Proof. We easily obtain

$$ \begin{align*} m^*(Q_{2^n})=2^{2n-2} \quad\mbox{for all } n\geq 3, \end{align*} $$

and

$$ \begin{align*} {\cal CD}(Q_{2^n})=\begin{cases} \{Q_8, \langle a\rangle, \langle b\rangle, \langle ab\rangle, \langle a^2\rangle\}&\hbox{if } n=3\\ \{\langle a\rangle\}&\hbox{if } n\geq 4.\end{cases} \end{align*} $$

These lead immediately to the desired conclusions.

Proof of Theorem 1.2

We divide the proof into two parts corresponding to the two parts of the theorem.

Part (a). Since $|{\cal CD}(G)|=|L(G)|-1$ , we have ${\cal CD}(G)=L(G)\setminus \{H_0\}$ , where $H_0\leq G$ . We infer that $|\mathrm { Im}(m_G)|=2$ and so G is a p-group with $|Z(G)|=p$ by Theorem 1.4. Then $m_G(1)<m_G(Z(G))$ , implying that

$$ \begin{align*} H_0=1\quad \mbox{and}\quad m^*(G)=m_G(Z(G))=p^{n+1}, \end{align*} $$

where $|G|=p^n$ .

Assume that there exists $H\leq G$ with $|H|=p$ and $H\neq Z(G)$ . Then $H\notin {\cal CD}(G)$ , which shows that $H=1$ and this contradicts the hypothesis. Thus, G has a unique subgroup of order p and Theorem 1.3 leads to

(2.1) $$ \begin{align} G\cong\mathbb{Z}_{p^n} \quad\mbox{or}\quad G\cong Q_{2^n} \quad\mbox{for some } n\geq 3. \end{align} $$

In the first case, we easily get $n=1$ , that is, $G\cong \mathbb {Z}_p$ , while in the second one, we get $G\cong Q_8$ by Lemma 2.1(a).

Part (b). The condition $|{\cal CD}(G)|=|L(G)|-2$ means ${\cal CD}(G)=L(G)\setminus \{H_1,H_2\}$ , where $H_1,H_2\leq G$ . Then $|\mathrm {Im}(m_G)|\leq 3$ . Recall that we cannot have $|\mathrm {Im}(m_G)|=1$ .

If $|\mathrm {Im}(m_G)|=2$ , then

$$ \begin{align*} m_G(H_1)=m_G(H_2)\neq m^*(G) \end{align*} $$

and again G is a p-group with $|Z(G)|=p$ . It is clear that one of the two subgroups $H_1$ and $H_2$ must be trivial, say $H_1=1$ . Then G has at most two subgroups of order p, namely $Z(G)$ and possibly $H_2$ . This implies that it has exactly one subgroup of order p because the number of subgroups of order p in a finite p-group is congruent to $1 \ (\mathrm {mod}\, p)$ . Consequently, one obtains again (2.1). For $G\cong \mathbb {Z}_{p^n}$ , we easily get $n=2$ , that is, $G\cong \mathbb {Z}_{p^2}$ , while for $G\cong Q_{2^n}$ , we get no solution by Lemma 2.1(b).

If $|\mathrm {Im}(m_G)|=3$ , then $m_G(H_1)$ , $m_G(H_2)$ and $m^*(G)$ are distinct. Also, (1.1) becomes

$$ \begin{align*} 3\geq 1+\sum_{p}n_p. \end{align*} $$

Since $n_p\geq 1$ for all p, we have the two possibilities described in Cases 1 and 2.

Case 1: $|G|=p^n$ and $|Z(G)|\in \{p,p^2\}$ .

