On the Action of a Subgroup of the Modular Group on Imaginary Quadratic Number Fields

Abstract

Consider the modular group $\mathrm{PSL}(2,\mathbb{Z})=⟨x,y|x^2=y^3=1⟩$ generated by the transformations $x:z\mapsto -1/z$ and $y:z\mapsto (z-1)/z$. Let H be the proper subgroup $⟨y,v|y^3=v^3=1⟩$ of $\mathrm{PSL}(2,\mathbb{Z})$, where $v=xyx$. For a positive square-free integer n, this article studies the action of H on the subset $\{{\displaystyle \frac{a+\sqrt{-n}}{c}}|a,b={\displaystyle \frac{a^2+n}{c}},c\in \mathbb{Z},c\ne 0\}$ of the imaginary quadratic number field $\mathbb{Q}\left(\sqrt{-n}\right)$ where, in particular, the accurate estimate of the number of orbits arising from this action is given, correcting the estimate given in some of the relevant literature.

1. Introduction

The study of the action of the modular group $G=\mathrm{PSL}(2,\mathbb{Z})$ and its subgroups on various commutative algebraic structures has been common since the works of Graham Higman and his collaborators in the 1980s ([1,2,3]). Various tracks of applications of the action of the modular group have been evolving recently. For instance, Alolaiyan et al. [4] presented a graph-theoretic application of the action of the modular group on certain structures by highlighting a connection of such an action to the study of homomorphic copies of closed graphs. On another front, some S-box cryptographic applications of the coset diagrams of the action of the modular group on various discrete structures were introduced in refs. [5,6,7]. This indeed suggests a promising future for the applications of this seemingly purely mathematical topic.

Stothers [8] showed that the modular group G has the finite presentation $\langle x,y|x^2=y^3=1\rangle$, where x and y are, respectively, the linear fractional transformations $z\mapsto -1/z$ and $z\mapsto (z-1)/z$. For a square-free integer m, the action of G and its subgroups on certain subsets of the quadratic number field $\mathbb{Q}\left(\sqrt{m}\right)$ has been extensively studied (see, for instance, refs. [3,9,10,11,12,13,14,15,16]).

For a positive square-free integer n, consider the following subset of the imaginary quadratic number field $\mathbb{Q}\left(\sqrt{-n}\right)$:

$${\mathbb{Q}}^{*}\left(\sqrt{-n}\right):=\left\{{\displaystyle \frac{a+\sqrt{-n}}{c}}\right|a,b={\displaystyle \frac{a^2+n}{c}},c\in \mathbb{Z},c\ne 0\}.$$

Deajim and Aslam, in ref. [10], studied the action of G on ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, in which the main result is that of counting the orbits resulting from this action. Ashiq and Mushtaq (in ref. [17]) studied the proper subgroup H of G generated by the transformations y and $v=xyx:z\mapsto -1/(z+1)$, where they showed that $y^3=v^3=1$ are defining relations for H, so $H=\langle y,v|y^3=v^3=1\rangle$. Then, in ref. [18], Ashiq and Mushtaq studied the action of H on ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, where they suggested a formula that counts the orbits resulting from this action. However, the author of this article has found some technical/mathematical mistakes in [18], which has resulted in calls for revision of the findings of [18], with the primary aim of providing a more accurate estimate of the number of orbits.

In Section 3, the errors of [18] are listed along with the reasoning for such a judgment. Section 4 aims at correcting the main result of [18], namely, ref. [18] (Theorem 9), by giving, in Theorem 1, the accurate formula for the number of orbits arising from the action of H on ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$. Moreover, some lemmas that were used to prove Theorem 1 also help in correcting some of the mistakes of [18] mentioned in Section 3. Finally, a conclusion containing a summary of the main findings of the article is given in Section 5.

2. Terminology

Upon a wise suggestion of a reviewer, this section was added for the sake of completeness and more readability.

2.1. Quadratic Number Fields

The information in this subsection can be found in many abstract algebra or algebraic number theory books, such as [19,20].

If K and L are fields such that $K\subseteq L$, we say that L is a field extension of K and express this as $L/K$. In this case, L can also be looked at as a vector space over K whose dimension is called the degree of the extension $L/K$. A field extension $L/K$ is said to be finite if its degree is finite. In particular, if the degree of the extension $L/K$ is 2, we say that the field extension is quadratic. If $p\left(x\right)\in K\left[x\right]$ is an irreducible polynomial of degree n, then there is a field extension of K containing a root $\alpha$ of $p\left(x\right)$. In particular, the field

$$L=K\left(\alpha \right)=\{a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1}|a_i\in K\}$$

is the field extension over K of degree n generated by $\alpha$ with a basis $\{1,\alpha ,\dots ,{\alpha }^{n-1}\}$.

In the case $K=\mathbb{Q}$, any finite field extension of $\mathbb{Q}$ is called a number field. If an integer n is not a perfect square, the polynomial $x^2-n\in \mathbb{Q}\left[x\right]$ is irreducible, so $\mathbb{Q}\left(\sqrt{n}\right)$ is called a quadratic number field (as it is a number field that is a quadratic extension of $\mathbb{Q}$) with a basis $\{1,\sqrt{n}\}$. So, an element of $\mathbb{Q}\left(\sqrt{n}\right)$ can be uniquely written in the form $p+q\sqrt{n}$ for some $p,q\in \mathbb{Q}$. For another integer m that is not a perfect square either, we have $\mathbb{Q}\left(\sqrt{n}\right)=\mathbb{Q}\left(\sqrt{m}\right)$ if and only if $n=k^{2}m$ for an integer k. So, we can always assume that n is square-free. For a square-free integer $n>0$ (resp. $n<0$), $\mathbb{Q}\left(\sqrt{n}\right)$ is said to be a real (resp. imaginary) quadratic number field.

2.2. The Modular Group and Its Action on ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$

A suitable reference for the modular group and its relation to linear fractional transformations is ref. [21]. A reader who is interested in finding, in one place, some more detailed information about the modular group, its properties, and its applications may consult [22] and the references therein.

It is known from group theory that

$$\mathrm{SL}(2,\mathbb{Z})=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|a,b,c,d\in \mathbb{Z},ad-bc=1\right\}$$

is a nonabelian multiplicative group whose center is $\{\pm I\}$. Moding $\mathrm{SL}(2,\mathbb{Z})$ out by its center $\{\pm I\}$ yields the modular group $G=\mathrm{PSL}(2,\mathbb{Z})$. So, an element of G is a coset $A\{\pm I\}=\{\pm A\}$ for $A\in \mathrm{SL}(2,\mathbb{Z})$. It should not be confusing though that a representative of a coset is more commonly used in the literature instead of the coset itself (keeping in mind that the two matrices A and $-A$ are identified).

