On the Norm of the Abelian p-Group-Residuals

Abstract

Let G be a group. $D_p\left(G\right)={\bigcap }_{H\le G}{N}_G\left(H^{\prime }\left(p\right)\right)$ is defined and, the properties of $D_p\left(G\right)$ are investigated. It is proved that $D_p\left(G\right)=P\left[A\right]$, where $P=D\left(P\right)$ is the Sylow p-subgroup and $A=N\left(A\right)$ is a Hall $p^{\prime }$-subgroup of $D_p\left(G\right)$, respectively. Furthermore, it is proved in a group G that (1) $D_p\left(G\right)=1$ if and only if $C_G\left(G^{\prime }\left(p\right)\right)=1$; (2) $O_{p^{\prime }}\left(D_p\left(G\right)\right)\le Z_{\infty }\left(O^p\left(G\right)\right)$ and (3) if $Z\left(G^{\prime }\left(p\right)\right)=1$, then $C_G\left(G^{\prime }\left(p\right)\right)=D_p\left(G\right)$.

1. Introduction

All groups considered in this paper are finite. The reader is referred to [1] for notation and terminology. Recall that the norm $N\left(G\right)$ of a group G, introduced by Baer in [2], is the intersection of the normalizers of all subgroups of G (cf. [2]). A closely related subgroup was introduced and studied by Wielandt in [3]. It is defined as the intersection of the normalizers of all subnormal subgroups of G and called the Wielandt subgroup of G (see [4]).

Some generalisations of Baer and Wielandt’s subgroups were considered and a lot of interesting results have been obtained (see [5,6,7,8,9]). The idea behind these investigations is to consider a set S of subgroups of G and consider the intersection of the normalizers of all subgroups in S.

If S is the set of the commutators of all subgroups of a group G, the intersection $D\left(G\right)$ of their normalisers was studied in [5].

Our main goal in this paper is to study a local version of $D\left(G\right)$: the intersection of the normalizers of the residuals of all subgroups of G with respect to the class of all abelian p-groups, p a prime.

Given a prime p and a group G, let $G^{\prime }\left(p\right)$ denote the residual of G with respect to the class of all abelian p-groups; it is know that $G^{\prime }\left(p\right)$ is the unique smallest normal subgroup of G for which the corresponding factor group is an abelian p-group. There is, of course, a relationship between $G^{\prime }\left(p\right)$ and $O^p\left(G\right)$ which is the unique smallest normal subgroup of G whose factor group is a (not necessarily abelian) p-group. In fact, $O^p\left(G\right)\le G^{\prime }\left(p\right)$ and $G^{\prime }\left(p\right)/O^p\left(G\right)$ is the commutator subgroup of $G/O^p\left(G\right)$. Therefore $G^{\prime }\left(p\right)=O^p\left(G\right)G^{\prime }$. This subgroup plays an important role in group theory because it is the kernel of the transfer homomorphism from G to $P/P^{\prime }$, where P is a Sylow p-subgroup of G ([10], 10.1.5).

Definition 1.

Let p a prime. The norm $D_p\left(G\right)$ of the abelian p-group-residuals is the subgroup

$$D_p\left(G\right)=\bigcap _{H\le G}{N}_G\left(H^{\prime }\left(p\right)\right).$$

Note that $D_p\left(G\right)\ne D\left(G\right)$ in general (it is enough to consider the alternating group of deree 4).

We prove:

Theorem 1.

Let G be a group. Then $D_p\left(G\right)=P\left[A\right]$, where P is the Sylow p-subgroup and A is a Hall $p^{\prime }$-subgroup of $D_p\left(G\right)$. Moreover, $P=D\left(P\right)$ and $A=N\left(A\right)$. In particular, ${\cap }_{p\left|\right|G|}{D}_p\left(G\right)=N\left(G\right)$.

Theorem 2.

Let G be a group. Then

(1) 

$D_p\left(G\right)=1$ if and only if $C_G\left(G^{\prime }\left(p\right)\right)=1$;

(2) 

$O_{p^{\prime }}\left(D_p\left(G\right)\right)\le Z_{\infty }\left(O^p\left(G\right)\right)$;

(3) 

if $Z\left(G^{\prime }\left(p\right)\right)=1$, then $C_G\left(G^{\prime }\left(p\right)\right)=D_p\left(G\right)$.

