with no p-elements in case p > 0. Let $\pi=\pi(G/A)$; in particular, π is finite. Then the  $\pi'$-component of $A/P$ is G-paranilpotent by Lemma 11.7 of [Reference Wehrfritz34]. We only need to show that the π-component of $A/P$ is G-paranilpotent.

Let $U/P$ be any Sylow q-subgroup of $A/P$ for some prime $q\in\pi\setminus\{p\}$. By Theorem 9.15 of [Reference Wehrfritz34], there is a q-subgroup Q of U such that U = QP. Since G is residually paranilpotent and Q satisfies the minimal condition on subgroups, there is a normal subgroup N of G such that $G/N$ is paranilpotent and $N\cap Q=\{1\}$. Let D be the finite residual of Q. Since DP is locally supersoluble and q ≠ p, we have that $[D,P]=\{1\}$ by Lemma 2.10. Thus, D is a characteristic subgroup of $DP\trianglelefteq G$, and hence $D\trianglelefteq G$. It follows that $D\simeq_G DN/N$, so D is G-paranilpotent. Therefore,  $U/DP$ is G-paranilpotent because G is hypercyclic. Thus, $U/P$ is G-paranilpotent.

The arbitrariness of $U/P$ shows that the π-component of $A/P$ is G-paranilpotent and completes the proof.

Proof. Let P be a non-trivial normal Sylow p-subgroup of N for some prime p. Then $P/\operatorname{Frat}(P)$ is elementary abelian, so it is X-paracentral by Remark 3.6. Since N is supersoluble, there is a finite N-invariant series

\begin{equation*}\operatorname{Frat}(P)=P_0\leqslant P_1\leqslant \cdots\leqslant P_n=P\end{equation*}

with cyclic factors (see Remark 2.2). Thus, this series is also X-invariant, so it is actually a G-invariant series, and hence P is G-supersoluble by Corollary 2.8. What we have just proved is clearly enough to prove the statement in case N is finite. In general, this only enables us to assume N has no non-trivial elements of odd order. Suppose T is a finite non-trivial G-invariant subgroup of N. Then T is a 2-group. Since N is supersoluble, we have that T contains a normal subgroup of N of order 2, which is, therefore, central in N. Then the Sylow 2-subgroup U of Z(N) is G-invariant, non-trivial and finite. Let V be a Sylow 2-subgroup of N containing U. If U is cyclic, then $N/U$ is G-supersoluble by induction, and so also N is G-supersoluble. Assume U is not cyclic, so in particular V is not cyclic. Then X normalizes V (see Remark 3.6), and the argument we employed above with P yields that V is X-supersoluble. It follows that Z(N) contains a cyclic subgroup that is normalized by X and so is also normal in G. Induction yields again that N is G-supersoluble. This argument shows that we may assume N has no non-trivial finite G-invariant subgroup.

Let A be a characteristic subgroup of N that is free abelian of rank $m\geqslant 1$ (every infinite polycyclic group has such a subgroup). For each prime number p, we have that  $A/A^p$ is finite, so also $X/C_X(A/A^p)$ is finite. Now, if p does not belong to the finite set $\pi(N/A)$, the hypotheses yield that every subgroup of  $A/A^p$ is X-invariant, so $A/A^p$ is X-paracentral and hence $X/C_X(A/A^p)$ is abelian. The arbitrariness of p yields that  $X/C_X(A)$ is abelian. Moreover, by induction, $N/A$ is G-supersoluble, so N is Xʹ-supersoluble and consequently $X'/C_{X'}(N)$ is soluble by Lemma 2.5. Then  $X/C_X(N)$ is soluble, and so even polycyclic by Lemma 2.5. Replacing G by  $G/C_X(N)$, we may assume that X is polycyclic, and so that also G is polycyclic.

We claim that $Y=[X,N]$ is nilpotent. Let K be a G-invariant subgroup of finite index of Y. We need to show that $Y/K$ is nilpotent. Since Y is infinite (being normal in G), then also K is infinite, and induction yields that $N/K$ is G-supersoluble, so  $Y/K$ is nilpotent by Lemma 2.4. Now, since G is polycyclic, then every finite-index subgroup of Y contains a G-invariant finite-index subgroup of Y. Thus, Y is nilpotent (see [Reference Segal30, Theorem 2, p. 18]) and the claim is proved.

Since Nʹ is nilpotent, it follows from Fitting’s theorem that $N'Y$ is nilpotent and (consequently) torsion-free. Let $Z=Z(N'Y)$, which is free abelian. Again, if p is a prime that is not contained in $\pi=\pi(N/Z)$, then X normalizes every subgroup of  $Z/Z^p$, so if we set

\begin{equation*}D=\bigcap_{p\in\pi'} C_X(Z/Z^p),\end{equation*}

then $X/D$ is abelian, and consequently $X'\leqslant D$ centralizes Z. Thus, $G'=X'N'Y$ centralizes Z. Let A be a normal subgroup of G contained in Z of minimal Hirsch length, and put $\widetilde{G}=G/C_G(A)$; in particular, $\widetilde{G}$ is abelian and A is a rationally irreducible  $\widetilde{G}$-module.

By [Reference Segal30, Proposition 1, p. 23], there is an algebraic number field $\mathcal{K}$, with ring of integers $\mathcal{O}$, and embeddings

\begin{equation*}\varphi: A\rightarrow \mathcal{O}^+\quad\text{and}\quad\psi:\widetilde{G}\rightarrow\mathcal{O}^\times\end{equation*}

such that:

• • $\big(a^g\big)^\varphi=a^\varphi\cdot g^\psi$ for all $a\in A$ and $g\in\widetilde{G}$, and

• • the indices $\big|\mathcal{O}^+:A^\varphi\big|$ and $\big|\mathcal{O}^+:\mathbb{Z}[G^\psi]\big|$ are both finite (where $\mathbb{Z}[G^\psi]$ denotes the subring of $\mathcal{O}$ generated by G ^ψ).