The set of complex-valued linear fractional transformations $f_{(a,b,c,d)}:z\mapsto {\displaystyle \frac{az+b}{cz+d}}$, with $a,b,c,d\in \mathbb{Z}$ and $ad-bc=1$, forms a nonabelian group under the composition of maps. Since $f_{(a,b,c,d)}=f_{(-a,-b,-c,-d)}$, a transformation $f_{(a,b,c,d)}$ can be identified with $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in G$, as an element of G is identified with its negative (see the paragraph above). This identification map can be shown to be a group isomorphism. So, the terminology of “modular group” usually refers to either of these two groups, and the elements of this group can either be written as matrices or as linear fractional transformations. As was mentioned in Section 1, the modular group has the finite presentation $G=\langle x,y|x^2=y^3=1\rangle$, where x and y are, respectively, the linear fractional transformations $z\mapsto -1/z$ and $z\mapsto (z-1)/z$.

For a square-free positive integer n and $\alpha =(a+\sqrt{-n})/c\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, we sometimes use the notation $a_{\alpha }$, $b_{\alpha }$, and $c_{\alpha }$ for a, b, and c, respectively. Throughout the article and without mention, we shall make use of the following

Table 1. The action of $x,y,y^2,v,$ and $v^2$.

t	${\mathit{a}}_{\mathit{t}\left(\mathit{\alpha }\right)}$	${\mathit{b}}_{\mathit{t}\left(\mathit{\alpha }\right)}$	${\mathit{c}}_{\mathit{t}\left(\mathit{\alpha }\right)}$
x	$-a$	c	b
y	$-a+b$	$-2a+b+c$	b
$y^2$	$-a+c$	c	$-2a+b+c$
v	$-a-c$	c	$2a+b+c$
$v^2$	$-a-b$	$2a+b+c$	b

, which shows the effect of the action of $t\in \{x,y,y^2,v,v^2\}$ on $\alpha$ and can be verified through straightforward computations.

For $y:z\mapsto (z-1)/z$ and $v:z\mapsto -1/(z+1)$, let $H=\langle y,v|y^3=v^3=1\rangle$, which is a subgroup of G. As G acts on the quadratic number field $\mathbb{Q}\left(\sqrt{-n}\right)$, the subgroup H acts on $\mathbb{Q}\left(\sqrt{-n}\right)$ as well. So, to show that H acts on the subset ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, it suffices to show that ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ is invariant under this action (in fact, it is enough to show that $y\left(\alpha \right),v\left(\alpha \right)\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ for any $\alpha \in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ since $\{y,v\}$ is a complete set of generators of H). For ${\displaystyle \alpha =(a+\sqrt{-n})/c\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)}$, we see from Table 1 that $\{a_{y\left(\alpha \right)},b_{y\left(\alpha \right)},c_{y\left(\alpha \right)},a_{v\left(\alpha \right)},b_{v\left(\alpha \right)},c_{v\left(\alpha \right)}\}\subseteq \mathbb{Z}$, $c_{y\left(\alpha \right)}\ne 0$, and $c_{v\left(\alpha \right)}\ne 0$. Notice that the statements $c_{y\left(\alpha \right)}\ne 0$ and $c_{v\left(\alpha \right)}\ne 0$ follow, respectively, from the facts that ${\displaystyle b_{y\left(\alpha \right)}{c}_{y\left(\alpha \right)}=a_{y\left(\alpha \right)}^2+n>0}$ and $b_{v\left(\alpha \right)}{c}_{v\left(\alpha \right)}=a_{v\left(\alpha \right)}^2+n>0$.

3. A Summary of Errors in Ashiq [18]

Throughout this article, n denotes a positive square-free integer.

An element $\alpha =(a+\sqrt{-n})/c\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ is said to be totally positive if $ac>0$, totally negative if $ac<0$, and of norm zero if $\left|\right|\alpha \left|\right|:=\left|a\right|=0$. For $b=(a^2+n)/c$, it is obvious that $bc>0$, so b and c are always of the same sign.

For every $\alpha \in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, denote the H-orbit $\left\{h\right(\alpha \left)\right|h\in H\}$ by ${\alpha }^H$ and the G-orbit $\left\{g\right(\alpha \left)\right|g\in G\}$ by ${\alpha }^G$. It is clear that ${\alpha }^H\subseteq {\alpha }^G$. Denote the set of H-orbits (resp. the set of G-orbits) in ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ by ${\mathcal{O}}_{-n}^H$ (resp. ${\mathcal{O}}_{-n}^G$).

As usual, we denote by $\lfloor .\rfloor$ the floor function; by $d\left(k\right)$ the number of positive divisors of k; and by $d_{\le i}\left(k\right)$, for $i,k\in \mathbb{N}$ with $i\le k$, the number of positive divisors of k that do not exceed i. For example, $d\left(15\right)=4$ and $d_{\le 5}\left(15\right)=3$. Recall that d is a multiplicative function and, for a prime p, $d\left(p^t\right)=t+1$ for $t\in \mathbb{N}$.

We summarize below the main errors in ref. [18] along with some comments.

Claim 1

([18] line 1 of the proof of Theorem 6). Let α be a totally positive imaginary quadratic number. Then, by Theorem 3 (i), $y\left(\alpha \right)$ or $y^2\left(\alpha \right)$ is a totally negative imaginary quadratic number.

Comments: Claim 1 is not necessarily true. We show in Lemma 1 that if $\alpha$ is totally positive, then one of $y\left(\alpha \right)$ or $y^2\left(\alpha \right)$ is totally positive and the other is either totally positive, totally negative, or of norm zero. For instance, consider the totally positive element $\alpha =(2+\sqrt{-5})/3$. Then, both $y\left(\alpha \right)=(1+\sqrt{-5})/3$ and $y^2\left(\alpha \right)=(1+\sqrt{-5})/2$ are totally positive as well. Another example is given by $\beta =(1+\sqrt{-1})/2$; then, $y\left(\beta \right)=\sqrt{-1}$ is of norm zero and $y^2\left(\beta \right)=1+\sqrt{-1}$ is totally positive. On the other hand, it should also be noted that [18] (Theorem 3 (i)) states that $y\left(\alpha \right)$ and $y^2\left(\alpha \right)$ are totally positive when $\alpha$ is totally negative, which is true, but Claim 1 above misquotes it. Note that Claim 1 being false renders the proof of [18] (Theorem 6) incorrect.

