Proof. $\Rightarrow $ Let us suppose that n is almost abelian. It follows that

(2.2) $$ \begin{align} \text{ for any divisor }d\text{ of }n, \text{ there is at most one nonabelian group of order }d. \end{align} $$

First, suppose that there exists $r\in \{1,\ldots ,k\}$ such that $n_r\geq 3$ . This contradicts (2.2), since there are at least two nonabelian groups of order $p_r^{n_r}$ when $n_r\geq 3$ .

Since n is not abelian, the only remaining possibility is that

(2.3) $$ \begin{align} \begin{array}{l} \text{there exist } i, j\in\{1,\ldots,k\} \text{ such that }p_i \mid p_j^{n_j}-1 \\ \text{and } n_r\leq 2 \quad \text{ for all } r\in\{1,\ldots,k\}. \end{array} \end{align} $$

Let us analyse which of the exponents can be $2$ . From (2.3), it follows that there exists a nontrivial semi-direct product $(C_{p_j})^{n_j}\rtimes C_{p_i}$ . If $n_r=2$ , for $r\in \{1,2,\ldots ,k\}\setminus \{i,j\}$ , it follows that there are two nonabelian groups of order $p_ip_j^{n_j}p_r^2$ , namely,

$$ \begin{align*} ((C_{p_j})^{n_j}\rtimes C_{p_i})\times (C_{p_r})^2 \quad\text{and}\quad ((C_{p_j})^{n_j} \rtimes C_{p_i})\times C_{p_r^2},\end{align*} $$

which gives a contradiction. If $n_i=2$ , then we will have two nonabelian groups of order $p_i^2p_j^{n_j}$ ,

$$ \begin{align*}(C_{p_j})^{n_j} \rtimes C_{p_i^2} \quad\text{and}\quad ((C_{p_j})^{n_j}\rtimes C_{p_i})\times C_{p_i},\end{align*} $$

which again gives a contradiction.

It follows that $n_r=1 \text { for all } r\neq j$ . If in (2.3) there are two distinct pairs $(i,j)$ and $(\alpha ,\beta )$ such that $p_{i} \mid p_j^{n_j}-1$ and $p_{\alpha } \mid p_{\beta }^{n_\beta }-1,$ then there are two distinct nonabelian groups of order $p_i p_j^{n_j}p_{\alpha }p_{\beta }^{n_{\beta }}$ ,

$$ \begin{align*}((C_{p_{\beta}})^{n_{\beta}}\rtimes C_{p_{\alpha}})\times C_{p_{i}p_{j}^{n_{j}}} \quad\text{and}\quad C_{p_{\alpha} p_{\beta}^{n_{\beta}}}\times ((C_{p_{j}})^{n_j}\rtimes C_{p_i}),\end{align*} $$

which is a contradiction. Consequently, there is at most one pair.

If $n_j=1$ , then (1.1) is satisfied. If $n_j=2$ , let us assume that $p_i\mid p_j-1$ . It follows that there are two distinct nonabelian groups of order $p_ip_j^2$ ,

$$ \begin{align*} (C_{p_j})^2\rtimes C_{p_i} \quad\text{and}\quad C_{{p_j}^2}\rtimes C_{p_i},\end{align*} $$

which contradicts our hypothesis. Consequently, $p_i\nmid p_j-1$ and since $p_i \mid p_j^2-1$ , we have $p_i \mid p_j+1$ , which gives (2.1).

$\Leftarrow $ We will prove by induction over n that

(2.4) $$ \begin{align} \begin{array}{l} \text{if }n\text{ has at least two nonprime factors and satisfies either }(1.1)\text{ or }(2.1),\\ \text{then there is a unique nonabelian group of order }n. \end{array} \end{align} $$

The base case is $n=6$ . There is just one nonabelian group of order $6$ , the symmetric group $S_3$ .

Let us proceed to the inductive step. Assume that (2.4) holds for any positive integer with at least two factors $n'<n$ . Let G be a nonabelian group of order n. We can assume that $p_1<p_2<\cdots <p_k$ . It follows that $j\geq 2$ . Indeed, if $j=1$ , there are two possibilities. If $p_i \mid p_1-1$ , then $p_i<p_1$ , which is a contradiction. If $p_i \mid p_1+1$ , then since $p_i>p_1$ , we have $p_i=p_1+1$ which implies $p_1=2$ and $p_i=p_2=3$ so that $p_j \mid p_i-1$ , which is a contradiction.