Then there exists a non-zero $t\in\mathbb{Z}$ such that

\begin{equation*}t\mathcal{O}\subseteq A^\varphi\quad\text{and}\quad t\mathcal{O}\subseteq\mathbb{Z}[G^\psi].\end{equation*}

Now, only finitely many prime ideals of $\mathcal{O}$ contain the integer t. Let $\mathfrak{p}$ be a non-zero prime ideal of $\mathcal{O}$ such that $t\not\in\mathfrak{p}$. Then $\mathfrak{p}+t\mathcal{O}=\mathcal{O}$, so also $\mathfrak{p}+A^\varphi=\mathcal{O}$ and

\begin{equation*}A^\varphi/\big(\mathfrak{p}\cap A^\varphi\big)\simeq_{\mathbb{Z}[G^\psi]}\mathcal{O}/\mathfrak{p}.\end{equation*}

Since $\mathcal{O}=\mathfrak{p}+\mathbb{Z}[G^\psi]$, we have that the $\mathbb{Z}[G^\psi]$-submodules of $\mathcal{O}/\mathfrak{p}$ are precisely the $\mathcal{O}$-submodules. Thus, the fact that $\mathcal{O}/\mathfrak{p}$ is simple as an $\mathcal{O}$-module implies that $A^\varphi/\big(\mathfrak{p}\cap A^\varphi\big)$ is simple as a $\mathbb{Z}[G^\psi]$-module. Therefore

\begin{equation*}A/\big(\mathfrak{p}\cap A^\varphi\big)^{\varphi^{-1}}\end{equation*}

is a simple G-module, so is cyclic (by induction on the Hirsch length of N), and this means that $\mathfrak{p}$ has degree 1. Finally, [Reference Segal30, Theorem 2, p. 55] yields that $\mathcal{K}$ is the ring of rational numbers, so  $\mathcal{O}^+=\mathbb{Z}$ and consequently A is cyclic. Since $N/A$ is G-supersoluble, we have that N is G-supersoluble.

Finally, the fact that $X/C_X(N)$ is polycyclic follows from Lemma 2.5.

Proof. Assume first N is paranilpotent. If p > 0 and $O_p(N)=\{1\}$, put s = 0; otherwise, put s = p. The structure of soluble periodic linear groups yields that  $N_{ \gt s}$ contains a characteristic (see [Reference de Giovanni and Trombetti10]) finite-index abelian subgroup A of finite rank with no elements of order p if p > 0. Let D be the finite residual of A. Then D is X-paracentral by Remark 3.7, and so is even G-paranilpotent (recall that N is paranilpotent). Replacing G by  $G/D$, we may assume A (and so $N_{ \gt s}$) is residually finite. Now, let q be a suitable prime number such that $B=N_{ \gt q}$ is a finite-index subgroup of A. For each prime r > q, the Sylow r-subgroup R of B (which coincides with the Sylow r-subgroup of N) is X-paranilpotent (by Corollary 2.7 and Remark 3.6); in particular,  $X/C_X(R)$ is soluble by Lemma 2.5. It follows that the residually finite group $X/C_X(B)$ is locally soluble, and so even soluble by Corollary 9.21 of [Reference Wehrfritz34]. Moreover,  $XB/C_X(B)$ contains no non-trivial normal p-subgroups if p > 0 (they would centralize $BC_X(B)/C_X(B)\simeq B$), so $XB/C_X(B)$ has an abelian subgroup  $W/C_X(B)$ of finite index. Replacing q by a suitable larger prime (if needed), we may assume  $B\leqslant W$, so $X/C_X(B)$ must now be finite. On the other hand,  $\pi(N/B)$ is finite, and $O_p(N)\leqslant C_N(B)$ if p > 0, so the residually finite group  $N/C_N(B)$ is finite. Consequently, $G/C_G(B)$ is finite. Finally, it easily follows from the hypotheses and Lemma 2.9 that every Sylow subgroup of B is G-hypercyclic, so B itself is G-hypercyclic. Replacing q by a suitable larger prime in such a way that $\pi(G/C_G(B))\cap\pi(B)=\emptyset$, we may employ Lemma 11.7 of [Reference Wehrfritz34] to derive that B is G-paranilpotent. Since $N_{ \gt s}/B$ is finite, and $N_{\geqslant s}/N_{ \gt s}$ is either trivial or (if p > 0) a nilpotent p-group of finite exponent, an application of the hypotheses, Lemma 2.9 and Corollary 2.7, yields that  $N_{\geqslant s}/B$ is G-paranilpotent, so  $N_{\geqslant s}$ is G-paranilpotent.

The structure of soluble periodic linear groups shows that the above takes care of all cases and completes the proof of the paranilpotent case.

Now, assume N is locally supersoluble, so N is also hypercyclic by Theorem 2.12. The argument to show that N is G-hypercyclic is essentially the one we employed in the paranilpotent case, but easier. In fact, in the above notation, we do not need to reduce to the case in which  $G/C_G(B)$ is finite, but we only need to note that R is G-hypercyclic by hypotheses and Lemma 2.9.

In order to show that $O_p(N)$ is G-paranilpotent if p > 0, we start proving that $\overline{U}=O_p\big(N/N_{ \gt p}\big)$ (which is the unique Sylow p-subgroup of $N/N_{ \gt p}$) is G-paranilpotent. By Remark 3.6, $\overline{U}/\operatorname{Frat}(\overline{U})$ is X-paracentral, so $\overline{U}/\operatorname{Frat}(\overline{U})$ is G-paranilpotent and consequently $\overline{U}$ is G-paranilpotent by Corollary 2.7. Then

\begin{equation*}O_p(N)N_{ \gt p}/N_{ \gt p}\simeq_G O_p(N)\end{equation*}

is G-paranilpotent and the statement is completely proved.

Proof. Of course, replacing G by $G/N^{\mathbf L\mathfrak U}$, we may assume that N is locally supersoluble. Suppose now that $\pi(N)$ is infinite and that we are able to prove the statement in case $\pi(N)$ is finite. Then $G^{\mathbf L\mathfrak U}N_{ \gt p}=X^{\mathbf L\mathfrak U}N_{ \gt p}$ for all  $p\in\pi(N)$. Thus,

\begin{equation*}G^{\mathbf L\mathfrak U}\leqslant \bigcap_{p\in\pi(N)}\big(G^{\mathbf L\mathfrak U}N_{ \gt p}\big) =\bigcap_{p\in\pi(N)}\big(X^{\mathbf L\mathfrak U}N_{ \gt p}\big)=X^{\mathbf L\mathfrak U}.\end{equation*}

Since obviously $X^{\mathbf L\mathfrak U}\leqslant G^{\mathbf L\mathfrak U}$, then $X^{\mathbf L\mathfrak U}=G^{\mathbf L\mathfrak U}$ and the statement is proved. Therefore, we may assume $\pi(N)$ is finite.