Claim 2

([18] Theorem 8). (v) If $\alpha =(1+\sqrt{-n})/c_1$ ($n\ne 3$), where $1+n=c_1{c}_2$, $c_1\ne 1$ or $n+1$, then α is the only element of norm 1 in ${\alpha }^H$.

Comments: It is noted that in the proof of Theorem 8(v) mentioned above, something contradicting the statement itself was mentioned. It was stated explicitly that if $\alpha =(1+\sqrt{-n})/2$, then $y^2\left(\alpha \right)=(1+\sqrt{-n})/{\displaystyle \frac{(n+1)}{2}}$ lies in the same orbit. But then, as $y^2\left(\alpha \right)$ is of norm 1 and is different from $\alpha$ (as $n\ne 3$), this shows that the statement of (v) is false. For instance, consider $\alpha =(1+\sqrt{-5})/2$. Then, $\alpha$ is of norm 1 and satisfies the assumptions of Theorem 8(v) mentioned above. However, $y^2\left(\alpha \right)=(1+\sqrt{-5})/3$ is also of norm 1 belonging to ${\alpha }^H$ and is different from $\alpha$.

As for the other parts of [18] (Theorem 8), it is also noted that the arguments in the proof of parts (iii) and (iv) of [18] (Theorem 8) are not sufficient, although the statements are correct. In fact, parts (iii) and (iv) specify certain cases where an H-orbit contains only one element of norm zero. Part of the proof of Theorem 1 shows that this is always the case for any n and for any H-orbit containing an element of norm zero.

Claim 3

([18] Theorem 9). If $n\ne 3$, then the total number of orbits of ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ under the action of H are:

(i) 

$2\left[d\right(n)+2d(n+1)-6]$ if n is odd, and

(ii) 

$2\left[d\right(n)+2d(n+1)-4]$ if n is even.

Comments: This is the main result of [18], which is inaccurate. Its proof relies on ([18] Theorem 8), which has some gaps (see Claim 2). The correct estimate of the number of orbits is given in Theorem 1 below. We briefly give here some counter-examples to [18] (Theorem 9). For odd integers, consider $n=1$ and $n=21$, for instance. For $n=1$, the above estimate gives $|{\mathcal{O}}_{-1}^H|=-2$, which is absurd. Our estimate (see Theorem 1) gives $|{\mathcal{O}}_{-1}^H|=2$. For $n=21$, the above estimate gives $|{\mathcal{O}}_{-21}^H|=12$, whereas our estimate gives $|{\mathcal{O}}_{-21}^H|=16$. In fact, according to Theorem 1, the number of orbits in this case has to be congruent to zero modulo 8. For even integers, consider $n=26$, for instance. The above estimate gives $|{\mathcal{O}}_{-26}^H|=16$, whereas our estimate gives $|{\mathcal{O}}_{-26}^H|=24$. The details of the computations of these counter-examples are given just after the proof of Theorem 1.

4. Lemmas and Main Result

The main aim of the article is to prove Theorem 1, which gives the precise estimate for the number of H-orbits $|{\mathcal{O}}_{-n}^H|$, correcting [18] (Theorem 9). Besides serving to prove Theorem 1, some of the lemmas below also contribute by correcting some of the errors of [18] mentioned in Section 3.

Lemma 1.

Let $\alpha =(a+\sqrt{-n})/c\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$.

(1) 

If α is totally positive, then one of $y\left(\alpha \right)$ or $y^2\left(\alpha \right)$ is totally positive and the other is either totally positive, totally negative, or of norm zero, whereas both $v\left(\alpha \right)$ and $v^2\left(\alpha \right)$ are totally negative.

(2) 

If α is totally negative, then one of $v\left(\alpha \right)$ or $v^2\left(\alpha \right)$ is totally negative and the other is either totally negative, totally positive, or of norm zero, whereas both $y\left(\alpha \right)$ and $y^2\left(\alpha \right)$ are totally positive.

(3) 

If α is of norm zero, then both $y\left(\alpha \right)$ and $y^2\left(\alpha \right)$ are totally positive and both $v\left(\alpha \right)$ and $v^2\left(\alpha \right)$ are totally negative.

Proof of Lemma 1.

Once and for all, note here and elsewhere that Table 1 is used without mention.

(1)

Assume that $a,b,c>0$ (the case $a,b,c<0$ is proved similarly). We show first that at least one of $y\left(\alpha \right)$ or $y^2\left(\alpha \right)$ is totally positive. However, suppose that neither of them is totally positive. If $y\left(\alpha \right)$ is totally negative, then (as $c_{y\left(\alpha \right)}=b>0$) $a_{y\left(\alpha \right)}=-a+b<0$. As $b_{y^2\left(\alpha \right)}=c>0$ and as $b_{y^2\left(\alpha \right)}$ and $c_{y^2\left(\alpha \right)}$ have the same sign, we have $-a+b-a+c=c_{y^2\left(\alpha \right)}>0$, implying that $a_{y^2\left(\alpha \right)}=-a+c>0$ (as $-a+b<0$). So, $y^2\left(\alpha \right)$ is totally positive, a contradiction. Similarly, if $y^2\left(\alpha \right)$ is totally negative, then $y\left(\alpha \right)$ is totally positive, a contradiction. On the other hand, if $y\left(\alpha \right)$ is of norm zero, then $a_{y\left(\alpha \right)}=-a+b=0$. So, $c_{y^2\left(\alpha \right)}=-a+b-a+c=-a+c=a_{y^2\left(\alpha \right)}$, implying that (as $c_{y^2\left(\alpha \right)}\ne 0$) $a_{y^2\left(\alpha \right)}{c}_{y^2\left(\alpha \right)}>0$ and thus $y^2\left(\alpha \right)$ is totally positive, a contradiction. Similarly, if $y^2\left(\alpha \right)$ is of norm zero, then $y\left(\alpha \right)$ is totally positive, a contradiction. This argument leads to the assertion that at least one of $y\left(\alpha \right)$ or $y^2\left(\alpha \right)$ is totally positive. Suppose that it is $y\left(\alpha \right)$ that is totally positive (if it were $y^2\left(\alpha \right)$, then the proof would be similar). It then follows that $y^2\left(\alpha \right)$ is either totally positive, totally negative, or of norm zero depending, respectively, on whether $a_{y^2\left(\alpha \right)}>0$, $a_{y^2\left(\alpha \right)}<0$, or $a_{y^2\left(\alpha \right)}=0$. As for $v\left(\alpha \right)$ and $v^2\left(\alpha \right)$, note that having $a_{v\left(\alpha \right)}=-a-c<0$ and $c_{v\left(\alpha \right)}=2a+b+c>0$ implies that $v\left(\alpha \right)$ is totally negative. Also, having $a_{v^2\left(\alpha \right)}=-a-b<0$ and $c_{v^2\left(\alpha \right)}=b>0$ implies that $v^2\left(\alpha \right)$ is totally negative.

