Proof. First assume that $R \leq D$ . Then

$$ \begin{align*} (DR/R)\cap (AR/R)^{s(G/R)}= (D/R)\cap (A^{sG}R/R) & =(D\cap A^{sG}R)/R \\ & =R(D\cap A^{sG})/R \leq R(D\cap A_{sG})/R. \end{align*} $$

However,

$$ \begin{align*}R(D\cap A_{sG})/R\leq (D\cap (AR)_{sG})/R = (D/R)\cap (AR)_{sG}/R =(DR/R)\cap (AR/R)_{s(G/R)}.\end{align*} $$

Therefore, $(DR/R)\cap (AR/R)^{s(G/R)}\leq (DR/R)\cap (AR/R)_{s(G/R)} $ and hence

$$ \begin{align*}(DR/R)\cap (AR/R)^{s(G/R)} =(DR/R)\cap (AR/R)_{s(G/R)}, \end{align*} $$

so $DR/R$ avoids the pair $((AR/R)^{s(G/R)}, (AR/R)_{s(G/R)})$ .

Now assume that $R \leq A$ . Then

$$ \begin{align*} (DR/R)\cap (AR/R)^{s(G/R)} & = (DR/R)\cap (A^{sG}/R)=(DR\cap A^{sG})/R=R(D\cap A^{sG})/R \\ & \leq R(D\cap A_{sG})/R \\ & \leq (DR/R) \cap (A_{sG}/R)= (DR/R)\cap (A/R)_{s(G/R)}. \end{align*} $$

Hence, $DR/R$ avoids $((AR/R)^{s(G/R)},(AR/R)_{s(G/R)})$ .

Proof. We have $A_{sG}\leq A_{sE}\leq A\leq A^{sE}\leq A^{G}$ by Lemma 2.2, and so from $A^{sG}\cap D=A_{sG}\cap D$ and Lemma 2.3(2), it follows that $E^{\mathfrak {N}}\cap A^{sG}\leq E^{\mathfrak {N}}\cap A_{sG}$ , where $E^{\mathfrak {N}}\cap A^{sE}\leq E^{\mathfrak {N}}\cap A^{sG}$ and $ E^{\mathfrak {N}}\cap A_{sG}\leq E^{\mathfrak {N_{\sigma }}}\cap A_{sE}$ .

Consequently, $E^{\mathfrak {N}}\cap A^{sE}\leq E^{\mathfrak {N}}\cap A_{sE}\leq E^{\mathfrak {N}}\cap A^{sE}$ and $E^{\mathfrak {N}}\cap A^{sE}= E^{\mathfrak {N}}\cap A_{sE}$ . Hence, $E^{\mathfrak {N}}$ avoids the pair $(A^{sEG}, A_{sE})$ . The lemma is proved.

Proof. Let $V/K $ be the normal Hall $\pi $ -subgroup of $H/K$ . Let D be a Hall $\pi '$ -subgroup of K. Then D is a normal Hall $\pi '$ -subgroup of V since K is nilpotent, so V has a Hall $\pi $ -subgroup, E say, by the Schur–Zassenhaus theorem. It is clear that V is $\pi '$ -soluble, so any two Hall $\pi $ -subgroups of V are conjugated in V by the Hall–Chunikhin theorem on $\pi $ -soluble groups.

Assume that $K\leq \Phi (G)$ . By a generalised Frattini argument, $G=VN_{G}(E)=DEN_{G}(E)=DN_{G}(E)=N_{G}(E)$ since $D\leq K\leq \Phi (G)$ . Thus, E is normal in H, that is, H is $\pi $ -closed since E is a Hall $\pi $ -subgroup of H.

Finally, assume that $K\leq Z_{\infty }(H)$ and then $D\leq Z_{\infty }(V)$ , so $V=D\rtimes E=D\times E$ . Hence, E is characteristic in V and so normal in H. Thus, H is $\pi $ -closed. The lemma is proved.

Proof. Assume that p divides $|D|$ . Then D has a maximal subgroup M such that $|D:M|=p$ and M is normal in G. It follows that $C_{G}(D/M)=G$ , that is, $D/M\leq Z(G/M)$ since $(p-1, |G|)=1$ . However, $G/D$ is nilpotent. Therefore, $G/M$ is nilpotent by Lemma 2.5 and hence $D\leq M < D$ , which is a contradiction. Therefore, the smallest prime in $\pi (G)$ belongs to $\pi (|G:D|)$ . In particular, $|D|$ is odd and so D is abelian since D is a Dedekind group by hypothesis. The lemma is proved.

Proof. It is clear that $DN/N$ is a normal Hall subgroup of $G/N$ and if $A/N\leq DN/N$ , then $A=N(A\cap D)$ , where $A\cap D$ is normal in G, so $A/N$ is normal in $G/N$ , that is, every element of $G/N$ induces a power automorphism in $DN/N$ . The lemma is proved.