2. Elementary Properties on ${\mathit{D}}_{\mathit{p}}\left(\mathit{G}\right)$

In this section, we list some elementary properties of $D_p\left(G\right)$ that will be used in the proofs of the main results.

Lemma 1.

Let G be a group. Then

 (1)

If $H\le G$, then $H^{\prime }\left(p\right)\le G^{\prime }\left(p\right)$;

 (2)

if $N⊴G$ and $N\le H\le G$, then ${(H/N)}^{\prime }\left(p\right)=H^{\prime }\left(p\right)N/N$;

 (3)

$G^{\prime }\left(p\right)$ is nilpotent if and only if ${(G/\Phi \left(G\right))}^{\prime }\left(p\right)$ is nilpotent;

 (4)

if $G=MN$, where $M\le G$ and $N⊴G$, then $G^{\prime }\left(p\right)\le M^{\prime }\left(p\right)N$. In particular, ${(M\times N)}^{\prime }\left(p\right)=M^{\prime }\left(p\right)\times N^{\prime }\left(p\right)$.

Proof. 

(1)

Let $H\le G$. Since $H/(H\cap O^p\left(G\right))\cong H{O}^p\left(G\right)/O^p\left(G\right)\le G/O^p\left(G\right)$, $H/(H\cap O^p\left(G\right))$ is a p-group and so $H^{\prime }\left(p\right)=H^{\prime }{O}^p\left(H\right)\le H^{\prime }(H\cap O^p\left(G\right))\le G^{\prime }{O}^p\left(G\right)=G^{\prime }\left(p\right)$.

(2)

Let $O^p(H/N)=R/N$. $O^p\left(H\right)N\le R$ Since $(H/N)/O^p(H/N)=(H/N)/(R/N)\cong H/R$. Conversely, $H/O^p\left(H\right)N\cong (H/O^p\left(H\right)/(O^p\left(H\right)N/O^p\left(H\right))$ and $H/O^p\left(H\right)N\cong (H/N)/(O^p\left(H\right)N/N)$, so $R/N\le O^p\left(H\right)N/N$. Hence $O^p(H/N)=O^p\left(H\right)N/N$. Then ${(H/N)}^{\prime }\left(p\right)={(H/N)}^{\prime }{O}^p(H/N)$

$=(H^{\prime }N/N)(O^p\left(H\right)N/N)=H^{\prime }{O}^p\left(H\right)N/N=H^{\prime }\left(p\right)N/N$.

(3)

Clearly, $G^{\prime }\left(p\right)$ is nilpotent if and only if $G^{\prime }\left(p\right)\Phi \left(G\right)/\Phi \left(G\right)\cong G^{\prime }\left(p\right)/G^{\prime }\left(p\right)\cap \Phi \left(G\right)$ is nilpotent. So (3) follows from (2).

(4)

By $G^{\prime }\le M^{\prime }N$ and $O^p\left(G\right)\le O^p\left(M\right)N$, we get $G^{\prime }\left(p\right)\le M^{\prime }\left(p\right)N$. If $M⊴G$, then $G^{\prime }\left(p\right)\le N^{\prime }\left(p\right)M$. Thus $G^{\prime }\left(p\right)\le M^{\prime }\left(p\right)N\cap N^{\prime }\left(p\right)M=M^{\prime }\left(p\right)(N\cap N^{\prime }\left(p\right)M)=M^{\prime }\left(p\right)N^{\prime }\left(p\right)(M\cap N)$. Hence, if $G=M\times N$, then $G^{\prime }\left(p\right)=M^{\prime }\left(p\right)\times N^{\prime }\left(p\right)$ by (1).

□

Proposition 1.