(2)

Similar to the proof of part (1).

(3)

Suppose that $a=0$. Then, $a_{y\left(\alpha \right)}{c}_{y\left(\alpha \right)}=b^2>0$, $a_{y^2\left(\alpha \right)}{c}_{y^2\left(\alpha \right)}=bc+c^2>0$ (b and c are of the same sign), so $y\left(\alpha \right)$ and $y^2\left(\alpha \right)$ are totally positive. On the other hand, $a_{v\left(\alpha \right)}{c}_{v\left(\alpha \right)}=-bc-c^2<0$, $a_{v^2\left(\alpha \right)}{c}_{v^2\left(\alpha \right)}=-b^2<0$, so $v\left(\alpha \right)$ and $v^2\left(\alpha \right)$ are totally negative.

 □

We have the following obvious observations (see [10] (Lemmas 3.2 and 3.3)).

Lemma 2

([10] Lemmas 3.2 and 3.3). For $\alpha =(a+\sqrt{-n})/c\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, we have

(i) 

α is totally positive if and only if $x\left(\alpha \right)$ is totally negative.

(ii) 

α has norm zero if and only if $x\left(\alpha \right)$ has norm zero.

For $\alpha \in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, the sets $\stackrel{y}{\widehat{\alpha }}=\{\alpha ,y\left(\alpha \right),y^2\left(\alpha \right)\}$ and $\stackrel{v}{\widehat{\alpha }}=\{\alpha ,v\left(\alpha \right),v^2\left(\alpha \right)\}$ are called, respectively, the y-cycle and the v-cycle of $\alpha$ (or just cycles when $\alpha$, y, and v are clear in the context). We call a cycle totally positive (resp. totally negative) if all its elements are totally positive (resp. if all its elements are totally negative).

Lemma 3.

Under the action of H on ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, the following holds for $\alpha =(a+\sqrt{-n})/c\in {\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$:

(i) 

[10] (Lemma 3.3) The cycle $\stackrel{y}{\widehat{\alpha }}$ is totally positive if and only if either ($a>0$, $a<b$, $a<c$) or ($a<0$, $a>b$, $a>c$).

(ii) 

The cycle $\stackrel{v}{\widehat{\alpha }}$ is totally negative if and only if either ($a>0$, $-a>b$, $-a>c$) or ($a<0$, $-a<b$, $-a<c$).

(iii) 

The cycle $\stackrel{y}{\widehat{\alpha }}$ is totally positive if and only if the cycle $\stackrel{v}{\widehat{x\left(\alpha \right)}}$ is totally negative.

(iv) 

If the cycle $\stackrel{y}{\widehat{\alpha }}$ is totally positive, then ${\alpha }^H\ne x{\left(\alpha \right)}^H$ (i.e., α and $x\left(\alpha \right)$ belong to distinct H-orbits).

(v) 

If $n\ne 1$ and α has norm zero, then ${\alpha }^H\ne x{\left(\alpha \right)}^H$.

Proof of Lemma 3.

(i) This is [10] (Lemma 3.3).

(ii)

Suppose that $\stackrel{v}{\widehat{\alpha }}$ is totally negative. As $\alpha$ is totally negative, either ($a>0$ and $b,c<0$) or ($a<0$ and $b,c>0$). Assume that $a>0$ and $b,c<0$. Since $b_{v\left(\alpha \right)}=c<0$, we obtain $c_{v\left(\alpha \right)}<0$. As further $v\left(\alpha \right)$ is totally negative, we must then have $-a-c=a_{v\left(\alpha \right)}>0$ so $-a>c$. Also, since $c_{v^2\left(\alpha \right)}=b<0$ and $v^2\left(\alpha \right)$ is totally negative, $a_{v^2\left(\alpha \right)}=-a-b>0$ so $-a>b$. A similar argument yields that if $a<0$ and $b,c>0$ then $-a<b$ and $-a<c$.

Conversely, suppose that $a>0$, $-a>b$, and $-a>c$ (the case when $a<0$, $-a<b$, and $-a<c$ is handled similarly). $a>0$ and $-a>c$, $c<0$, so $\alpha$ is totally negative. As $a_{v\left(\alpha \right)}=-a-c>0$ and $b_{v\left(\alpha \right)}=c<0$ and $b_{v\left(\alpha \right)},c_{v\left(\alpha \right)}$ are of the same sign, $v\left(\alpha \right)$ is totally negative too. As $a_{v^2\left(\alpha \right)}=-a-b>0$ and $c_{v^2\left(\alpha \right)}=b<0$, $v^2\left(\alpha \right)$ is totally negative as well. Hence, the cycle $\stackrel{v}{\widehat{\alpha }}$ is totally negative.

(iii)

Suppose that $\stackrel{y}{\widehat{\alpha }}$ is totally positive. By (i), if $a>0$, $a<b$, and $a<c$, then it follows from (ii) that $\stackrel{v}{\widehat{x\left(\alpha \right)}}$ is totally negative since $a_{x\left(\alpha \right)}=-a<0$, $-a_{x\left(\alpha \right)}=a<c=b_{x\left(\alpha \right)}$, and $-a_{x\left(\alpha \right)}=a<b=c_{x\left(\alpha \right)}$. The case when $a<0$, $a>b$, and $a>c$ is similar.

Conversely, suppose that $\stackrel{v}{\widehat{x\left(\alpha \right)}}$ is totally negative. Then, by (ii), if $a_{x\left(\alpha \right)}=-a>0$, $-a_{x\left(\alpha \right)}=a>b_{x\left(\alpha \right)}=c$, and $-a_{x\left(\alpha \right)}=a>c_{x\left(\alpha \right)}=b$, then it follows from (i) that $\stackrel{y}{\widehat{\alpha }}$ is totally positive. The case when $a_{x\left(\alpha \right)}<0$, $-a_{x\left(\alpha \right)}<b_{x\left(\alpha \right)}$, and $-a_{x\left(\alpha \right)}<c_{x\left(\alpha \right)}$ is similar.