Let q be the largest prime in N, and let Q be the unique Sylow q-subgroup of N. We know that either Q is an infinite Černikov group or it is nilpotent of finite exponent. In the former case, let D be the finite residual of Q, so D is X-paracentral by Remark 3.7. In the latter case, $Q/\operatorname{Frat}(Q)$ is N-paranilpotent by Theorem 2.12 and X-paracentral by Remark 3.6, so it is also G-paranilpotent; then Corollary 2.7 implies that Q is G-paranilpotent, and we let D be the first term of a G-paranilpotent series of Q. Now, an easy induction (either on the size of Q or on length of a G-paranilpotent series) shows that  $M=G^{\mathbf L\mathfrak U}D=X^{\mathbf L\mathfrak U}D$. Since X acts as a group of power automorphisms of D, we have that $X^{\mathbf L\mathfrak U}\leqslant X'\leqslant C_X(D)$, so $M=X^{\mathbf L\mathfrak U}\times D$, and hence M ^ʹ is a normal subgroup of G contained in $X^{\mathbf L\mathfrak U}$. Replacing G with $G/M'$, we may assume M is abelian. Also, by Corollary 3.12, we may further assume $X^{\mathbf L\mathfrak U}\neq\{1\}$.

Now, suppose X is finite, so $X^{\mathbf L\mathfrak U}=X^{\mathfrak U}$. Theorem IV.5.18 of [Reference Doerk and Hawkes16] yields that $X=K \lt imes X^{\mathfrak U}$ for some K < X. Let Z = KN. By Corollary 3.12, we know that Z is locally supersoluble (recall that $K\simeq X/X^{\mathfrak U}$ is supersoluble). Let E be any finite X-invariant subgroup of N, and note that $Z_E=KE$ is supersoluble and that $XE=(M\cap (XE))Z_E$. Then Proposition III.3.23 of [Reference Doerk and Hawkes16] yields that Z_E is contained in a unique $\mathfrak{U}$-projector R of XE. On the other hand, we have that the $\mathfrak U$-residual of XE is contained in $M\cap (XE)$, so is abelian, and hence Theorem IV.5.18 of [Reference Doerk and Hawkes16] gives that  $(XE)^{\mathfrak U}\cap R=\{1\}$. Since $XE=X^{\mathfrak U}Z_E$, we have that $X^{\mathfrak U}=(XE)^{\mathfrak U}$. Then  $X^{\mathfrak U}$ is normalized by E. The arbitrariness of E shows that  $X^{\mathfrak U}$ is normalized by N and is therefore normal in G. By Corollary 3.12,  $G/X^{\mathfrak U}$ is locally supersoluble, so $X^{\mathfrak U}=G^{\mathfrak U}$ and the statement is proved in the special case in which X is finite.

Assume now that X is arbitrary, and let $\mathcal{F}$ be the set of all finite subgroups of X. For each $F\in\mathcal{F}$, put $G_F=F \lt imes N$, and note that, by Remark 3.6, G_F is the very weak direct product of F and N, so the first part of the proof yields that $G_F^{\mathbf L\mathfrak U}=F^{\mathbf L\mathfrak U}=F^{\mathfrak U}$. Since $F^{\mathfrak U}\leqslant E^{\mathfrak U}$ when E is a finite subgroup of X containing F, we have that

\begin{equation*}U=\bigcup_{F\in\mathcal{F}}G_F^{\mathbf L\mathfrak U}=\bigcup_{F\in\mathcal{F}}F^{\mathfrak U}\end{equation*}

is a G-invariant subgroup of X such that $G/U$ is locally supersoluble; in particular, $X^{\mathbf L\mathfrak U}\leqslant G^{\mathbf L\mathfrak U}\leqslant U$. On the other hand, $F^{\mathfrak U}\leqslant X^{\mathbf L\mathfrak U}$ for all $F\in\mathcal{F}$, so $U=X^{\mathbf L\mathfrak U}$ is normal in G. Finally, by Corollary 3.12, $G/X^{\mathbf L\mathfrak U}\in\mathbf L\mathfrak U$, and so $X^{\mathbf L\mathfrak U}=G^{\mathbf L\mathfrak U}$.

Proof. By Theorems 2.15 and 3.13, we may replace G by $G/X^{\mathbf L\mathfrak U}N^{{\mathbf{P}}{\mathcal{N}}}$, and so assume that G is locally supersoluble and that N is paranilpotent. Thus, G is hypercyclic and N is G-paranilpotent by Theorem 3.10.

Suppose first that $\pi(N)$ is infinite and that we are able to prove the statement in case  $\pi(N)$ is finite. Then $G^{{\mathbf{P}}{\mathcal{N}}}N_{ \gt p}=X^{{\mathbf{P}}{\mathcal{N}}}N_{ \gt p}$ for all $p\in\pi(N)$, so

\begin{equation*}G^{{\mathbf{P}}{\mathcal{N}}}\leqslant \bigcap_{p\in\pi(N)}\big(G^{{\mathbf{P}}{\mathcal{N}}}N_{ \gt p}\big) =\bigcap_{p\in\pi(N)}\big(X^{{\mathbf{P}}{\mathcal{N}}}N_{ \gt p}\big)=X^{{\mathbf{P}}{\mathcal{N}}}.\end{equation*}

Since obviously $X^{{\mathbf{P}}{\mathcal{N}}}\leqslant G^{{\mathbf{P}}{\mathcal{N}}}$, then $X^{{\mathbf{P}}{\mathcal{N}}}=G^{{\mathbf{P}}{\mathcal{N}}}$ and the statement is proved. Therefore, we may assume $\pi(N)$ is finite.

Let q be the largest prime in N, and let Q be the unique Sylow q-subgroup of N. Let A be a G-paracentral subgroup of Q such that $Q/A$ has smaller G-paranilpotent length. Then induction (either on the order of $\pi(N)$ or on the G-paranilpotent length of Q) yields that $L=G^{{\mathbf{P}}{\mathcal{N}}}A=X^{{\mathbf{P}}{\mathcal{N}}}A$. Since A is G-paracentral, $X^{{\mathbf{P}}{\mathcal{N}}}\leqslant C_X(A)$, so actually $L=X^{{\mathbf{P}}{\mathcal{N}}}\times A$ and hence L ^ʹ is a G-invariant subgroup of $X^{{\mathbf{P}}{\mathcal{N}}}$. Replacing G by  $G/L'$, we may assume $L\geqslant X^{{\mathbf{P}}{\mathcal{N}}}$ is abelian. Since L is normal in G, we may also assume that $X^{{\mathbf{P}}{\mathcal{N}}}$ has no non-trivial Sylow r-subgroups for any prime r ≠ q; in particular, $X^{{\mathbf{P}}{\mathcal{N}}}\leqslant L$ is a q-group.

