Proof. Obviously, the projection of the nonnormal subgroups of $G/N$ onto G are still nonnormal, and hence $n_c(G/N)\leq n_c(G)$ . If there exists a nonnormal subgroup of G whose order is not divisible by $|N|$ , then $n_c(G/N)<n_c(G)$ . This completes the proof.

Proof. Let G be a p-group and assume that $n_c(G)=t$ . If G has no elementary abelian normal subgroup of order $p^2$ , then, by Lemma 2.1, G is either a cyclic group or a $2$ -group of maximal class. It is easy to see that $s_p(G')\leq n_c(G)+1$ and the result follows.

Now, suppose that there exists an elementary abelian normal subgroup N of order $p^2$ . In this case, we perform induction on t. If $t=0$ , clearly, G is Dedekind and $s_p(G')\leq 1$ , as required. Next, suppose that $t\geq 1$ . We consider the factor group $G/N$ . Assume that M is a nonnormal subgroup of minimal order of G. Then M is cyclic by Lemma 2.2. Let $|M|=p^m$ . We claim that $n_c(G/N)\leq t-1$ . If $p^m\leq p^2$ , it follows from Lemma 2.3 that $n_c(G/N)\leq t-1$ . Conversely, if $p^m>p^2$ , then $G/N$ has no nonnormal subgroups of order $p^{m-2}$ . Otherwise, there exists a noncyclic nonnormal subgroup of order $p^m$ of G, which contradicts the minimality of M. Thus, according to Lemma 2.3, we have $n_c(G/N)\leq t-1$ , as claimed. Here, by induction on t, it follows that $s_p((G/N)')\leq 2(t-1)+1$ . Therefore,

$$ \begin{align*}s_p(G')\leq s_p(N)+s_p((G/N)')\leq 2t+1.\end{align*} $$

The proof is complete.

Proof. Let $P_i\in \mathrm {Syl}_{p_i}(G)$ and assume that $G=P_1\times P_2\times \cdots \times P_k$ with $n_c(G)=t$ . If $k=1$ , the result is trivial by Lemma 2.4. Now, let $k\geq 1$ . We assume that $G=H\times P_{k}$ . Since $n_c(G)=t$ , we have $n_c(H)< t/2$ and $n_c(P_k)\leq t/2$ . By induction on k, it follows that $s_p(H')<t+1$ and $s_p({P_k}')<t+1$ . Therefore, $s_p(G')\leq 2t+1$ .

Proof of Theorem 1.1.

Let G be a p-group and assume that $n_c(G)=t$ . By Lemma 2.4, we see that $|G'|\leq p^{2t+1}$ and thus $c(G)\leq 2t+2$ . It suffices to show that $G^{(i)}\leq G_{2^i}$ for $ i\geq 1$ since, by induction on i,

$$ \begin{align*}G^{(i)}=[G^{(i-1)},G^{(i-1)}]\leq [G_{2^{i-1}},G_{2^{i-1}}]\leq G_{2^{i}}.\end{align*} $$

Note that $1=G_{2t+3}=G^{(dl(G))}\leq G_{2^{dl(G)}}$ . Consequently, $2^{dl(G)}\leq 2t+3$ , that is, $dl(G)\leq \lceil \log _2(2t+3)\rceil $ . This completes the proof.

Proof. By Lemma 2.3, $n_c(G/N)\leq n_c(G)$ . First, we claim that there exists a minimal normal subgroup N of G such that $n_c(G/N)<n_c(G)$ . Let $P_i\in \mathrm {Syl}_{p_i}(G)$ . Noting that G is nonnilpotent, we may assume that $P_1$ is a nonnormal Sylow subgroup of G. If, for $i\geq 2$ , there exists a Sylow subgroup $P_i$ such that $P_i$ is nonnormal, we may assume that $P_2$ is nonnormal. Then $n_c(G/N)<n_c(G)$ is always true for any minimal normal subgroup $N\neq 1$ . Otherwise, by Lemma 2.3, the orders of both $P_1$ and $P_2$ are divisible by the order of N, so that $N=1$ , which is a contradiction. On the other hand, if $P_i\unlhd G$ for all $i\geq 2$ , we may take $N\leq P_2$ . According to Lemma 2.3 again, $n_c(G/N)<n_c(G)$ since the order of $P_1$ is not divisible by the order of N. This proves the claim.

Since $ N $ is a minimal normal subgroup of G, it follows that N is an elementary abelian p-group and proper subgroups of N are nonnormal subgroups of G. There are $s_p(N)-1$ nonnormal subgroups of G contained by N. Thus,

$$ \begin{align*}n_c(G/N)\leq n_c(G)-(s_p(N)-1).\end{align*} $$

Here, if $n_c(G/N)=n_c(G)-s_p(N)+1$ , then, similarly, both the orders of $P_1$ and $P_2$ are divisible by p, which is a contradiction. Hence, $n_c(G/N)\leq n_c(G)-s_p(N)$ and the proof is complete.