(iv)

Suppose that $\stackrel{y}{\widehat{\alpha }}$ is totally positive. By (iii), $\stackrel{v}{\widehat{x\left(\alpha \right)}}$ is totally negative. To show that ${\alpha }^H\ne x{\left(\alpha \right)}^H$, we need to show that $h\left(\alpha \right)\ne x\left(\alpha \right)$ for any $h\in H$. It can be seen that an arbitrary non-identity element of H takes one and only one of the following forms:

$$\begin{array}{cc}\hfill h_1& =y^{ϵ},\mathrm{for}ϵ\in \{1,2\}\hfill \\ \hfill h_2& =v^{\delta },\mathrm{for}\delta \in \{1,2\}\hfill \\ \hfill h_3& =y^{{ϵ}_k}{v}^{{\delta }_k}\dots y^{{ϵ}_2}{v}^{{\delta }_2}{y}^{{ϵ}_1}{v}^{{\delta }_1},\mathrm{for}k\ge 1,{ϵ}_i,{\delta }_i\in \{1,2\},1\le i\le k\hfill \\ \hfill h_4& =v^{{\delta }_k}{y}^{{ϵ}_{k-1}}{v}^{{\delta }_{k-1}}\dots y^{{ϵ}_2}{v}^{{\delta }_2}{y}^{{ϵ}_1}{v}^{{\delta }_1},\mathrm{for}k\ge 2,{ϵ}_i,{\delta }_j\in \{1,2\},1\le i\le k-1,1\le j\le k\hfill \\ \hfill h_5& =v^{{\delta }_k}{y}^{{ϵ}_k}\dots v^{{\delta }_2}{y}^{{ϵ}_2}{v}^{{\delta }_1}{y}^{{ϵ}_1},\mathrm{for}k\ge 1,{ϵ}_i,{\delta }_i\in \{1,2\},1\le i\le k\hfill \\ \hfill h_6& =y^{{ϵ}_k}{v}^{{\delta }_{k-1}}{y}^{{ϵ}_{k-1}}\dots v^{{\delta }_2}{y}^{{ϵ}_2}{v}^{{\delta }_1}{y}^{{ϵ}_1},\mathrm{for}k\ge 2,{ϵ}_i,{\delta }_j\in \{1,2\},1\le i\le k,1\le j\le k-1\hfill \end{array}$$

The h₁ case:

Since there is no $z\in \mathbb{C}$ such that $y^{ϵ}\left(z\right)=x\left(z\right)$, we obtain $h_1\left(\alpha \right)=y^{ϵ}\left(\alpha \right)\ne x\left(\alpha \right)$. Alternatively, $y^{ϵ}\left(\alpha \right)\ne x\left(\alpha \right)$ since $y^{ϵ}\left(\alpha \right)$ is totally positive and $x\left(\alpha \right)$ is totally negative (by Lemma 2).

The h₂ case:

Since there is no $z\in \mathbb{C}$ such that $v^{\delta }\left(z\right)=x\left(z\right)$, we obtain $h_2\left(\alpha \right)=v^{\delta }\left(\alpha \right)\ne x\left(\alpha \right)$. Alternatively, $v^{\delta }\left(\alpha \right)\ne x\left(\alpha \right)$ since if equality holds then $x{y}^{\delta }x\left(\alpha \right)=x\left(\alpha \right)$ would imply that $x\left(\alpha \right)=y^{-\delta }\left(\alpha \right)$, a contradiction (see the $h_1$ case above).

The h₃ case:

As $\alpha$ is totally positive, it follows from Lemma 1 (1) that $v^{{\delta }_1}\left(\alpha \right)$ is totally negative so, by Lemma 1 (2), $y^{{ϵ}_1}{v}^{{\delta }_1}\left(\alpha \right)$ is totally positive. Repeating this argument for $i=1,\dots ,k$, we obtain that $h_3\left(\alpha \right)$ is totally positive. Since $x\left(\alpha \right)$ is totally negative, $h_3\left(\alpha \right)\ne x\left(\alpha \right)$.

The h₄ case:

If we let $h_4^{*}=v^{-{\delta }_k}{h}_4=y^{{ϵ}_{k-1}}{v}^{{\delta }_{k-1}}\dots y^{{ϵ}_2}{v}^{{\delta }_2}{y}^{{ϵ}_1}{v}^{{\delta }_1}$, then by an argument similar to that in the $h_3$ case, $h_4^{*}\left(\alpha \right)$ is totally positive. If $h_4\left(\alpha \right)=x\left(\alpha \right)$, then $h_4^{*}\left(\alpha \right)=v^{-{\delta }_k}x\left(\alpha \right)\in \stackrel{v}{\widehat{x\left(\alpha \right)}}$, so $h_4^{*}\left(\alpha \right)$ is totally negative, a contradiction.

The h₅ case:

If we let $h_5^{*}=v^{-{\delta }_k}{h}_5=y^{{ϵ}_k}\dots v^{{\delta }_2}{y}^{{ϵ}_2}{v}^{{\delta }_1}{y}^{{ϵ}_1}$, then as $y^{{ϵ}_1}\left(\alpha \right)$ is totally positive, it by an argument similar to that in the $h_3$ case that $h_5^{*}\left(\alpha \right)$ is totally positive. If $h_5\left(\alpha \right)=x\left(\alpha \right)$, then $h_5^{*}\left(\alpha \right)=v^{-{\delta }_k}x\left(\alpha \right)$, which is impossible since $h_5^{*}\left(\alpha \right)$ is totally positive, while $v^{-{\delta }_k}x\left(\alpha \right)$ is totally negative (see the $h_4$ case).

The h₆ case:

If we let $h_6^{*}=y^{-{ϵ}_k}{h}_6=v^{{\delta }_{k-1}}{y}^{{ϵ}_{k-1}}\dots v^{{\delta }_2}{y}^{{ϵ}_2}{v}^{{\delta }_1}{y}^{{ϵ}_1}$, then as $y^{{ϵ}_1}\left(\alpha \right)$ is totally positive, it is by an argument similar to that in the $h_3$ case that $h_6^{*}\left(\alpha \right)$ is totally negative. If $h_6\left(\alpha \right)=x\left(\alpha \right)$, then $h_6^{*}\left(\alpha \right)=y^{-{ϵ}_k}x\left(\alpha \right)$, which is impossible since $h_6^{*}\left(\alpha \right)$ is totally negative, while $y^{-{ϵ}_k}x\left(\alpha \right)$ is totally positive by Lemma 1 (2).

Now, by having just shown the impossibility of all cases above, we conclude that there is no $h\in H$ such that $h\left(\alpha \right)=x\left(\alpha \right)$, and therefore $\alpha$ and $x\left(\alpha \right)$ belong to distinct H-orbits as claimed.