Proof. Since $A_{G}\leq A_{sG}\leq A\leq A^{sG}\leq A^{G}$ by Lemma 2.2, it is enough to show that D avoids the pair $(A^{G}, A_{G})$ . Assume this is false and let G be a counterexample of minimal order.

First we prove that $A\cap D=1$ . Indeed, assume that $N:=A\cap D\ne 1$ . Then $N\leq A_{G}$ and $D/N=(G/N)^{\mathfrak {N}}$ is a special subgroup of $G/N$ by Lemma 2.8, and $A/N$ is an S-permutable subgroup of $G/N$ by [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Lemma 1.2.7]. Therefore, $(G/N)^{\mathfrak {N}}$ avoids the pair $ ((A/N)^{G/N}, (A/N)_{G/N})$ by the choice of G, that is,

$$ \begin{align*}(G/N)^{\mathfrak{N}}\cap (A/N)^{G/N}= (G/N)^{\mathfrak{N}}\cap (A/N)_{G/N}.\end{align*} $$

However, $(A/N)^{G/N}=A^{G}/N$ and $(A/N)_{G/N}=A_{G}/N$ , so

$$ \begin{align*}(G/N)^{\mathfrak{N}}\cap (A/N)^{G/N}=(D/N)\cap (A^{G}/N)=(D\cap A^{G})/N\end{align*} $$

and

$$ \begin{align*}(G/N)^{\mathfrak{N}}\cap (A/N)_{G/N}=(D/N)\cap (A_{G}/N)=(D\cap A_{G})/N.\end{align*} $$

Consequently, $D\cap A^{G}=D\cap A_{G}$ . Hence, D avoids the pair $(A^{G}, A_{G})$ , which is a contradiction.

Therefore, $A\cap D=1$ , so $AD/D\simeq A= P_{1}\times \cdots \times P_{t}$ , where $P_{i}$ is the Sylow $p_{i}$ -subgroup of A for all i. Then $P_{i}$ is S-permutable in G by Lemma 2.9 and so $D\leq N_{G}(P_{i})$ for all i by [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Lemma 1.2.16]. Therefore, $D\leq N_{G}(A)$ .

Let $\pi =\pi (D)$ . Then G is $\pi $ -soluble since every subgroup of D is normal in G by hypothesis. Moreover, D has a complement M in G since D is a Hall $\pi $ -subgroup of G and for some $x\in G$ , we have $A\leq M^{x}$ by the Chunikhin–Hall theorem [Reference Huppert9, VI, Hauptsatz 1.7]. Finally, $D\leq N_{G}(A)$ and hence $A^{G}=A^{DM^{x}}=A^{M^{x}}\leq M_{G}\leq M$ , so $A^{G}\cap D=1$ . Therefore, D avoids $(A^{G}, A_{G})=(A^{G}, 1)$ , contrary to the choice of G. The lemma is proved.

Claim 1. If R is a minimal normal subgroup of G, then $G/R$ is a ${PST}$ -group.

In view of the choice of G, it is enough to show that the hypothesis holds for $G/R$ . First note that $DR/R=(G/R)^{\mathfrak {N}}$ by Lemma 2.3 and if $A/R$ is a subnormal subgroup of $G/R$ , then A is subnormal in G, so D avoids the pair $(A^{sG}, A_{sG})$ by hypothesis. Therefore, $DR/R$ avoids the pair $((A/R)^{s(G/R)}, (A/R)_{s(G/R)})$ by Lemma 2.1. This proves Claim 1.

Claim 2. If E is a proper subnormal subgroup of G, then E is a ${PST}$ -group.

Every subnormal subgroup A of E is subnormal in G, so D avoids the pair $(A^{sG}, A_{sG})$ by hypothesis. However, then $E^{\mathfrak {N}}$ avoids the pair $(A^{sE}, A_{sE})$ by Lemma 2.4. Hence, the hypothesis holds for E, so Claim 2 holds by the choice of G.

Claim 3. D is nilpotent and every subgroup of D is S-permutable in G. Hence, every chief factor of G below D is cyclic.

First we show that if $L\leq D$ , where L is a minimal normal subgroup of G, then L is cyclic. Since G is soluble, $L\leq G_{p}$ for some Sylow subgroup $G_{p}$ of G and then some maximal subgroup V of L is normal in $G_{p}$ and V is subnormal in G. Assume that V is not S-permutable in G. Then $V\ne 1$ and $V^{sG}=L$ , so $V^{sG}\cap D=L=V_{sG}\cap D < V< L,$ which is a contradiction. Hence, V is S-permutable in G, so $G=G_{p}O^{p}(G)\leq N_{G}(V)$ by [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Lemma 1.2.16]. Therefore, $V=1$ , so $|L|=p$ .