Proof. Assume that $n_c(G)=t$ . The proof will be done by induction to t. If $t=1$ , then, by Lemma 3.1,

$$ \begin{align*}G=\langle a\rangle\rtimes\langle b\rangle,\end{align*} $$

where $o(a)={p_2}$ , $o(b)={p_1}^{n_1}$ and $p_1, p_2$ are different primes. Since $G/\langle a\rangle $ is cyclic, we have $s_p(G')=1$ .

Now, let $t\geq 2$ . According to the proof of Lemma 3.3, it suffices to show that there exists a minimal normal subgroup N such that $n_c(G/N)<t$ .

Case 1: $G/N$ is nonnilpotent.

In this case, since $n_c(G/N)<t$ , it follows that $s_p((G/N)')\leq n_c(G/N)$ by induction on t. In addition, $|G'|=|G'\cap N||(G/N)'|$ because $(G/N)'\cong G'/(G'\cap N)$ . Hence, $|N||(G/N)'|$ is divisible by $|G'|$ . Therefore,

$$ \begin{align*}s_p(G')\leq s_p(N)+s_p((G/N)')\leq s_p(N)+n_c(G/N).\end{align*} $$

By Lemma 3.3, $n_c(G/N)\leq n_c(G)-s_p(N)$ , and hence

$$ \begin{align*}s_p(G')\leq s_p(N)+n_c(G/N)\leq n_c(G)=t.\end{align*} $$

This completes the proof in Case 1.

Case 2: $G/N$ is nilpotent. In this case, we consider the following two situations.

Case 2a: there exists a minimal normal subgroup M such that $M\neq N$ .

Since G is a nonnilpotent group, it follows that $G/M$ is also nonnilpotent. Otherwise, since $G/(M\cap N) \lesssim G/M \times G/N$ , we see that $G/(M\cap N)$ is nilpotent. However, $G/(M\cap N)\cong G$ is nonnilpotent, which is a contradiction. Now, assume that $|M|=p^m$ and $|N|=q^n$ , where $p,q$ are different primes. We consider two cases, namely, $m\geq 2$ and $m=1$ . If $m\geq 2$ , since $N_1M_1\ntrianglelefteq G$ for all $1<M_1<M$ and $1\leq N_1\leq N$ , then

$$ \begin{align*}n_c(G/M)\leq n_c(G)-(m-1)(n+1)\leq n_c(G)-m.\end{align*} $$

Here, it follows easily by induction that $s_p((G/M)')\leq n_c(G/M)$ . This condition is similar to Case 1 and it follows that

$$ \begin{align*}s_p(G')\leq s_p(M)+n_c(G/M)\leq n_c(G).\end{align*} $$

Now suppose that $m=1$ , that is, $|M|=p$ . If there exists a nonnormal subgroup H such that $|H|$ is not divisible by p, then $n_c(G/M)\leq n_c(G)-1$ from Lemma 2.3, and so $s_p((G/M)')\leq n_c(G/M)$ by induction. As before, the result holds. On the other hand, if, for every subgroup H of $G$ whose order is not divisible by p, H is always normal, then we may assume that $G=KP$ , where K is a Hall $p'$ -subgroup of G. Obviously, all subgroups of K are normal and P is nonnormal. We consider the following two cases.

(i) If there exists a minimal normal subgroup T of G contained in K satisfying $T\neq N$ , then $G/T$ is nonnilpotent. It suffices to show that $n_c(G/T)\leq n_c(G)-1$ by Lemma 2.3, and thus $s_p((G/T)')\leq n_c(G/T)$ by induction. As before, the result holds.

(ii) If N is a unique minimal normal subgroup of G contained in K, then K is a group of prime power order. It follows from Lemma 2.1 that K is either a cyclic group or a $2$ -group of maximal class. In addition, since every subgroup of K is a normal subgroup of G, it follows that K is either a cyclic group or a quaternion group $Q_8$ . We claim that K is cyclic. Otherwise, $K\cong Q_8$ . Note that $N\leq Z(G)\cap Q_8$ and $G/N$ is nilpotent. According to Lemma 3.2, G is nilpotent, which is a contradiction. Now, let K be a cyclic group of order $q^r$ with $r\geq 2$ . For $1\leq K_1\leq K$ , it follows that $K_1P_1$ is nonnormal as $P_1\leq P$ and $P_1\ntrianglelefteq G$ . Also, there exists a maximal subgroup M of P that is normal in P, but $MK_1$ is a nonnormal subgroup of G for $1\leq K_1<K$ . Hence,

$$ \begin{align*}n_c(G/K)(r+1)+r\leq t.\end{align*} $$

By Lemma 2.4, $s_p((G/K)')\leq 2(t-r)/(r+1)+1$ . Note that $n_c(G)=t\geq 2r+1$ and $r\geq 2$ . Therefore,

$$ \begin{align*} s_p(G')\leq s_p(K)+s_p((G/K)')&\leq r+\frac{2(t-r)}{r+1}+1 \\ &\leq \frac{r(r+1)+r(t-r)+(r+1)}{r+1} \leq \frac{r(t+1)+t-r}{r+1} =t. \notag \end{align*} $$

Case 2b: N is a unique minimal normal subgroup of G.