(v)

Remark first that we excluded the case $n=1$ since the only elements of $\mathbb{C}$ fixed by x are $\pm \sqrt{-1}\in {\mathbb{Q}}^{*}\left(\sqrt{-1}\right)$ (i.e., $x(\pm \sqrt{-1})=\pm \sqrt{-1}$); see [10] (Lemma 3.4). Let $n\ne 1$, and suppose that $\alpha$ has norm zero. Then, $x\left(\alpha \right)$ has norm zero too. We need to show that $h\left(\alpha \right)\ne x\left(\alpha \right)$ for any $h\in H$. By Lemma 1 (3), both $y\left(\alpha \right)$ and $y^2\left(\alpha \right)$ are totally positive, and both $v\left(\alpha \right)$ and $v^2\left(\alpha \right)$ are totally negative. Now, for $h_1,\dots ,h_6$ as in (iv), and since none of $y\left(\alpha \right),y^2\left(\alpha \right),v\left(\alpha \right),v^2\left(\alpha \right)$ has norm zero, it is easily seen that $h_i\left(\alpha \right)$ is either totally negative or totally positive for every $i=1,\dots ,6$ (see the argument in (iv)). Thus, $h_i\left(\alpha \right)\ne x\left(\alpha \right)$ for any $i=1,\dots ,6$. This settles the claim.

 □

Example 1.

For $n=7$ and $\alpha =(-1+\sqrt{-7})/(-4)$, it can be checked that $y\left(\alpha \right)=(-1+\sqrt{-7})/(-2)$, $y^2\left(\alpha \right)=(-3+\sqrt{-7})/(-4)$, so α, $y\left(\alpha \right)$, and $y^2\left(\alpha \right)$ are all totally positive. On the other hand, $x\left(\alpha \right)=(1+\sqrt{-7})/(-2)$, $vx\left(\alpha \right)=(1+\sqrt{-7})/(-4)$, and $v^{2}x\left(\alpha \right)=(3+\sqrt{-7})/(-4)$, which are all totally negative.

Lemma 4

([10] Corollary 3.1). Under the action of G on ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, we have the following:

(i) 

Every G-orbit in ${\mathbb{Q}}^{*}\left(\sqrt{-1}\right)$ contains a unique element of norm zero.

(ii) 

Every G-orbit in ${\mathbb{Q}}^{*}\left(\sqrt{-2}\right)$ contains a unique pair of distinct elements of norm zero.

(iii) 

Every G-orbit in ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$, for $n\ge 3$, contains either a unique pair of distinct elements of norm zero or a unique totally positive cycle, but not both.

The aim of our main result below is to compute the cardinality $|{\mathcal{O}}_{-n}^H|$ correcting the estimate given in [18] (Theorem 9) (see Claim 3 in Section 3). In particular, we utilize the above results to simply show that

$$|{\mathcal{O}}_{-1}^H|=|{\mathcal{O}}_{-1}^G|, \mathrm{and} |{\mathcal{O}}_{-n}^H|=2|{\mathcal{O}}_{-n}^G| \mathrm{for} n\ge 2,$$

and then make use of the precise estimate of $|{\mathcal{O}}_{-n}^G|$ given in [10] (Theorem 2.1) for all $n\ge 1$.

Theorem 1.

Under the action of H, the number of orbits in ${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)$ is 

$$\begin{array}{ll}|{\mathcal{O}}_{-n}^H|& =\left\{\begin{array}{cc}|{\mathcal{O}}_{-1}^G|\hfill & ,if n=1\hfill \\ \\ 2|{\mathcal{O}}_{-n}^G|\hfill & ,if n\ge 2\hfill \end{array}\right.\\ & =\left\{\begin{array}{cc}2\hfill & ,if n=1\hfill \\ 4\hfill & ,if n=2\hfill \\ 8\hfill & ,if n=3\hfill \\ 2d\left(n\right)+{\displaystyle \frac{4}{3}}\sum _{i=1}^{\lfloor (n-1)/2\rfloor }[d(i^2+n)-2d_{\le i}(i^2+n)]\hfill & ,if n>3.\hfill \end{array}\right.\end{array}$$

Moreover, $|{\mathcal{O}}_{-n}^H|\equiv 0(mod8)$ for $n\ge 3$.

Proof of Theorem 1.

We begin with the case $n\ge 2$, where we show that $|{\mathcal{O}}_{-n}^H|=2|{\mathcal{O}}_{-n}^G|$. We show this equality by showing that every G-orbit splits into two distinct H-orbits. Let ${\alpha }^G\in {\mathcal{O}}_{-n}^G$. We know from Lemma 4 (ii, iii) that ${\alpha }^G$ either contains a unique pair of distinct elements of norm zero or a unique totally positive cycle, but not both. We proceed by dealing with these two possibilities.

Firstly, suppose that ${\alpha }^G$ contains a unique pair of distinct elements $\beta$ and $x\left(\beta \right)$ of norm zero. Since $\beta \ne x\left(\beta \right)$ in this case, we obtain from Lemma 3 (v) that it cannot be the case that both elements belong to ${\alpha }^H$. We show that ${\alpha }^G$ splits into two distinct H-orbits (${\alpha }^H$ and ${\beta }^H$) or (${\alpha }^H$ and $x{\left(\beta \right)}^H$) by showing that ${\alpha }^H$ must contain one and only one of these two norm-zero elements, while the other belongs to ${\alpha }^G-{\alpha }^H$. If $\beta \in {\alpha }^H$, then (by Lemma 3 (v)) ${\alpha }^G$ is the disjoint union of ${\alpha }^H$ and $x{\left(\beta \right)}^H$ and we are done. Similarly, if $x\left(\beta \right)\in {\alpha }^H$, then ${\alpha }^G$ is the disjoint union of ${\alpha }^H$ and ${\beta }^H$. Now, it remains to show that one of $\beta$ or $x\left(\beta \right)$ must belong to ${\alpha }^H$. Assume that $\beta \notin {\alpha }^H$; we show that $x\left(\beta \right)\in {\alpha }^H$. Since $\alpha \in {\alpha }^G-{\alpha }^H$, there is some $g\in G-H$ such that $\beta =g\left(\alpha \right)$. In order to show that $x\left(\beta \right)\in {\alpha }^H$, we argue that $xg\in H$, which then would imply that $x\left(\beta \right)=xg\left(\alpha \right)\in {\alpha }^H$ as desired. Keeping in mind the forms $h_0=1$ and $h_1,\dots ,h_6$ that elements of H can take (see the proof of Lemma 3 (iv)), we see that g, as an element of G but not an element of H, can take one of the forms $x{h}_i$, $h_{i}x$, $y^{\gamma }x{h}_i$, or $h_{i}x{y}^{\gamma }$ for an appropriate $h_i$, $i=0,\dots ,6$, $\gamma \in \{1,2\}$. We present the following argument, which exhausts all possible cases.

If $g=x{h}_i$, then $xg=h_i\in H$ for $i=0,\dots ,6$.