Now, if q = p, then $X^{{\mathbf{P}}{\mathcal{N}}}$ is X-paranilpotent by Theorem 2.12, so X is paranilpotent and consequently G is paranilpotent by Corollary 3.11. Therefore, $X^{{\mathbf{P}}{\mathcal{N}}}=G^{{\mathbf{P}}{\mathcal{N}}}=\{1\}$ and the statement is proved.

Assume q ≠ p, so L is Černikov. Let D be the finite residual of $X^{{\mathbf{P}}{\mathcal{N}}}$. Since G is hypercyclic, it follows that $X^{{\mathbf{P}}{\mathcal{N}}}/D$ is X-supersoluble, so $X/D$ is paranilpotent and consequently $D=X^{{\mathbf{P}}{\mathcal{N}}}$. If D were the finite residual of L, then D would be normal in G, and $G/D$ would be paranilpotent by Corollary 3.11, so D would coincide also with  $G^{{\mathbf{P}}{\mathcal{N}}}$ and the statement would be proved. Therefore, we may assume D is not the finite residual of L, which means that A is infinite. Since N is G-paranilpotent, we can now replace A by a G-invariant Prüfer subgroup of Q (induction on the size), so L is now divisible.

Suppose $X^{{\mathbf{P}}{\mathcal{N}}}N=LN$ has an abelian subgroup B of finite index. Then $L\leqslant B$, so  $N/C_N(L)$ is finite. Let

\begin{equation*}C_N(L)=M_0\leqslant M_1\leqslant \cdots\leqslant M_\ell=N\end{equation*}

be a G-invariant series with cyclic factors $M_{i+1}/M_i=\langle g_iM_i\rangle$ (recall that G is hypercyclic). We prove by induction that $[M_i,X^{{\mathbf{P}}{\mathcal{N}}}]=\{1\}$. This is certainly true if i = 0. Suppose the claim holds for i and let’s prove it for i + 1. Since $[N,X^{{\mathbf{P}}{\mathcal{N}}}]\leqslant A$, the map

\begin{equation*}\varphi_i:\, a\in X^{{\mathbf{P}}{\mathcal{N}}}\mapsto [g_i,a]\in A\end{equation*}

turns out to be a homomorphism. Suppose $[M_{i+1},X^{{\mathbf{P}}{\mathcal{N}}}]\neq\{1\}$. Then

\begin{equation*}X^{{\mathbf{P}}{\mathcal{N}}}/\operatorname{Ker}(\varphi_i)\simeq A\end{equation*}

is a Prüfer group. Also, $C_X(M_{i+1})\geqslant \operatorname{Ker}(\varphi_i)$. Thus, $X/C_{X^{{\mathbf{P}}{\mathcal{N}}}}(M_{i+1})$ is paranilpotent and so $X^{{\mathbf{P}}{\mathcal{N}}}\leqslant C_{X}(M_{i+1})$, a contradiction. This means that $[M_{i+1},X^{{\mathbf{P}}{\mathcal{N}}}]=\{1\}$ and completes the proof of the claim. Now, $X^{{\mathbf{P}}{\mathcal{N}}}$ is normalized by $M_\ell=N$ and is consequently normal in G; the statement then follows by applying Corollary 3.11 to $G/X^{{\mathbf{P}}{\mathcal{N}}}$.

Finally, suppose LN is not abelian-by-finite, so p > 0 and $O_p(N)$ is non-trivial. The previous paragraph yields that $G^{{\mathbf{P}}{\mathcal{N}}}\leqslant X^{{\mathbf{P}}{\mathcal{N}}} \lt imes O_p(N)$ so

\begin{equation*}G^{{\mathbf{P}}{\mathcal{N}}}=\big(X^{{\mathbf{P}}{\mathcal{N}}} \lt imes O_p(N)\big)\cap\big(X^{{\mathbf{P}}{\mathcal{N}}} \lt imes A\big)=X^{{\mathbf{P}}{\mathcal{N}}}\end{equation*}

and the statement is proved.

Proof. By Corollary 3.12, we may assume G is locally supersoluble, so the finite residual of N coincides with $N^{\mathfrak U}$, and hence we can assume N is residually finite. If N is finite, then the finite residual of G coincides with the finite residual of X, that in turns coincides with $X^{\mathfrak U}$. Thus, we may assume $\pi(N)$ is infinite. Now, for each $q\in\pi(N)$, $X^{\mathfrak U}N_{ \gt q}/N_{ \gt q}$ is the finite (and so supersoluble) residual of $G/N_{ \gt q}$ (note that $N/N_{ \gt q}$ is finite). Then $X^{\mathfrak U}$ is the supersoluble residual of G, and the statement is proved.

Proof. A well-known theorem of Baer (see [Reference Baer3, Satz 3.1]) shows that a polycyclic group is supersoluble provided that every finite homomorphic image is such. Thus, by Remark 4.2, we only need to prove the assertion for a finite group G. Suppose the statement is false and let G be a minimal counterexample. Then G has only one minimal normal subgroup, say M; of course, M is an abelian p-group for some prime p. If $M\leqslant \operatorname{Frat}(G)$, then $G/\operatorname{Frat}(G)$ is supersoluble, so also G is supersoluble, a contradiction. Thus, $M\cap\operatorname{Frat}(G)=\{1\}$ and hence $\operatorname{Frat}(G)=\{1\}$. Since Gʹ is nilpotent, it follows that Gʹ is a p-group, so G has a normal Sylow p-subgroup P. Then P is elementary abelian and it must coincide with M (by Maschke’s theorem); in particular,  $P=C_G(P)$ and $P=M\leqslant N$.