Thus, $j\geq 2$ and therefore $n_1=1$ . Hence, the $p_1$ -Sylow subgroups of G are cyclic of order $p_1$ . By the Burnside normal p-complement theorem, G has a $p_1$ -normal complement, that is, $\text {there exists } H\triangleleft G$ with $|H|=n/p_1$ such that $G=H\rtimes C_{p_1}$ . We identify two cases.

Case 1: n satisfies (1.1). If $i=1$ , it follows that H is cyclic so that

$$ \begin{align*}G=C_{{n}/{p_1}}\rtimes C_{p_1}\cong (C_{p_j}\rtimes C_{p_1})\times C_{{n}/{p_1p_j}}.\end{align*} $$

If $i\geq 2$ , there are again two possibilities.

• • If H is abelian, then H is cyclic which is analogous to the case $i=1$ .

• • If H is nonabelian, then $H\cong (C_{p_j}\rtimes C_{p_i})\times C_{{n}/{p_1p_ip_j}}$ which implies

  $$ \begin{align*}G\cong ((C_{p_j}\rtimes C_{p_i})\times C_{{n}/{p_1p_ip_j}})\rtimes C_{p_1} \cong ((C_{p_j}\rtimes C_{p_i})\rtimes C_{p_1})\times C_{{n}/{p_1p_ip_j}}.\end{align*} $$
  Since $\lvert \operatorname {\mathrm {Aut}}(C_{p_j}\rtimes C_{p_i})\rvert =p_j(p_j-1)$ and $p_1\nmid p_j(p_j-1)$ , the semidirect product $(C_{p_j}\rtimes C_{p_i})\rtimes C_{p_1}$ is trivial; therefore,
  $$ \begin{align*}G\cong ((C_{p_j}\rtimes C_{p_i})\times C_{p_1})\times C_{{n}/{p_1p_ip_j}} \cong (C_{p_j}\rtimes C_{p_i})\times C_{{n}/{p_ip_j}}.\end{align*} $$
  It follows that in this case, there is a single nonabelian group of order n.

Case 2: n satisfies (2.1). If $i=1$ , then H is abelian and there are two possibilities.

• • If $H\cong C_{p_j^2}\times C_{{n}/{p_1p_j^2}}$ , then $G\cong (C_{p_j^2}\rtimes C_{p_1})\times C_{{n}/{p_1p_j^2}}$ . Since $p_1\nmid \lvert \operatorname {\mathrm {Aut}}((C_{p_j^2})\rvert =p_j(p_j-1)$ , it follows that $C_{p_j^2}\rtimes C_{p_1}=C_{p_j^2}\times C_{p_1}$ . Therefore, $G\cong C_n$ , which is a contradiction.

• • If $H\cong (C_{p_j})^2\times C_{{n}/{p_1p_j^2}}$ , then $G\cong ((C_{p_j})^2\rtimes C_{p_1})\times C_{{n}/{p_1p_j^2}}$ , which is nonabelian.

If $i\geq 2$ , then we again have two cases.

• • If $H\kern1.5pt{\cong}\kern1.5pt ((C_{p_j})^2\kern1.5pt{\rtimes}\kern1.5pt C_{p_i})\kern1.5pt{\times}\kern1.5pt C_{{n}/{p_1p_ip_j^2}}$ , then $G\kern1.5pt{\cong}\kern1.5pt (((C_{p_j})^2\kern1.5pt{\rtimes}\kern1.5pt C_{p_i})\kern1.5pt{\rtimes}\kern1.5pt C_{p_1})\kern1.5pt{\times}\kern1.5pt C_{{n}/{p_1p_ip_j^2}}$ . From [Reference Campedel, Caranti and Del Corso2], it follows that $\lvert \operatorname {\mathrm {Aut}}((C_{p_j})^2\rtimes C_{p_i}\rvert =2(p_j^2-1)p_j^2\not \mid p_1$ . Therefore,

  $$ \begin{align*}G\cong (((C_{p_j})^2\rtimes C_{p_i})\times C_{p_1})\times C_{{n}/{p_1p_ip_j^2}} \cong ((C_{p_j})^2\rtimes C_{p_i})\times C_{{n}/{p_ip_j^2}}.\end{align*} $$