Let G be a group. Then

 (1)

$N\left(G\right)C_G\left(G^{\prime }\left(p\right)\right)\le D_p\left(G\right)\le D\left(G\right)$;

 (2)

$D_p\left(G\right)$ is soluble;

 (3)

if $M\le G$, then $M\cap D_p\left(G\right)\le D_p\left(M\right)$;

 (4)

if $N⊴G$, then $D_p\left(G\right)N/N\le D_p(G/N)$;

 (5)

if $G=A\times B$, where $A,B\le G$ and $\left(\right|A|,|B\left|\right)=1$, then $D_p\left(G\right)=D_p\left(A\right)\times D_p\left(B\right)$.

Proof. 

(1)

Since $H^{\prime }\le H^{\prime }\left(p\right)\le H$ and $H^{\prime },H^{\prime }\left(p\right)$ are characteristic subgroups of H, we have $N_G\left(H^{\prime }\right)\ge N_G\left(H^{\prime }\left(p\right)\right)\ge N_G\left(H\right)$, that is, $N\left(G\right)\le D_p\left(G\right)\le D\left(G\right)$.

If $x\in C_G\left(G^{\prime }\left(p\right)\right)$, then x is a normalizer of $H^{\prime }\left(p\right)$ for all $H\le G$ by Lemma 1 (1). Hence $x\in D_p\left(G\right)$ and so, $C_G(G^{\prime }\left(p\right)\le D_p\left(G\right)$.

(2)

It follows from (1) and $D\left(G\right)$ is soluble in ([9], Proposition 2.4).

(3)

It is easy to see that $M\cap D_p\left(G\right)\le {\bigcap }_{H\le M}{N}_M\left(H^{\prime }\left(p\right)\right)=D_p\left(M\right)$.

(4)

If $x\in D_p\left(G\right)$, then $xN$ normalizes ${(H/N)}^{\prime }\left(p\right)$ for all $H/N\le G/N$ by Lemma 1 (2). Hence $D_p\left(G\right)N/N\le D_p(G/N)$.

(5)

$H=(H\cap A)\times (H\cap B)$ for all $H\le G$ by the hypotheses. It follows from Lemma 1 (4) that $H^{\prime }\left(p\right)={(H\cap A)}^{\prime }\left(p\right)\times {(H\cap B)}^{\prime }\left(p\right)$. Hence

$$\begin{array}{ccc}{N}_G\left(H^{\prime }\left(p\right)\right)\hfill & =\hfill & N_G\left({(H\cap A)}^{\prime }\left(p\right)\right)\cap N_G\left({(H\cap B)}^{\prime }\left(p\right)\right)\hfill \\ & =\hfill & (N_A\left({(H\cap A)}^{\prime }\left(p\right)\right)\times B)\cap (A\times N_B\left({(H\cap B)}^{\prime }\left(p\right)\right))\hfill \\ & =\hfill & N_A\left({(H\cap A)}^{\prime }\left(p\right)\right)\times N_B\left({(H\cap B)}^{\prime }\left(p\right)\right),\hfill \end{array}$$

which implies that $D_p\left(G\right)=D_p\left(A\right)\times D_p\left(B\right)$.

□

Proposition 2.

Let $G\ne 1$ be a group. Then

 (1)

If $G^{\prime }\left(p\right)$ is nilpotent, then $D_p\left(G\right)>1$.

 (2)

If $G^{\prime }\left(p\right)$ is a minimal normal subgroup of G and $D_p\left(G\right)$ is nilpotent, then $C_G\left(G^{\prime }\left(p\right)\right)=D_p\left(G\right)$.

Proof. 

(1)

If $G^{\prime }\left(p\right)=1$, then G is abelian p-group and $G=D_p\left(G\right)>1$. If $G^{\prime }\left(p\right)\ne 1$, then $D_p\left(G\right)\ge C_G\left(G^{\prime }\left(p\right)\right)\ge Z\left(G^{\prime }\left(p\right)\right)>1$ by Proposition 1 (1).