Now, if G = XP, then $X\cap P=\{1\}$ (otherwise $P=M\leqslant X$ and G = X is supersoluble); in particular, G is the very weak direct product of X and P, so G is supersoluble by Corollary 3.9. Assume G ≠ XP. By minimality, XP is supersoluble, so $XP/P$ is abelian of exponent dividing p − 1. Similarly, $N/P$ is supersoluble and so it is abelian of exponent dividing p − 1. Thus,  $G/P$ is abelian of exponent dividing p − 1. It follows from [Reference Robinson28, 9.4.3] that $G/P$ is cyclic of order dividing p − 1, so the same result yields that P is cyclic. Therefore, G is supersoluble, and we have reached the final contradiction.

Proof. First we observe that the statement holds if N is abelian. In fact, in this case, $X\cap N$ is G-hypercyclic and $XN/(X\cap N)$ is the very weak direct product of  $X/(X\,\cap\, N)$ and $N/(X\,\cap\, N)$, so it is locally supersoluble by Corollary 3.12. Therefore, G is hypercyclic and we are done. Next we observe that the statement holds if N is finite. In fact, in this case, $C_X(N)$ is G-hypercyclic, so G is locally supersoluble since $G/C_X(N)$ is finite supersoluble by Theorem 4.3. In the following, we may, therefore, assume N is infinite non-abelian.

In order to prove the general case, we will do as many reductions as possible. If Gʹ is not primary, then G is residually with a primary commutator subgroup. Thus, we may assume Gʹ is a q-subgroup for some prime q; in particular, G contains a normal Sylow q-subgroup Q such that $G^{\mathbf L\mathfrak U}\leqslant G'\leqslant Q$.

Since N is locally supersoluble, for every prime $r\geqslant q$, we have that $N_{ \gt r}$ is a normal subgroup of G having trivial intersection with Q. Thus, we may assume $\pi(N)$ is finite. By Theorem 2.12, N has a characteristic nilpotent subgroup M of finite index. If $G^{\mathbf L\mathfrak U}\leqslant O_{q'}(M)$, then $G^{\mathbf L\mathfrak U}\leqslant O_{q'}(M)\cap Q=\{1\}$ and so G is locally supersoluble. It is, therefore, enough to prove the statement for $G/O_{q'}(M)$, so we may simply assume M is a q-group; in particular, $N\cap Q$ has finite index in N, and it is consequently infinite.

Suppose either p = 0 or p > 0 and $O_p(N)=\{1\}$. Then N is Černikov, and it follows from [Reference Robinson27, Theorem 3.29] that $X/C_X(N)$ is Černikov as well. Since $C_X(N)$ is G-hypercyclic, we may assume G is Černikov. Also, similarly to what we did above with M, we may assume the finite residual of G is a q-group, so it coincides with the finite residual of Q. Let D be the smallest infinite G-invariant subgroup of  $N\cap Q$; in particular, D is divisible abelian. Now, XD is the very weak normal product of X and D, and D is abelian, so XD is locally supersoluble. Let  $C=C_G\big(\Omega_1(Z(Q)\cap D)\big)$. Then  $G/C=XN/C$ is an abelian q’-group. Since N is locally supersoluble, we have that  $NC/C$ has exponent dividing q − 1. Similarly, $XC/C$ has exponent dividing q − 1. Thus,  $G/C$ has exponent dividing q − 1, and this implies (by Maschke’s theorem) that  $\Omega_1(Z(Q)\cap D)$ contains a cyclic G-invariant subgroup (see also [Reference Robinson27, 9.4.3]). This argument shows that D is G-hypercyclic. By induction,  $G/D$ is hypercyclic, so G is hypercyclic and we are done.

Suppose p > 0 and $O_p(N)\neq\{1\}$. If p ≠ q, then the above shows that $G^{\mathbf L\mathfrak U}\leqslant O_p(N)$, so $G^{\mathbf L\mathfrak U}\leqslant O_p(N)\cap Q=\{1\}$ and the statement is proved. Assume, therefore, that p = q, so $Q=O_p(G)$. By Lemma 2.9, we only need to show that $G/\operatorname{Frat}(Q)$ is hypercyclic, so without loss of generality we may assume Q is elementary abelian. Using the hypotheses, we see that  $X(Q\cap N)$ is the very weak normal product of X and the abelian subgroup $Q\cap N$, so it is locally supersoluble. Thus, by Theorem 2.12, $X(Q\cap N)$ is nilpotent-by-finite, so also G is nilpotent-by-finite; in particular, G has a characteristic nilpotent subgroup F of finite index.

Assume now $\pi(G)$ is finite. If F ≠ Q, then G contains a non-trivial normal r-subgroup R for some prime $r\neq p=q$, and hence  $G/R$ is locally supersoluble by induction either on the number of primes in $\pi(G)$ or on the size of a Sylow r-subgroup. Thus G embeds in $G/R\times G/Q$ and is locally supersoluble. Therefore, F = Q. Now, let $C=C_G(Q\cap N)$. Then $G/C$ is a finite abelian q ^ʹ-group. Since $X(Q\cap N)$ is locally supersoluble, we have that $XC/C$ has exponent dividing q − 1; similarly, $NC/C$ has exponent dividing q − 1, so $G/C$ has exponent dividing q − 1 and consequently $Q\cap N$ is G-hypercyclic. Since $N/(Q\cap N)$ is finite, we have that $G/(Q\cap N)$ is hypercyclic, so G is hypercyclic too. The proof is complete in this case.

Finally, assume $\pi(G)$ is infinite. Then $\pi(F)$ is infinite, and for each prime $r\in\pi(F)$, we have that $G/F_{ \gt r}$ is locally supersoluble. Since $\bigcap_{r\in\pi(F)}F_{ \gt r}=\{1\}$, it follows that G is locally supersoluble and the statement is completely proved.

Proof. It follows from Theorem 4.4 that G is locally supersoluble, so it contains a nilpotent normal subgroup F of finite index (see Theorem 2.12). Since we only need to show that $G^{{\mathbf{P}}{\mathcal{N}}}\leqslant O_{q'}(F)$ for every $q\in\pi(F)$, we may assume F is a q-subgroup for some prime q. If q = p, then F is G-paranilpotent by Theorem 2.12, so G is paranilpotent since $G/F$ is finite supersoluble. Assume q ≠ p, so G is Černikov.