• • Finally, if H is abelian, then we identify two possibilities. The first possibility is $H\cong C_{p_2}\times \cdots \times C_{p_j}\times \cdots \times C_{p_k}\cong C_{{n}/{p_1}}$ , which implies that G cyclic, and this is a contradiction. The second possibility is $H\cong C_{p_2}\times \cdots C_{p_j}^2\times \cdots \times C_{p_k}\cong (C_{p_j})^2\times C_{{n}/{p_1p_j^2}}$ . We observe that

  (2.5) $$ \begin{align} p_1\nmid \lvert\operatorname{\mathrm{Aut}}((C_{p_j})^2 \times C_{{n}/{p_1p_j^2}})\rvert=(p_j^2-1)(p_j^2-p_j)\cdot \varphi\bigg(\frac{n}{p_1p_j^2}\bigg). \end{align} $$
  Thus, $G\cong ((C_{p_j})^2\times C_{{n}/{p_1p_j^2}})\rtimes C_{p_1}\cong ((C_{p_j})^2\times C_{{n}/{p_1p_j^2}})\times C_{p_1}{\cong } (C_{p_j})^2\times C_{{n}/{p_j^2}}$ , where the final congruence comes from (2.5). This means G is abelian, which is a contradiction.

Therefore, also in this case, there is only one nonabelian group of order n.

Proof. $\Leftarrow $ The conclusion follows from Theorem 2.1 since all the groups constructed in the proof are nonnilpotent.

$\Rightarrow $ Let us assume n is almost nilpotent. It follows that $\text { for all } d \mid n, \text { there exists } $ at most one nonnilpotent group of order d. Since n is nonnilpotent, there are integers $(i,j)\in \{1,\ldots ,k\}^2$ and $d_j$ with $1\leq d_j\leq n_j$ such that $p_i \mid p_j^{d_j}-1$ . It follows that $\alpha _j=n_j$ . Otherwise, there would be two nonnilpotent nonisomorphic groups of order $p_j^{n_j}p_i$ , namely

$$ \begin{align*}((C_{p_j})^{\alpha_j}\rtimes C_{p_i})\times C_{p_j^{n_j-d_j}} \quad\text{and}\quad (C_{p_j})^{n_j}\rtimes C_{p_i}.\end{align*} $$

Furthermore, the pair $(i,j)$ is unique. Otherwise, if there were two pairs $(i',j')\neq (i,j)$ such that $p_{i'} \mid p_{j'}^{n_j}-1$ , again there would be two nonnilpotent nonisomorphic groups of order n,

$$ \begin{align*}((C_{p_j})^{n_j}\rtimes C_{p_i})\times C_{{n}/{p_ip_j^{n_j}}} \quad\text{and}\quad ((C_{p_{j'}})^{n_{j'}}\rtimes C_{p_{i'}})\times C_{{n}/{p_{i'}p_{j'}^{n_{j'}}}}.\end{align*} $$

Let us observe that $n_r=1 \text { for all } r\neq i, j$ . Otherwise, there would exist at least two distinct groups $P_r$ and $Q_r$ of order $p_r^{n_r}$ , which would give two nonnilpotent, nonisomorphic groups of order $p_j^{n_j}p_ip_r^{n_r}$ ,

$$ \begin{align*}((C_{p_j})^{n_j}\rtimes C_{p_i})\times P_r \quad\text{and}\quad ((C_{p_j})^{n_j}\rtimes C_{p_i})\times Q_r.\end{align*} $$

Analogously, we can show that $n_i=1$ .

If $n_j=1$ , then (1.1) holds. If $n_j\geq 2$ , then $n_j=2$ , since otherwise, we would have two nonnilpotent, nonisomorphic groups of order $p_j^{n_j}p_i$ . Thus,

(2.6) $$ \begin{align} p_i \mid p_j^2-1. \end{align} $$

In addition, if $p_1 \mid p_j-1$ , there are two nonnilpotent, nonisomorphic groups of order $p_j^{2}p_i$ ,

$$ \begin{align*}(C_{p_j})^2\rtimes C_{p_i} \quad\text{and}\quad C_{p_j^2}\rtimes C_{p_i}.\end{align*} $$

It follows that $p_i\nmid p_j-1$ and, by (2.6), $p_i \mid p_j+1$ . This yields (2.1) and concludes the proof.