Now we show that D is nilpotent. Assume that this is false and let R be a minimal normal subgroup of G. Then $G/R$ is a ${PST}$ -group by Claim 1.

Note also that $(G/R)^{\frak {N}}=RD/R \simeq D/(D\cap R)$ by Lemma 2.3, where $(G/R)^{\frak {N}}$ is abelian by Theorem 1.1, so $R\leq D$ and if N is a minimal normal subgroup of G, then $N=R$ since otherwise $D\simeq D/1=D/(N\cap R)$ is abelian. Moreover, $|R|=p$ for some prime p and $R\nleq \Phi (G)$ by Lemma 2.5, so for some maximal subgroup M of G, we have $G=R\rtimes M$ and $C_{G}(R)\cap M$ is a normal subgroup of G, so $C_{G}(R)\cap M=1$ . Therefore, $C_{G}(R)=R(C_{G}(R)\cap M)=R$ and then $G/R=G/C_{G}(R)$ is cyclic. Hence, $R=D$ is nilpotent. This contradiction shows that D is nilpotent. So, for every subgroup A of D,

$$ \begin{align*} A^{sG}=D\cap A^{sG}=D\cap A_{sG}=A_{sG}. \end{align*} $$

Therefore, every subgroup of D is S-permutable in G.

By Theorem 1.1 and Claim 1, every chief factor of G between R and D is cyclic, so every chief factor of G below D is cyclic by the Jordan–Hölder theorem for the chief series. Hence, Claim 3 holds.

Claim 4. D is a Hall subgroup of G.

Suppose that this is false and let P be a Sylow p-subgroup of D such that ${1 < P < G_{p}}$ , where $G_{p}\in \text {Syl}_{p}(G)$ .

(a) $D=P$ is a minimal normal subgroup of G and $|D|=p$ . Hence, $D\leq Z(G_{p})$ and $G_{p}$ is normal in G.

Let R be a minimal normal subgroup of G contained in D. Then R is a q-group for some prime q and $D/R=(G/R)^{\mathfrak {N}}$ is a Hall subgroup of $G/R$ by Claim 1 and Theorem 1.1.

Suppose that $PR/R \ne 1$ . Then $PR/R \in \text {Syl}_{p}(G/R)$ . If $q\ne p$ , then $P \in \text {Syl}_{p}(G)$ . This contradicts the fact that $P < G_{p}$ . Hence, $q=p$ , so $R\leq P$ and therefore, $P/R \in \text {Syl}_{p}(G/R)$ and again $P \in \text {Syl}_{p}(G)$ . This contradiction shows that $PR/R=1$ , which implies that $R=P$ is the unique minimal normal subgroup of G contained in D. Since D is nilpotent, a $p'$ -complement E of D is characteristic in D and so it is normal in G. Hence, $E=1$ , which implies that $R=D=P$ . Claim 3 implies that $|D|=p$ , so $D\leq Z(G_{p})$ . Finally, since $G/D$ is nilpotent and $D \leq G_{p}$ , $G_{p}$ is normal in G.

(b) $D\nleq \Phi (G)$ . Hence, $G=D\rtimes M$ for some maximal subgroup M of G and $C_{G}(D)=D\times (C_{G}(D)\cap M$ ).

This follows from part (a) since G is not nilpotent.

(c) If G has a minimal normal subgroup $L\ne D$ , then $G_{p}=D\times L$ . Hence, $O_{p'}(G)=1$ .

Indeed, $DL/L\simeq D$ is a Hall subgroup of $G/L$ by Theorem 1.1 and Claim 1. Hence, $G_{p}L/L=DL/L$ , so $G_{p}=D\times (L\cap G_{p})=D\times L$ since $G_{p}$ is normal in G by part (a). Thus, $O_{p'}(G)=1$ .

(d) $G_{p}\cap M\ne 1$ is normal in G.

Observe that $V\kern1.3pt{:=}\kern1.3pt G_{p}\cap M$ is normal in M by part (a). Also from $G_{p}\kern1.3pt{=}\kern1.3pt G_{p}\kern1.3pt{\cap}\kern1.3pt D\rtimes M= D(G_{p}\cap M)$ , where $D\leq Z(G_{p})$ by part (a), it follows that V is normal $G_{p}$ . Therefore, V is normal in G and $V\ne 1$ since $D < G_{p}$ .

(e) If $L\leq G_{p}\cap M$ , where L is a minimal normal subgroup of G, then $L= G_{p}\cap M$ and so $G_{p}=D\times L$ is abelian.

This follows from parts (c) and (d).