(2)

Since $G^{\prime }\left(p\right)$ is a minimal normal subgroup of G, $G^{\prime }\left(p\right)\cap F\left(G\right)=G^{\prime }\left(p\right)$ or 1. If $G^{\prime }\left(p\right)$∩$F\left(G\right)=1$, then $G^{\prime }\left(p\right)F\left(G\right)=[G^{\prime }\left(p\right)\times F\left(G\right)]$ and so, $F\left(G\right)\le C_G\left(G^{\prime }\left(p\right)\right)$. If $G^{\prime }\left(p\right)$∩$F\left(G\right)\ne 1$, then $G^{\prime }\left(p\right)\le F\left(G\right)$ and $[G^{\prime }\left(p\right),F\left(G\right)]\le G^{\prime }\left(p\right)$. However, $F\left(G\right)$ is nilpotent and hence $[G^{\prime }\left(p\right),F\left(G\right)]<G^{\prime }\left(p\right)$. Thus, $[G^{\prime }\left(p\right),F\left(G\right)]=1$ and we have $F\left(G\right)\le C_G\left(G^{\prime }\left(p\right)\right)$.

By Proposition 1 (1), $F\left(G\right)\le C_G\left(G^{\prime }\left(p\right)\right)\le D_p\left(G\right)$. The nilpotency of $D_p\left(G\right)$ implies that $F\left(G\right)=C_G\left(G^{\prime }\left(p\right)\right)=D_p\left(G\right)$.

□

3. Proofs of Theorems 1 and 2

Proof of Theorem 1. 

(1) By Proposition 1 (2), $D_p\left(G\right)$ is soluble. Then $D_p\left(G\right)$ has a Hall $p^{\prime }$-subgroup, denoted by A. Let P be a Sylow p-subgroup of $D_p\left(G\right)$. Then $D_p\left(G\right)=PA$.

Firstly, A is a Dedekind group. Case 1. A is a q-group.

For a subgroup H of A. Since $A\le D_p\left(G\right)$, we have A normalizes $H^{\prime }\left(p\right)$. It follows from $H^{\prime }\left(p\right)=H$ that H is normal in A, that is, $A=N\left(A\right)$ is a Dedekind group.

Case 2. A is not a q-group.

Let $A_q$ and $A_r$ be any Sylow q-subgroup and Sylow r-subgroup of A, respectively, $q\ne r$. Since $A_q$ and $A_r$ are subgroups of $D_p\left(G\right)$, $A_q$ normalizes $A_r^{\prime }\left(p\right)$ and $A_r$ normalizes $A_q^{\prime }\left(p\right)$. Then it follows from $A_r^{\prime }\left(p\right)=A_r$ and $A_q=A_q^{\prime }\left(p\right)$ that $[A_q,A_r]=1$, that is, A is nilpotent. For a subgroup H of $A_q$, by the same argument above, $A_q$ is Dedekind group, hence A is Dedekind group.

Secondly, $P=D\left(P\right)$ is a D-group.

For a subgroup K of P. Since $P\le D_p\left(G\right)$, we have P normalizes $K^{\prime }\left(p\right)$. It follows from $K^{\prime }\left(p\right)=K^{\prime }$ that $K^{\prime }$ is normal in P, that is, $P=D\left(P\right)$ is a D-group.

Finally, $D_p\left(G\right)=P\left[A\right]$.

Since P normalizes $A^{\prime }\left(p\right)$ and $A^{\prime }\left(p\right)=A$, we have $D_p\left(G\right)=P\left[A\right]$.

(2) By (1), the Hall $p^{\prime }$-subgroup of $D_p\left(G\right)$ is Dedekind group for any prime $p\in \pi \left(G\right)$, then ${\cap }_{p\left|\right|G|}{D}_p\left(G\right)\le N\left(G\right)$. Hence ${\cap }_{p\left|\right|G|}{D}_p\left(G\right)=N\left(G\right)$ by Proportion 1 (1). □

Suppose that a group H acts on a group G. We say that H acts hypercentrally on N if N has a subnormal series $1=N_0\le N_1\le \cdots \le N_s=N$ such that $[H,N_i]\le N_{i-1}$, for all $i=1,2,\cdots ,s$ (cf. [11]). Clearly, if N is a normal subgroup of H then H acts hypercentrally on N if and only if $N\le Z_{\infty }\left(H\right)$.

Lemma 2.

Let G be a $\{p,q\}$-group. Assume that N is a normal q-subgroup of G and H is a subgroup of G with $H=O^p\left(H\right)$. If $N\le D_p\left(G\right)$, then H acts hypercentrally on N.