In this case, $G/H$ is nilpotent for $1\neq H\unlhd G$ . We can assume that $G/N=P_1\rtimes P_2$ with $N\leq P_1$ . Let $|N|={p_1}^k$ . Then there are $k-1$ nonnormal subgroups of G contained in N. Clearly, if $NK$ is nonnormal in G for $K\leq G$ , then $K\ntrianglelefteq G$ . Note that $P_2N\unlhd G$ but $P_2$ is a nonnormal subgroup of G. Moreover, we can always find $gN\in Z(G/N)$ such that $g\in G-N$ and $g^p\in N$ since $G/N$ is nilpotent. Also, $\langle g\rangle N\unlhd G$ but $\langle g\rangle $ is nonnormal in G. Therefore,

$$ \begin{align*}2n_c(G/N)+(k-1)+1+1\leq t.\end{align*} $$

It follows that $n_c(G/N)\leq (t-k-1)/2$ and, by Lemma 2.5, $s_p((G/N)')\leq t-k$ . Hence,

$$ \begin{align*}s_p(G')\leq s_p(N)+s_p((G/N)') \leq k+t-k \leq t.\end{align*} $$

The proof is complete.

Proof. We prove the result by induction on n. If $ n=1$ , the result is trivially true. Assume that $n\geq 2$ . If $s_p(G/{G'})\geq 2$ , then $s_p(G')\leq n-2$ . It follows that $dl(G')\leq \lfloor (2n-2)/3\rfloor $ by the inductive hypothesis applied to $G'$ . Hence,

$$ \begin{align*}dl(G)\leq \lfloor(2n-2)/3\rfloor+1\leq \lfloor(2n+2)/3\rfloor.\end{align*} $$

In this case, the proof is complete.

Now, let $s_p(G/{G'})=1$ , that is, $s_p(G')=n-1$ . We may assume that $dl(G)=k+1$ where $k\geq 2$ . Then $G^{(k)}>1$ . Also, suppose that N is a maximal abelian normal subgroup of G containing $G^{(k)}$ . If $s_p(N)\geq 2$ , we see that $s_p(G/N)\leq n-2$ . Application of the inductive hypothesis to $G/N$ yields $dl(G/N)\leq \lfloor (2n-2)/3\rfloor $ . Thus,

$$ \begin{align*}dl(G)\leq \lfloor(2n-2)/3\rfloor+1\leq \lfloor(2n+2)/3\rfloor,\end{align*} $$

and the result follows.

The remaining case is where $s_p(N)=1$ , which implies that $N=G^{(k)}$ . Since $G/N=N_G(N)/C_G(N)\lesssim \mathrm {Aut}(N)$ is cyclic, it suffices to show that $N=G^{(k)}\leq Z(G')$ . Hence,

$$ \begin{align*}N=G^{(k)}\leq Z(G^{(k-1)}).\end{align*} $$

Now $G^{(k-1)}$ is nonabelian since $G^{(k)}\neq 1$ . We claim that $s_p(G^{(k-1)})\geq 3$ . Otherwise, $G^{(k-1)}$ is a nonabelian group of order $pq$ with $p\neq q$ . Since $G^{(k-1)}/{G^{(k)}}$ is cyclic, it suffices to show that $G^{(k-1)}$ is an abelian group, which is a contradiction. Hence, $s_p(G/G^{(k-1)})\leq n-3$ . Apply the inductive hypothesis to $G/{G^{(k-1)}}$ . Then $dl(G/{G^{(k-1)}})\leq \lfloor (2n-4)/3\rfloor $ . Therefore,

$$ \begin{align*}dl(G)\leq \lfloor(2n-4)/3\rfloor+2= \lfloor(2n+2)/3\rfloor.\end{align*} $$

The proof is complete.

Proof of Theorem 1.2.

Suppose that G is a solvable nonnilpotent group with $n_c(G)=t$ . From Lemma 3.4, $s_p(G')\leq t$ , and hence, by Lemma 3.5,

$$ \begin{align*}dl(G')\leq \lfloor(2t+2)/3\rfloor.\end{align*} $$

Hence, $dl(G)\leq \lfloor (2t+2)/3\rfloor +1$ . The proof is complete.

Proof. Assume that $\pi (G)\geq t+2$ . Since G is a solvable group, G possesses a Sylow system $\mathcal {S}$ . Suppose that $\mathcal {S}=\{P_1,P_2,\ldots ,P_{t+2},\ldots \}$ . Note that G is nonnilpotent and we may assume that $P_1$ is a nonnormal Sylow subgroup of G. Let

$$ \begin{align*}\mathcal{T}=\{P_1P_2,P_1P_3,P_1P_4,\ldots,P_1P_{t+2}\}.\end{align*} $$

Obviously, for $1\leq i\leq t+2$ , $P_1P_i$ is a subgroup of G. If, for the set $\mathcal {T}$ , there are two or more normal subgroups of G, then $P_1$ is a normal subgroup, which is a contradiction. Thus, at most one normal subgroup is contained in the set $\mathcal {T}$ and it follows that $n_c(G)\geq t+1$ . This contradicts the hypothesis and the proof is complete.