If N is finite, then $C_X(N)$ is G-paranilpotent, and $G/C_X(N)$ is finite supersoluble, so the statement is proved. Thus, we may assume N is infinite. Let P be a normal Prüfer subgroup of N; in particular, XP = PX. If $P\leqslant Z(N)$, then $P^X\cap X$ is G-paranilpotent; also, $XP/(P^X\cap X)$ is the very weak direct product of $X/(P^X\cap X)$ and (the divisible subgroup) $P^X/(P^X\cap X)$, so $P^X/(P^X\cap X)$ is X-paranilpotent (see Theorem 3.10), and hence P^X is G-paranilpotent. By induction on the size of G, we get that  $G/P^X$ is paranilpotent, which means that G is paranilpotent and we are done. It is therefore possible to assume $P\not\leqslant Z(G)$.

Suppose there is $x\in X$ such that $C_N(P)\neq C_N(P^x)$; in particular, $P\neq P^x$. Note that P^x is normal in $N^x=N$, and that $C_N(P)\cap C_N(P^x)$ is strictly contained in both $C_N(P)$ and $C_N(P^x)$ (since these latter subgroups have the same index in G). Let y be an element of N centralizing P but not P^x. Then y^x centralizes P^x, so $y^{-1}y^x$ does not centralize P^x. This means that $R=\langle [y,x],P^x\rangle$ is not nilpotent. On the other hand, $R\leqslant G'$, so we get a contradiction. This shows that $C_N(P)=C_N(P^x)$ for all $x\in X$. Now, if X does not centralize  $N/C_N(P)$, then again Gʹ contains a non-nilpotent subgroup, contradicting the hypotheses. Thus, X centralizes  $N/C_N(P)$ and we get that N acts as a group of power automorphisms on P^X. An application of Theorem 3.10 yields that the divisible subgroup P^X is X-paranilpotent, so also G-paranilpotent. By induction on the size of G, we get that  $G/P^X$ is G-paranilpotent. Consequently, G is paranilpotent and we are done.

Proof. Let

\begin{equation*}G=\langle a\rangle \lt imes\Big(\big(\langle b\rangle\times\langle c\rangle\big) \lt imes\big(D_1\times D_2\big)\Big),\end{equation*}

where D ₁ and D ₂ are Prüfer 2-groups, b (respectively, c) acts as the inversion on D ₁ (respectively, D ₂) and centralizes D ₂ (respectively, D ₁), a interchanges b with c, and D ₁ with D ₂.

Put $X=\langle a,bc, D_1,D_2\rangle$ and $N=\langle b,c,D_1,D_2\rangle$. Then X and N are paranilpotent, but G is not paranilpotent. Since X is normal in G, then clearly G is the (very) weak normal product of X and N.

Proof. First, note that G is the very weak normal product of X and N. Replacing G by $G/N^{\mathbf L\mathfrak U}$, we may assume N is locally supersoluble, so, by Theorem 4.4, we only need to show that $X^{\mathbf L\mathfrak U}$ is normal in G. Thus, without loss of generality, we may assume that  $X^{\mathbf L\mathfrak U}$ contains no non-trivial G-invariant subgroups. Since  $G^{\mathbf L\mathfrak U}\leqslant G'$, we have that  $G^{\mathbf L\mathfrak U}$ is locally nilpotent. In particular, $X^{\mathbf L\mathfrak U}$ is ascendant in G by Theorem 2.14 of [Reference de Giovanni, Trombetti and Wehrfritz12] (recall that every subgroup of a locally nilpotent group is serial).

Suppose we can prove the statement when either p = 0 or p > 0 and $O_p(G)=\{1\}$. Assume p > 0 and $O_p(G)\neq\{1\}$. Let Z be the socle of the centre of $O_p(G)$. Since $O_p(G)$ is nilpotent of finite exponent, it has a finite central series with elementary abelian factors. By induction on the length of such a series, we have that $U=G^{\mathbf L\mathfrak U}Z=X^{\mathbf L\mathfrak U}Z$ is a locally nilpotent normal subgroup of G. Since $X^{\mathbf L\mathfrak U}$ contains no non-trivial G-invariant subgroups, we have that $X^{\mathbf L\mathfrak U}$ is a p-group. Therefore, $X^{\mathbf L\mathfrak U}\leqslant O_p(G)$, so $X^{\mathbf L\mathfrak U}$ is centralized by Z. Now, the elementary abelian residual of U is contained in  $X^{\mathbf L\mathfrak U}$ and so is trivial. Thus, U (and in particular $X^{\mathbf L\mathfrak U}$) is an elementary abelian p-group.

Suppose the statement holds when Gʹ is a p-group. If $Q=O_{p'}(G')$, then $X^{\mathbf L\mathfrak U}\times Q$ is normal in G, so also $X^{\mathbf L\mathfrak U}$ is normal in G. We may, therefore, assume Gʹ is a p-group; in particular, G has a unique normal Sylow p-subgroup P.

We distinguish two cases. Suppose first N is a p-group, and let Y be a pʹ-subgroup of X such that $X=Y(P\cap X)$. By Theorem 3.10, we have that N is Y-paranilpotent. Thus, since P is nilpotent, we actually have that N is G-paranilpotent. Let M be the first term of a G-invariant series of N whose factors have exponent p and are G-paracentral; in particular, $M\leqslant Z(P)$. By induction on the length of such a series, we have that $G^{\mathbf L\mathfrak U}\leqslant X^{\mathbf L\mathfrak U}M$, so, if $G^{\mathbf L\mathfrak U} \gt X^{\mathbf L\mathfrak U}$, we can then find a G-invariant subgroup L of  $G^{\mathbf L\mathfrak U}\cap M$ such that $|L|=p$ and $L\not\leqslant X^{\mathbf L\mathfrak U}$. Now, let $\mathcal{F}$ be the set of all finite subgroups of G containing L. Since

\begin{equation*}G^{\mathbf L\mathfrak U}=\bigcup_{F\in{\mathcal{F}}}F^{\mathfrak U},\end{equation*}

we can find $E\in{\mathcal{F}}$ such that $L\leqslant E^{\mathfrak U}$. By [Reference Doerk and Hawkes16, Theorem V.3.2], L is contained in every  $\mathfrak U$-normalizer of E. Also, by Theorem V.4.2 of [Reference Doerk and Hawkes16], the $\mathfrak U$-normalizers of E coincide with the $\mathfrak U$-projectors. Since $E^{\mathfrak U}$ is abelian, Theorem IV.5.18 yields that every  $\mathfrak U$-projector is a complement of $E^{\mathfrak U}$ in E. Finally, we reach a contradiction because L is contained in $E^{\mathfrak U}$ and in any  $\mathfrak U$-projector.