Proof. 

Suppose that the lemma is not true. Let G be a counterexample of minimal order. Then

(1) $G=NH$.

If $NH<G$, then $NH$ satisfies the condition of the lemma by Proposition 1 (3) and the choice of G shows that H acts hypercentrally on N, a contradiction.

(2) Let T be a minimal supplement of $C_G\left(N\right)$ in G, then $O^p\left(T\right)=T$.

Since $H=O^p\left(H\right)$ and N is a normal q-subgroup, we have $G=HN=O^p\left(H\right)N=O^p\left(G\right)$. Let T be a minimal supplement of $C_G\left(N\right)$ in G. Then $G=C_G\left(N\right)T$. Assume that $O^p\left(T\right)<T$. Then $C_G\left(N\right)O^p\left(T\right)<G$ by the minimality of T. It is easy to see that $G/C_G\left(N\right)O^p\left(T\right)=C_G\left(N\right)T/C_G\left(N\right)O^p\left(T\right)\cong T/C_T\left(N\right)O^p\left(T\right)$ is a p-group, and then $O^p\left(G\right)\le C_G\left(N\right)O^p\left(T\right)<G$, a contradiction.

(3) $G=NT$, and $T⊴G$.

If $NT<G$, then $NT$ satisfies the condition of the lemma by Proposition 1 (3). By the choice of G, T acts hypercentrally on N. Let $T_p$ and $C_G{\left(N\right)}_p$ be Sylow p-subgroup of T and $C_G\left(N\right)$, respectively. Then $T_p$ acts trivially on N, and then $G_p=C_G{\left(N\right)}_p{T}_p$ acts trivially on N. Since G is a $\{p,q\}$-group, $G/C_G(N_i/N_{i-1})$ is a q-group for each G-chief factor $N_i/N_{i-1}$ of N. However, $O_q(G/C_G(N_i/N_{i-1}))=1$ by ([12], Lemma 1.7.11). It follows that $G/C_G(N_i/N_{i-1})=1$. This shows that G acts hypercentrally on N, and so does H, a contradiction. Thus, $G=NT$

Since N normalizes $T^{\prime }\left(p\right)$ and $T=T^{\prime }\left(p\right)$, we have $T⊴G=NT$.

(4) $G=RT$, where R is a nontrivial normal subgroup in G with $R\le N$.

If $RT<G$, then one can see that $RT$ satisfies the condition by Proposition 1 (3). Hence T acts hypercentrally on R by the choice of G. Since $N/R\le D_p(RT/R)$ and $O^p(RT/R)=RT/R$, then, by the choice of G, $RT/R$ acts hypercentrally on $N/R$. Then T acts hypercentrally on N, that is, $G=NT$ acts hypercentrally on N by ([12], Lemma 1.7.11), so does H, a contradiction.

(5) Final contradiction.

Since $G/R=TR/R$ acts hypercentrally on $N/R$, without generality, we can assume $R=N$ is minimal normal in G. Then, by the minimality of N and the normality of T, we have that $G=N\times T$ or $G=T$.

If $G=N\times T$, then $N\le Z\left(G\right)$, a contradiction.

Let $G=T$. Since T is the minimal supplement of $C_G\left(N\right)$ in G, we have that $T\cap C_G\left(N\right)\le \Phi \left(T\right)$ by ([12], Lemma 2.3.4). Thus, $C_G\left(N\right)\le \Phi \left(G\right)$. By the minimality of N and N, $O_q\left(G\right)\le C_G\left(N\right)\le \Phi \left(T\right)=\Phi \left(G\right)$. It follows that $O_{q,p}\left(G\right)$ is p-closed. Choose P to be a Sylow p-subgroup in $O_{q,p}\left(G\right)$. Then $P⊴G$ and so, $P\le C_G\left(N\right)\le \Phi \left(G\right)$. Therefore $O_{q,p}\left(G\right)\le \Phi \left(G\right)$, a contradiction. □

Proof of Theorem 2. 

(1) Since $C_G\left(G^{\prime }\left(p\right)\right)\le D_p\left(G\right)$, the necessity is clear.