If $g=y^{\gamma }x{h}_i$, then $xg=x{y}^{\gamma }x{h}_i=v^{\gamma }{h}_i\in H$ for $i=0,\dots ,6$.

If $g=h_{0}x=x$, then $xg=1\in H$.

If $g=h_{0}x{y}^{\gamma }$, then $xg=y^{\gamma }\in H$.

If $g=h_{1}x=y^{ϵ}x$, then $xg=x{y}^{ϵ}x=v^{ϵ}\in H$.

If $g=h_{1}x{y}^{\gamma }=y^{ϵ}x{y}^{\gamma }$, then $xg=x{y}^{ϵ}x{y}^{\gamma }=v^{ϵ}{y}^{\gamma }\in H$.

If $g=h_{2}x=v^{\delta }x=x{y}^{\delta }$, then $xg=y^{\delta }\in H$.

If $g=h_{2}x{y}^{\gamma }=v^{\delta }x{y}^{\gamma }=x{y}^{\delta +\gamma }$, then $xg=y^{\delta +\gamma }\in H$.

If $g=h_{3}x$, then $xg=x(y^{{ϵ}_k}{v}^{{\delta }_k}\dots y^{{ϵ}_1}{v}^{{\delta }_1})x=v^{{ϵ}_k}{y}^{{\delta }_k}\dots v^{{ϵ}_1}{y}^{{\delta }_1}\in H$.

If $g=h_{3}x{y}^{\gamma }$, then $xg=x(y^{{ϵ}_k}{v}^{{\delta }_k}\dots y^{{ϵ}_1}{v}^{{\delta }_1})x{y}^{\gamma }=v^{{ϵ}_k}{y}^{{\delta }_k}\dots v^{{ϵ}_1}{y}^{{\delta }_1+\gamma }\in H$.

If $g=h_{4}x$, then $xg=x(v^{{\delta }_k}{y}^{{ϵ}_{k-1}}{v}^{{\delta }_{k-1}}\dots y^{{ϵ}_1}{v}^{{\delta }_1})x=y^{{\delta }_k}{v}^{{ϵ}_{k-1}}{y}^{{\delta }_{k-1}}\dots v^{{ϵ}_1}{y}^{{\delta }_1}\in H$.

If $g=h_{4}x{y}^{\gamma }$, then $xg=x(v^{{\delta }_k}{y}^{{ϵ}_{k-1}}{v}^{{\delta }_{k-1}}\dots y^{{ϵ}_1}{v}^{{\delta }_1})x{y}^{\gamma }=y^{{\delta }_k}{v}^{{ϵ}_{k-1}}{y}^{{\delta }_{k-1}}\dots v^{{ϵ}_1}{y}^{{\delta }_1+\gamma }\in H$.

If $g=h_{5}x$, then $xg=x(v^{{\delta }_k}{y}^{{ϵ}_k}\dots v^{{\delta }_1}{y}^{{ϵ}_1})x=y^{{\delta }_k}{v}^{{ϵ}_k}\dots y^{{\delta }_1}{v}^{{ϵ}_1}\in H$.

If $g=h_{5}x{y}^{\gamma }$, then $xg=x(v^{{\delta }_k}{y}^{{ϵ}_k}\dots v^{{\delta }_1}{y}^{{ϵ}_1})x{y}^{\gamma }=y^{{\delta }_k}{v}^{{ϵ}_k}\dots y^{{\delta }_1}{v}^{{ϵ}_1}{y}^{\gamma }\in H$.

If $g=h_{6}x$, then $xg=x(y^{{ϵ}_k}{v}^{{\delta }_{k-1}}{y}^{{ϵ}_{k-1}}\dots v^{{\delta }_1}{y}^{{ϵ}_1})x=v^{{ϵ}_k}{y}^{{\delta }_{k-1}}{v}^{{ϵ}_{k-1}}\dots y^{{\delta }_1}{v}^{{ϵ}_1}\in H$.

If $g=h_{6}x{y}^{\gamma }$, then $xg=x(y^{{ϵ}_k}{v}^{{\delta }_{k-1}}{y}^{{ϵ}_{k-1}}\dots v^{{\delta }_1}{y}^{{ϵ}_1})x{y}^{\gamma }=v^{{ϵ}_k}{y}^{{\delta }_{k-1}}{v}^{{ϵ}_{k-1}}\dots y^{{\delta }_1}{v}^{{ϵ}_1}{y}^{\gamma }\in H$.

• This shows that $x\left(\beta \right)\in {\alpha }^H$. Similarly, if $x\left(\beta \right)\notin {\alpha }^H$, then $\beta \in {\alpha }^H$.

Secondly, suppose that ${\alpha }^G$ contains a unique totally positive cycle $\stackrel{y}{\widehat{\beta }}$. By Lemma 3 (iii), $\stackrel{v}{\widehat{x\left(\beta \right)}}$ is a totally negative cycle that is obviously contained in ${\alpha }^G$. Note also, by Lemma 3 (iv), that it cannot be the case that both cycles are contained in ${\alpha }^H$. Following exactly the same argument of the zero-norm elements above, we conclude that ${\alpha }^G$ is either the disjoint union of ${\alpha }^H$ and ${\beta }^H$ or is the disjoint union of ${\alpha }^H$ and $x{\left(\beta \right)}^H$, and therefore ${\alpha }^G$ splits into two distinct H-orbits in this case as well.

We have, so far, proved that $|{\mathcal{O}}_{-n}^H|=2|{\mathcal{O}}_{-n}^G|$ for $n\ge 2$. The proof would be complete in this case by applying [10] (Theorem 2.1), which precisely gives the value of $|{\mathcal{O}}_{-n}^G|$ as follows:

$$\begin{array}{cc}\hfill |{\mathcal{O}}_{-n}^G|& =\left\{\begin{array}{cc}2& \hfill ,\mathrm{if} n=2\\ 4& \hfill ,\mathrm{if} n=3\\ d\left(n\right)+{\displaystyle \frac{2}{3}}\sum _{i=1}^{\lfloor (n-1)/2\rfloor }[d(i^2+n)-2d_{\le i}(i^2+n)]& \hfill ,\mathrm{otherwise}.\end{array}\right.\hfill \end{array}$$

and, moreover, declares the congruence $|{\mathcal{O}}_{-n}^G|\equiv 0(mod4)$ for $n\ge 3$.