Suppose N is not a p-group. Then the previous paragraph shows that

\begin{equation*}\big(X(P\cap N)\big)^{\mathbf L\mathfrak U}=X^{\mathbf L\mathfrak U}.\end{equation*}

Since $X(P\cap N)$ contains Gʹ, it follows that $X(P\cap N)\trianglelefteq G$, so also $X^{\mathbf L\mathfrak U}\trianglelefteq G$ and we are done.

Actually, the above argument proving the statement when either p = 0 or p > 0 and  $O_p(G)\neq\{1\}$ can be used to prove the statement in the very special case in which G is finite. One just needs to use induction on the order of G and replace p by the largest prime in $\pi(G)$. Thus, we may assume G is infinite. Suppose now N is finite, and let F be any finite subgroup of X containing $N\cap X$. Then FN is a finite group satisfying the hypotheses of the statement, when F takes the place of X. This means that $(FN)^{\mathfrak U}=F^{\mathfrak U}$ is normalized by N. The arbitrariness of F yields that $X^{\mathbf L\mathfrak U}$ is normal in G. Therefore, we may assume N is infinite.

It remains to prove the statement when p = 0, and when p > 0 and $O_p(G)=\{1\}$. In these cases, G contains an abelian normal subgroup A of finite rank and of finite index. First, assume $\pi(A)$ is finite. Since G is infinite, there is a prime q such that A contains a G-invariant divisible q-group. An argument similar to the one we employed in the first half of the proof shows that we may assume the following holds for G:

• • $X^{\mathbf L\mathfrak U}$ and Gʹ are q-groups, so, in particular, G contains a unique Sylow q-subgroup Q;

• • N is a G-hypercyclic q-group;

• • the finite residual D of Q coincide with the finite residual of G.

Now, $G^{\mathbf L\mathfrak U}\leqslant X^{\mathbf L\mathfrak U}(D\cap N)$, so, if $G^{\mathbf L\mathfrak U} \gt X^{\mathbf L\mathfrak U}$, then $G^{\mathbf L\mathfrak U}\cap D\cap N$ contains a finite G-hypercyclic subgroup L such that $L\not\leqslant X^{\mathbf L\mathfrak U}$. Let again $\mathcal{F}$ be the set of all finite subgroups of G containing L. Since

\begin{equation*}G^{\mathbf L\mathfrak U}=\bigcup_{F\in{\mathcal{F}}}F^{\mathfrak U},\end{equation*}

we can find $E\in{\mathcal{F}}$ such that $L\leqslant E^{\mathfrak U}$. As in the first half of the proof, a combination of [Reference Doerk and Hawkes16, Theorems IV.5.18, V.3.2 and V.4.2] gives a contradiction.

Finally, suppose $\pi(A)$ contains infinitely many primes. Let π be any infinite subset of $\pi(A)\setminus\pi(G/A)$ such that $\pi(A)\setminus \pi$ is finite. Then $X^{\mathbf L\mathfrak U}O_\pi(A)$ is normal in G, so  $\pi(X^{\mathbf L\mathfrak U})\subseteq\pi$. The arbitrariness of π shows that $\pi(X^{\mathbf L\mathfrak U})=\emptyset$, so $X^{\mathbf L\mathfrak U}=\{1\}$ and the statement is proved.

Proof. Replacing G by $G/N^{{\mathbf{P}}{\mathcal{N}}}$, we may assume N is paranilpotent. Also, by Theorem 4.8, we may assume G is locally supersoluble (note that this is the only place in the proof where the additional hypothesis on the Sylow subgroups of X is employed). Since G is the very weak normal product of X and N, we only need to prove that $X^{{\mathbf{P}}{\mathcal{N}}}$ is normal in G (see Theorem 4.5). Now, the same arguments we employed in the proof of Theorem 4.8 enables us to assume that:

• • $X^{{\mathbf{P}}{\mathcal{N}}}$ is a q-group for some prime q ≠ p (otherwise X would be paranilpotent and the statement would follow from Theorem 4.5);

• • G is Černikov;

• • $N\leqslant G'$ and the finite residual of G are q-groups.

Since X is Černikov and hypercyclic, it follows that $X^{{\mathbf{P}}{\mathcal{N}}}$ is divisible. Thus, we may assume N is infinite. In fact, if N were finite, then $X^{{\mathbf{P}}{\mathcal{N}}}$ would be normalized by N and so normal in G.

Let D be finite residual of N. Since $XD\trianglelefteq G$, it is enough to show that $(XD)^{{\mathbf{P}}{\mathcal{N}}}=X^{{\mathbf{P}}{\mathcal{N}}}$. Thus, without loss of generality, we may assume D = N. However, in this case, $X^{{\mathbf{P}}{\mathcal{N}}}$ (being divisible) is normalized by both X and N, so is normal in G, and we are done.

Proof. Let D be the finite residual of G, and let D_X be the finite residual of $X\cap D$ (which coincides with the finite residual of X). Since $X^{\mathfrak U}\geqslant D_X$, $(X/D_X)^{\mathfrak U}=X^{\mathfrak U}/D_X$. Thus, it follows from Remark 4.13 that

\begin{equation*}\big(X/(D\cap X)\big)^{\mathfrak U}=X^{\mathfrak U}(D\cap X)/(D\cap X),\end{equation*}

so

\begin{equation*} (XD/D)^{\mathfrak U}=X^{\mathfrak U}D/D.\end{equation*}

Similarly, using the finite residual D_N of N, we get that

\begin{equation*} (ND/D)^{\mathfrak U}=N^{\mathfrak U}D/D.\end{equation*}

Now, Theorem 3.13 yields that $G^{\mathfrak U}/D=G^{\mathbf L\mathfrak U}D/D=X^{\mathfrak U}N^{\mathfrak U}D/D$. In order to complete the proof, we only need to show that $D\leqslant X^{\mathfrak U}N^{\mathfrak U}$. First, we claim that $D=D_XD_N$. To see this, we may let $D_N=\{1\}$, so N is residually finite and we need only show that $D=D_X$. Let A be the product of all the normal Sylow subgroups of N (each of which is finite); in particular, $|N:A| \lt \infty$. Suppose by contradiction that $D \gt D_X$, so $D=D_X\times E$ for some divisible subgroup E. Then $D_XA/A \lt DA/A$ and we may assume $A=\{1\}$. This means that N is finite, and so that the finite residual of G coincides with the finite residual of X, a contradiction. Therefore $D=D_XD_N$. Finally, $X^{\mathfrak U}\geqslant D_X$ and $N^{\mathfrak U}\geqslant D_N$, so $X^{\mathfrak U}N^{\mathfrak U}\geqslant D_XD_N=D$ as required.