Conversely, assume that $C_G\left(G^{\prime }\left(p\right)\right)=1$ and $D_p\left(G\right)>1$. It implies that $G^{\prime }\left(p\right)\cap D_p\left(G\right)>1$. Otherwise, $D_p\left(G\right)\le C_G\left(G^{\prime }\left(p\right)\right)$ and $C_G\left(G^{\prime }\left(p\right)\right)\ne 1$. By Proposition 1 (2), $D_p\left(G\right)$ is soluble. So G has a minimal normal subgroup N such that $N\le G^{\prime }\left(p\right)\cap D_p\left(G\right)$. Then N is elementary abelian.

$N\le Z\left(G^{\prime }\right)$.

Assume $G^{\prime }\cap D_p\left(G\right)=1$. Since $[G,D_p\left(G\right)]\le [G,G]=G^{\prime }$ and $[G,D_p\left(G\right)]\le D_p\left(G\right)$, $[G,D_p\left(G\right)]\le G^{\prime }\cap D_p\left(G\right)=1$. It follows that $D_p\left(G\right)\le Z\left(G\right)$, a contradiction and thus $G^{\prime }\cap D_p\left(G\right)\ne 1$. Since $G^{\prime }\left(p\right)\cap D_p\left(G\right)\ge G^{\prime }\cap D_p\left(G\right)$, we can assume that $N\le G^{\prime }\cap D_p\left(G\right)$.

Now, by the ([13], Theorem 2.3 (1)), we have $N\le G^{\prime }\cap D\left(G\right)\le Z_{\infty }\left(G^{\prime }\right)$. It follows from the minimality of N that $N\le Z\left(G^{\prime }\right)$.

$N\le C_G\left(O^p\left(G\right)\right)$.

Let N be q-group for some prime q and r a prime divisor of $\left|G\right|$ different to p and q. If R is a r-group. Then $N\le N_G\left(R\right)$ by $N\le D_p\left(G\right)$ and hence $[N,R]\le N\cap R=1$. Thus, $R\le C_G\left(N\right)$ and it follows from the choice of r that $G/C_G\left(N\right)$ is a $\{p,q\}$-group. Therefore, without generality, we can assume that G is a $\{p,q\}$-group.

If $q\ne p$, then, by Lemma 2, $N\le Z_{\infty }\left(O^p\left(G\right)\right)$. It follows from the minimality of N that $N\le Z\left(O^p\left(G\right)\right)$.

If N is a p-group, then $[N,Q]=[N,Q^{\prime }{O}^p\left(Q\right)]=1$ for any Sylow r-subgroup of G with $r\ne p$. Then $[N,O^p\left(G\right)]=1$, and $N\le C_G\left(O^p\left(G\right)\right)$.

Hence, one can see that $N\le C_G\left(G^{\prime }\left(p\right)\right)$, a contradiction.

(2) If $O_{p^{\prime }}\left(D_p\left(G\right)\right)=1$, the result is clear.

If $O_{p^{\prime }}\left(D_p\left(G\right)\right)\ne 1$, then G has a minimal normal subgroup N with $N\le O_{p^{\prime }}\left(D_p\left(G\right)\right)$.

For any Sylow r-subgroup R of G, we have $[N,R]=1$. Then $G/C_G\left(N\right)$ is a $\{p,q\}$-group, hence, without loss of generality, we assume that G is a $\{p,q\}$-group.

If N is a q-group, then, by Lemma 2, $N\le Z_{\infty }\left(O^p\left(G\right)\right)$. It follows from the minimality of N that $N\le Z\left(O^p\left(G\right)\right)$.

If N is a p-group, then $[N,Q]=[N,Q^{\prime }{O}^p\left(Q\right)]=1$ for any Sylow q-subgroup of G. Then $[N,O^p\left(G\right)]=1$, and $N\le Z\left(O^p\left(G\right)\right)$.

By induction, $O_{p^{\prime }}(D_p\left(G\right)/N)\le Z_{\infty }(O^p\left(G\right)/N)$, then $O_{p^{\prime }}\left(D_p\left(G\right)\right)\le Z_{\infty }\left(O^p\left(G\right)\right)$.