Obviously, if G is abelian, we get $G\cong \mathbb {Z}_{p^2}$ . Assume that G is not abelian. Since $m_G(1)<m_G(Z(G))=m_G(G)$ , we infer that one of the two subgroups $H_1$ and $H_2$ is trivial and that

$$ \begin{align*} m^*(G)=m_G(Z(G))=m_G(G). \end{align*} $$

If $|Z(G)|=p$ , then G has a unique subgroup of order p and so it is a generalised quaternion $2$ -group, contradicting Lemma 2.1(b). The same can also be said when $|Z(G)|=p^2$ because all subgroups of order p of G are outside of ${\cal CD}(G)$ .

Case 2: $|G|=p^nq^m$ and the Sylow p-subgroups and q-subgroups of G have centres of orders p and q, respectively.

Let P be a Sylow p-subgroup and Q be a Sylow q-subgroup of G. Since $P\subseteq C_G(Z(P))$ , we have

$$ \begin{align*} m_G(Z(P))=p\,|C_G(Z(P))|=p^{n+1}q^x \quad\mbox{for some } x \mbox{ with } 0\leq x\leq m, \end{align*} $$

and similarly,

$$ \begin{align*} m_G(Z(Q))=p^yq^{m+1} \quad\mbox{for some } y \mbox{ with } 0\leq y\leq n. \end{align*} $$

Also,

$$ \begin{align*} m_G(1)=p^nq^m \quad\mbox{and}\quad m_G(G)=p^nq^m|Z(G)|. \end{align*} $$

We observe that the measures $m_G(Z(P))$ , $m_G(Z(Q))$ and $m_G(1)$ are distinct and consequently they are all possible measures of the subgroups of G. We distinguish two subcases.

Subcase 2.1: $Z(G)=1$ .

Then $m^*(G)=m_G(1)=m_G(G)$ . Indeed, if $m^*(G)=m_G(Z(P))$ , then $1$ , G and $Z(Q)$ will be outside of ${\cal CD}(G)$ , and this contradicts the hypothesis. In the same way, we cannot have $m^*(G)=m_G(Z(Q))$ . Since $m_G(P)$ is divisible by $p^{n+1}$ and $m_G(Q)$ is divisible by $q^{m+1}$ , we infer that $m_G(P)=m_G(Z(P))$ and $m_G(Q)=m_G(Z(Q))$ . Thus, $P,Z(P),Q,Z(Q)\notin {\cal CD}(G)$ and our hypothesis implies that $P=Z(P)$ and $Q=Z(Q)$ , that is, G is a nonabelian group of order $pq$ . Assume that $p<q$ . Then ${\cal CD}(G)$ consists of the unique subgroup of order q of G and therefore we obtain $|L(G)|=3$ , and this contradicts the hypothesis.

Subcase 2.2: $Z(G)\neq 1$ .

Then $m_G(1)<m_G(G)$ , which shows that $m_G(G)$ equals either $m_G(Z(P))$ or $m_G(Z(Q))$ . Assume that $m_G(G)=m_G(Z(P))$ . Then $x=m$ and $|Z(G)|=p$ , implying that

(2.2) $$ \begin{align} Z(G)=Z(P). \end{align} $$

Note that we cannot have $m^*(G)=m_G(Z(Q))$ because in this case, $1$ , $Z(G)$ and G will be outside of ${\cal CD}(G)$ , and this contradicts the hypothesis. Consequently,

$$ \begin{align*} m^*(G)=m_G(Z(P))=m_G(P)=m_G(Z(G))=m_G(G). \end{align*} $$

It follows that $1$ , $Z(Q)$ and Q are not contained in ${\cal CD}(G)$ , which leads to $Q=Z(Q)$ . In other words, ${\cal CD}(G)=L(G)\setminus \{1,Q\}$ . Thus, ${\cal CD}(G)$ is the lattice interval

$$ \begin{align*} [G/Z(G)]=\{H\in L(G)\mid Z(G)\leq H\leq G\} \end{align*} $$

and [Reference Tărnăuceanu11, Corollary 2] shows that G is nilpotent. Then $G=P\times Q$ and it follows that $Z(G)=Z(P)\times Q$ , which contradicts (2.2).

This completes the proof.