Proof. Obviously, we may assume $n\geqslant 2$, otherwise G would be abelian and the statement automatically proved. A combination of 2.6 and Theorem 9.15 of [Reference Wehrfritz34] shows that $P=O_p(G)$ is nilpotent. Since homomorphic images of serial subgroups in locally finite groups are serial (see for instance [Reference de Giovanni, Trombetti and Wehrfritz12], Lemma 2.1), we may assume (by induction on the nilpotency class) that P is even abelian. Now, $X\cap P$ is normal in G, and hence Lemma 2.3 of [Reference de Giovanni, Trombetti and Wehrfritz12] shows that $X/C$ is a p-group, where $C=C_X(P/P\cap X)$. A further application of [Reference Wehrfritz34, Theorem 9.15] shows that X has a p-subgroup Q such that X = QC. Since QP is a p-group, it is nilpotent of class at most n − 1. In particular,  $[Q,_{n-1}P]=\{1\}$, so

\begin{equation*}[X,_{n-1}P]=[QC,_{n-1}P]\leqslant P\cap X\end{equation*}

and hence X is subnormal in G = XP of defect at most n − 1.

Proof. Suppose first N has a nilpotent normal Sylow q-subgroup Q of finite exponent for some prime q. Temporarily, put $\operatorname{Frat}(Q)=\{1\}$, so Q is elementary abelian. Then $X\cap Q\trianglelefteq XQ$, and $XQ/(X\cap Q)$ is the very weak direct product of $X/(X\cap Q)$ and $Q/(X\cap Q)$. This implies that  $Q/(X\cap Q)$ is X-hypercyclic (see Theorem 3.10), so also Q is X-hypercyclic (recall that X is hypercyclic). Turning back to the general situation, it follows that Q is X-hypercyclic by Lemma 2.9. Then Xʹ centralizes an ascending series of normal subgroups of Q and consequently Xʹ is ascendant in $X'Q$. Since either Q is finite or p = q, we have in both cases by Lemma 5.1 that Xʹ is subnormal in $X'Q$. This is enough to not only prove the statement in the special case in which N is finite but also enables us to assume $O_p(N)=\{1\}$.

Now, N has a finite-index abelian subgroup A of finite rank. Let D be the finite residual of N. Then $XD/(X\cap D)$ is the very weak direct product of $X/(X\cap D)$ and $D/(X\cap D)$, so $D/(X\cap D)$ is X-paranilpotent by Theorem 3.10. Since the action of Xʹ is trivial on X-paracentral groups, and since Xʹ is obviously normalized by $(X\cap D)$, it follows that

\begin{equation*}X'\trianglelefteq X'(X\cap D)\trianglelefteq\trianglelefteq XD.\end{equation*}

Replacing G by $G/D$, we may assume A is residually finite, so there is a prime q such that $M=N_{ \gt q}$ is abelian of finite index in N. Again, $XM/(X\,\cap\, M)$ is the very weak direct product of $X/(X\cap M)$ and $M/(X\cap M)$, and hence $M/(X\cap M)$ is X-paranilpotent by Theorem 3.10. As before, $X'\trianglelefteq\trianglelefteq X'M$, so we may replace G by $G/M$ and consequently assume N is finite.

Proof. Clearly, G is polycyclic. If M is any normal subgroup of G with finite index, then $X'M/M$ is subnormal by Theorem 5.2. It follows that Xʹ is subnormal in G (see [Reference Segal30, p. 82]) and we are done.

Proof. The set ${\mathcal{F}}$ of all finite subgroups E of G such that $E=(E\cap N)(E\cap M)$ is a local system of G. Since

\begin{equation*}G^{\mathbf L\mathfrak N}=\bigcup_{F\in{\mathcal{F}}}F^{\mathfrak N},\quad N^{\mathbf L\mathfrak N}=\bigcup_{F\in{\mathcal{F}}}(F\cap N)^{\mathfrak N}\quad\text{and}\quad M^{\mathbf L\mathfrak N}=\bigcup_{F\in{\mathcal{F}}}(F\cap M)^{\mathfrak N},\end{equation*}

the result follows at once from Lemma II.2.12 of [Reference Doerk and Hawkes16].

Proof. Clearly, $G^{\mathfrak N}\geqslant L=N^{\mathfrak N}M^{\mathfrak N}$. Also, $G/L$ is the product of the two nilpotent subgroups $NL/L$ and $ML/L$, so it is nilpotent by Fitting’s theorem. Thus, $G^{\mathfrak N}\leqslant L$ and hence $G^{\mathfrak N}=L$.

Proof. (1) By Theorem 4.8, $G/(G')^{\mathbf L\mathfrak N}$ is locally supersoluble, so $G^{\mathbf L\mathfrak U}\leqslant (G')^{\mathbf L\mathfrak N}$. Moreover, the derived subgroup of $G/G^{\mathbf L\mathfrak U}$ is locally nilpotent, so $(G')^{\mathbf L\mathfrak N}=G^{\mathbf L\mathfrak U}$.

Now, $G'=(X')^G(N')^G[X,N]$. Since X and N are locally supersoluble, it follows that Xʹ and N ^ʹ are locally nilpotent. Thus, $(X')^G$ is locally nilpotent, being Xʹ ascendant by Theorem 5.2. Clearly, $(N')^G$ is locally nilpotent, so also $(X')^G(N')^G$ is locally nilpotent. An application of Lemma 5.4 shows that $(G')^{\mathbf L\mathfrak N}=[X,N]^{\mathbf L\mathfrak N}$ and completes the proof.

(2) The proof is the same as that of (1). We only need to replace Theorem 4.8 by Theorem 4.10 and Lemma 5.4 by Lemma 5.5 and to recall that the Fitting subgroup of a homomorphic image of a periodic linear group is nilpotent (see Corollary 2.13).