(3) Note that $Z\left(G^{\prime }\left(p\right)\right)=1$ if and only if $D_p\left(G\right)\cap G^{\prime }\left(p\right)=1$ by (1). Then $[D_p\left(G\right),G^{\prime }\left(p\right)]\le D_p\left(G\right)\cap G^{\prime }\left(p\right)=1$, therefore $D_p\left(G\right)=C_G\left(G^{\prime }\left(p\right)\right)$ by Proposition 1 (1). □

4. Minimal Subgroups and ${\mathit{D}}_{\mathit{p}}\left(\mathit{G}\right)$

The main aim of this section is to to prove the following theorem.

Theorem 3.

Let q be a prime. Assume that every element of order q lies in $D_p\left(G\right)$, and in addition, if $q=2$ and the Sylow q-subgroup of G is nonabelian, then every element of order 4 lies in $D_p\left(G\right)$. Then G is q-soluble and $l_q\left(G\right)\le 1$.

Proof. 

Let $\Omega =\langle x\in O^p\left(G\right)\mid x^q=1\rangle$, if $q\ne 2$ or the Sylow q-subgroup of G is abelian; $\Omega =\langle x\in O^p\left(G\right)\mid x^4=1\rangle$, if $q=2$ and the Sylow q-subgroup of G is nonabelian. Then $\Omega \le O^p\left(G\right)\cap D_p\left(G\right)$ by hypothesis.

Assume $p\ne q$. By Theorem 1.3, $\Omega$ is a $p^{\prime }$-group and by Theorem 1.4, $\Omega \le Z_{\infty }\left(O^p\left(G\right)\right)$. If $O^p\left(G\right)$ is not q-nilpotent, then there exists a minimal non-q-nilpotent subgroup H of $O^p\left(G\right)$. By the structure of the minimal non-q-nilpotent groups, we have that $H=\left[Q\right]R$, where $Q=O_q\left(H\right)$ and exp$\left(Q\right)=q$ or 4 (if q = 2 and Q is non-abelian) and R is a cyclic r-group with $r\ne q$. However, $Q\le \Omega \le Z_{\infty }(O^p\left(G\right)$, so $Q\le H\cap Z_{\infty }(O^p\left(G\right)\le Z_{\infty }\left(H\right)$. It follows that H is nilpotent, a contradiction. This contradiction shows that $O^p\left(G\right)$ is q-nilpotent. Thus, G is q-soluble and $l_q\left(G\right)\le 1$ since $G/O^p\left(G\right)$ is a p-group.

Assume $p=q$. If $O^p\left(G\right)$ is of order $p^{\prime }$ then G is $p^{\prime }$-closed and so is p-nilpotent. In particular, G is p-soluble with $l_p\left(G\right)\le 1$. If $O^p\left(G\right)$ is not a $p^{\prime }$-group, then $\Omega \ne Ø$ and by Theorem 1.3, $O_{p^{\prime }}(\Omega )$ is the Hall $p^{\prime }$-subgroup of $\Omega$. Let T be any $p^{\prime }$-subgroup of G. Then $\Omega \le N_G\left(R\right)$. Since, clearly, $\Omega$ is normal in G, we see that $[\Omega ,T]\le \Omega \cap T\le O_{p^{\prime }}(\Omega )$. Since $O^p\left(G\right)=\langle T\le G\mid p\nmid \left|T\right|\rangle$, $[\Omega ,O^p\left(G\right)]\le O_{p^{\prime }}(\Omega )$. Now, considering on the quotient $O^p\left(G\right)/O_{p^{\prime }}(\Omega )$, we have that $\Omega /O_{p^{\prime }}(\Omega )\le Z(O^p\left(G\right)/O_{p^{\prime }}(\Omega ))$. By a same argument as above (or by Ito’s theorem), it can be obtained that $O^p\left(G\right)/O_{p^{\prime }}(\Omega )$ is p-nilpotent. Therefore, $O^p\left(G\right)$ is p-nilpotent and so is G. Thus, is p-soluble with $l_p\left(G\right)\le 1$. The proof is completed. □