Finally, for the case $n=1$, the only elements of norm zero in ${\mathbb{Q}}^{*}\left(\sqrt{-1}\right)$ are $\pm \sqrt{-1}$, and both are fixed by x (see the proof of Lemma 3 (v)). We also have by Lemma 4 (i) that ${\alpha }^G$ contains a unique element of norm zero, that is, ${\alpha }^G$ must contain either $\sqrt{-1}$ or $-\sqrt{-1}$ but not both. Suppose that $\sqrt{-1}\in {\alpha }^G$. If $\sqrt{-1}\notin {\alpha }^H$, then, by following the argument in the second paragraph of this proof, $x\left(\sqrt{-1}\right)\in {\alpha }^H$. But $x\left(\sqrt{-1}\right)=\sqrt{-1}$, a contradiction. Thus, $\sqrt{-1}\in {\alpha }^H$. Similarly, if $-\sqrt{-1}\in {\alpha }^G$, then $-\sqrt{-1}\in {\alpha }^H$. This shows that ${\alpha }^H={\alpha }^G$ and hence $|{\mathcal{O}}_{-1}^H|=|{\mathcal{O}}_{-1}^G|$, which in turn equals 2 (by [10] (Theorem 2.1)). □

Remark 1.

Below, we present some examples showing that the formulas given by [18] (Theorem 9) are incorrect in general (see Claim 3 in Section 3).

1. 

For $n=1$, the number of orbits $|{\mathcal{O}}_{-1}^H|$ given by [18] (Theorem 9) is

$$2\left[d\right(1)+2d(2)-6]=2[1+4-6]=-2,$$

which is absurd. By Theorem 1, however, $|{\mathcal{O}}_{-1}^H|=2$.

2. 

For $n=21$, the number of orbits $|{\mathcal{O}}_{-21}^H|$ given by [18] (Theorem 9) is

$$2\left[d\right(21)+2d(22)-6]=2[4+8-6]=12.$$

This estimate immediately appears to be wrong when tested using the congruence relation in Theorem 1. Nonetheless, Theorem 1 gives precisely the following estimate:

$$\begin{array}{cc}\hfill |{\mathcal{O}}_{-21}^H|& =2d\left(21\right)+{\displaystyle \frac{4}{3}}\sum _{i=1}^{10}\left[d(i^2+21)-2d_{\le i}(i^2+21)\right]\hfill \\ \hfill & =8+{\displaystyle \frac{4}{3}}\{d\left(22\right)-2d_{\le 1}\left(22\right)+d\left(25\right)-2d_{\le 2}\left(25\right)+d\left(30\right)-2d_{\le 3}\left(30\right)+d\left(37\right)\hfill \\ \hfill & -2d_{\le 4}\left(37\right)+d\left(46\right)-2d_{\le 5}\left(46\right)+d\left(57\right)-2d_{\le 6}\left(57\right)+d\left(70\right)-2d_{\le 7}\left(70\right)\hfill \\ \hfill & +d\left(85\right)-2d_{\le 8}\left(85\right)+d\left(102\right)-2d_{\le 9}\left(102\right)+d\left(121\right)-2d_{\le 10}\left(121\right)\}\hfill \\ \hfill & =8+{\displaystyle \frac{4}{3}}\{4-2+3-2+8-6+2-2+4-4+4-4+8-8+4-4+8\hfill \\ \hfill & -8+3-2\}\hfill \\ \hfill & =8+8\hfill \\ & =16.\hfill \end{array}$$

3. 

For $n=26$, the number of orbits $|{\mathcal{O}}_{-26}^H|$ given by ([18], Theorem 9) is

$$2\left[d\right(26)+2d(27)-4]=2[4+8-4]=16.$$

Theorem 1 gives precisely the following estimate:

$$\begin{array}{cc}\hfill |{\mathcal{O}}_{-26}^H|& =2d\left(26\right)+{\displaystyle \frac{4}{3}}\sum _{i=1}^{12}\left[d(i^2+26)-2d_{\le i}(i^2+26)\right]\hfill \\ \hfill & =8+{\displaystyle \frac{4}{3}}\{d\left(27\right)-2d_{\le 1}\left(27\right)+d\left(30\right)-2d_{\le 2}\left(30\right)+d\left(35\right)-2d_{\le 3}\left(35\right)+d\left(42\right)\hfill \\ \hfill & -2d_{\le 4}\left(42\right)+d\left(51\right)-2d_{\le 5}\left(51\right)+d\left(62\right)-2d_{\le 6}\left(62\right)+d\left(75\right)-2d_{\le 7}\left(75\right)\hfill \\ \hfill & +d\left(90\right)-2d_{\le 8}\left(90\right)+d\left(107\right)-2d_{\le 9}\left(107\right)+d\left(126\right)-2d_{\le 10}\left(126\right)+d\left(147\right)\hfill \\ \hfill & -2d_{\le 11}\left(147\right)+d\left(170\right)-2d_{\le 12}\left(170\right)\}\hfill \\ \hfill & =8+{\displaystyle \frac{4}{3}}\{4-2+8-4+4-2+8-6+4-4+4-4+6-6+12-10+2\hfill \\ \hfill & -2+12-12+6-6+8-8\}\hfill \\ \hfill & =8+16\hfill \\ & =24.\hfill \end{array}$$

5. Conclusions

Consider the subgroup $H=\langle y,v|y^3=v^3=1\rangle$ of $\mathrm{PSL}(2,\mathbb{Z})=\langle x,y|x^2=y^3=1\rangle$, where $x:z\mapsto -1/z$, $y:z\mapsto (z-1)/z$, and $v=xyx:z\mapsto -1/(z+1)$. For a square-free $n\in \mathbb{N}$, it is shown that the number of orbits in the set

$${\mathbb{Q}}^{*}\left(\sqrt{-n}\right)=\{{\displaystyle \frac{a+\sqrt{-n}}{c}}\in \mathbb{Q}\left(\sqrt{-n}\right)|a,b={\displaystyle \frac{a^2+n}{c}},c\in \mathbb{Z},c\ne 0\},$$

under the action of H, is given as follows:

$$\left\{\begin{array}{cc}2\hfill & ,\mathrm{if} n=1\hfill \\ 4\hfill & ,\mathrm{if} n=2\hfill \\ 8\hfill & ,\mathrm{if} n=3\hfill \\ 2d\left(n\right)+{\displaystyle \frac{4}{3}}\sum _{i=1}^{\lfloor (n-1)/2\rfloor }[d(i^2+n)-2d_{\le i}(i^2+n)]\hfill & ,\mathrm{if} n>3\hfill \end{array}\right.$$

and is congruent to 0 modulo 8 for all $n\ge 3$, where $d\left(k\right)$ denotes the number of positive divisors of k, $d_{\le i}\left(k\right)$ denotes the number of positive divisors of k that do not exceed i for $i,k\in \mathbb{N}$ with $i\le k$, and $\lfloor .\rfloor$ is the floor function.