(f) Every normal subgroup Z of G contained in $G_{p}$ with $1\ne Z\ne G_{p}$ is G-isomorphic to either L or D. In particular, Z is a minimal normal subgroup of G and either $Z\in \{D, L\}$ or $D\simeq _{G}Z\simeq _{G}L$ , and so $C_{G}(D)=C_{G}(Z)=C_{G}(L)$ .

Assume that $D\ne Z\ne L$ . If $Z\cap L\ne 1$ , then $L\leq Z$ and so $Z=L(Z\cap D)=L$ since $1\ne Z\ne G_{p}=LD,$ which is a contradiction. Hence, $Z\cap L=1$ and $Z\cap D=1$ . Therefore, $G_{p}=D\times Z =D\times L$ and so the G-isomorphisms ${L\kern1.3pt{\simeq}\kern1.3pt LD/D\kern1.3pt{=}\kern1.3pt G_{p}/D \kern1.3pt{=}\kern1.3pt DZ/D\kern1.3pt{\simeq}\kern1.3pt Z }$ and $D\simeq DL/L= G_{p}/D = LZ/L \simeq Z $ yield $D\simeq _{G}Z\simeq _{G}L$ . In particular, Z is a minimal normal subgroup of G and $C_{G}(D)=C_{G}(Z)=C_{G}(L)$ .

(g) If $N=\langle ab \rangle $ , where $D=\langle a \rangle $ and b is an element of order p in L, then $|N|=p$ and $N\cap D=N\cap L=1.$

Since $G_{p}=D\times L$ is abelian by part (e) and $|D|=p$ by part (a), $|ab|=|N|=p$ . Hence, $N\cap D=N\cap L=1$ since $a\not \in L$ and $b\not \in D$ .

(h) N is a minimal normal subgroup of G.

First we show that N is normal in G. In view of [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Lemma 1.2.16] and part (e), it is enough to show that $N=N^{sG}$ is S-permutable in G. Assume that $N < N^{sG}$ . Then $|N^{sG}|> p$ . Since $G_{p}=DL$ by part (f) and $|D|=p$ by part (a),

$$ \begin{align*}|G_{p}:L|=p=|N^{sG}L/L|=|N^{sG}/(N^{sG}\cap L)|,\end{align*} $$

so $N^{sG}\cap L\ne 1$ . However, $N^{sG}\cap L$ is S-permutable in G by [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Theorem 1.2.19] and so $N^{sG}\cap L$ is normal in G by [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Lemma 1.2.16] and part (e). Hence, $L\leq N^{sG}$ by the minimality of L. Then $N^{sG} = N^{sG}\cap G_{p}=L(N^{sG}\cap D).$ However, N is subnormal in G and so $N^{sG}\cap D=N_{sG}\cap D=1$ . Hence, $N^{sG}=L$ and then $N\cap L\ne 1$ , in contrast to part (g). Hence, $N = N^{sG}$ and so N is normal in G. Therefore, N is a minimal normal subgroup of G since $|N|=p$ . This proves part (h).

(i) The final contradiction to prove Claim 4.

In view of parts (f), (g) and (h), $C_{G}(D)=C_{G}(N)=C_{G}(L)$ . However, $C_{G}(L)=G$ by part (e) since $G/D\simeq M$ is nilpotent and $L\leq M$ . Therefore, $D\leq Z(G)$ and so G is nilpotent. This contradiction proves Claim 4.

Claim 5. Every subgroup A of D is normal in G. Hence, every element of $ G$ induces a power automorphism in D.

Since D is nilpotent by Claim 3, it is enough to consider the case when A is a p-group for some prime p. Moreover, A is S-permutable in G by Claim 3 and the Sylow p-subgroup $D_{p}$ of D is a Sylow p-subgroup of G by Claim 4. Therefore, ${G=D_{p}O^{p}(G)=DO^{p}(G)\leq N_{G}(A)}$ by [Reference Ballester-Bolinches, Esteban-Romero and Asaad3, Lemma 1.2.16]. This proves Claim 5.

Claim 6. D is an abelian group of odd order.

This follows from Lemma 2.6 and Claim 5.

Claim 7. The final contradiction.

From Claims 3–6, it follows that G is a ${PST}$ -group by Theorem 1.1, in contrast to the choice of G. Hence, there is no minimal counterexample and G is a ${PST}$ -group.

Finally, given that G is a ${PST}$ -group, we show that D avoids the pair $(A^{sG}, A_{sG})$ for every subnormal subgroup A of G. There is a series $A=A_{0} \trianglelefteq A_{1} \trianglelefteq \cdots \trianglelefteq A_{n}=G$ , so A is S-permutable in G since G is a ${PST}$ -group. However, $D=G^{\mathfrak {N}}$ is a special subgroup of G by Theorem 1.1 and so D avoids the pair $(A^{sG}, A_{sG})$ by Lemma 2.10.

The theorem is proved.