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# Connectedness of a suborbital graph for congruence subgroups

Journal of Inequalities and Applications20132013:117

https://doi.org/10.1186/1029-242X-2013-117

• Received: 5 December 2012
• Accepted: 1 March 2013
• Published:

## Abstract

In this paper, we give necessary and sufficient conditions for the graph ${H}_{u,n}$ to be connected and a forest.

MSC:20H10, 20H05, 05C05, 05C20.

## Keywords

• modular groups
• congruence subgroups
• suborbital graphs

## 1 Introduction

Let $\stackrel{ˆ}{\mathbb{Q}}=\mathbb{Q}\cup \left\{\mathrm{\infty }\right\}$ be the extended rationals and $\mathrm{\Gamma }=PSL\left(2,\mathbb{Z}\right)$ be the modular group acting on $\stackrel{ˆ}{\mathbb{Q}}$ as with the upper half-plane $\mathcal{H}=\left\{z\in \mathbb{C}:Imz>0\right\}$:
$g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right):z=\frac{x}{y}\to \frac{az+b}{cz+d}=\frac{ax+by}{cx+dy},$

where a, b, c, and d are rational integers and $ad-bc=1$.

Jones, Singerman, and Wicks  used the notion of the imprimitive action  for a Γ-invariant equivalence relation induced on $\stackrel{ˆ}{\mathbb{Q}}$ by the congruence subgroup ${\mathrm{\Gamma }}_{0}\left(n\right)=\left\{g\in \mathrm{\Gamma }:c\equiv 0\phantom{\rule{0.25em}{0ex}}\left(modn\right)\right\}$ to obtain some suborbital graphs and examined their connectedness and forest properties. They left the forest problem as a conjecture, which was settled down by the second author in .

In this paper we introduce a different Γ-invariant equivalence relation by using the congruence subgroup ${\mathrm{\Gamma }}_{1}\left(n\right)$ instead of ${\mathrm{\Gamma }}_{0}\left(n\right)$ and obtain some results for the newly constructed subgraphs ${H}_{u,n}$. In Section 4 we will prove our main theorems on ${H}_{u,n}$ which give conditions for ${H}_{u,n}$ to be connected or to be a forest, and we work out some relations between the lengths of circuits in ${H}_{u,n}$ and the elliptic elements of the group ${\mathrm{\Gamma }}_{1}\left(n\right)$. As Γ only has finite order elements of orders 2 and 3, the same is true for ${\mathrm{\Gamma }}_{1}\left(n\right)$.

Here, it is worth noting that these concepts are very much related to the binary quadratic forms and modular forms in  and [7, 8] respectively.

## 2 Preliminaries

Let ${\mathrm{\Gamma }}_{1}\left(n\right)=\left\{g\in \mathrm{\Gamma }:a\equiv d\equiv 1\phantom{\rule{0.25em}{0ex}}\left(modn\right),c\equiv 0\phantom{\rule{0.25em}{0ex}}\left(modn\right)\right\}$, which is one of the congruence subgroups of Γ. Then ${\mathrm{\Gamma }}_{\mathrm{\infty }}<{\mathrm{\Gamma }}_{1}\left(n\right)⩽\mathrm{\Gamma }$ for each n, where ${\mathrm{\Gamma }}_{\mathrm{\infty }}$ is the stabilizer of ∞ generated by the element $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, and second inclusion is strict if $n>1$.

Since, by , Γ acts transitively on $\stackrel{ˆ}{\mathbb{Q}}$, any reduced fraction $\frac{r}{s}$ in $\stackrel{ˆ}{\mathbb{Q}}$ equals $g\left(\mathrm{\infty }\right)$ for some $g\in \mathrm{\Gamma }$. Hence, we get the following imprimitive Γ-invariant equivalence relation on $\stackrel{ˆ}{\mathbb{Q}}$ by ${\mathrm{\Gamma }}_{1}\left(n\right)$:

where $g=\left(\begin{array}{cc}r& \ast \\ s& \ast \end{array}\right)$ and h is similar.

Here, as in , the imprimitivity means that the above relation is different from the identity relation ($a\sim b$ if and only if $a=b$) and the universal relation ($a\sim b$ for all $a,b\in \stackrel{ˆ}{\mathbb{Q}}$).

From the above, we can easily verify that

The equivalence classes are called blocks and the block containing $\frac{x}{y}$ is denoted by $\left[\frac{x}{y}\right]$.

Here we must point out that the above equivalence relation is different from the one in . This is because we take the group ${\mathrm{\Gamma }}_{1}\left(n\right)$ instead of ${\mathrm{\Gamma }}_{0}\left(n\right)$. The main reason of changing the equivalence relation lies in the fact that in the case of ${\mathrm{\Gamma }}_{1}\left(n\right)$, as we will see below, the elliptic elements do not necessarily correspond to circuits of the same order. It was the case in .

## 3 Subgraphs ${H}_{u,n}$

The modular group Γ acts on $\stackrel{ˆ}{\mathbb{Q}}×\stackrel{ˆ}{\mathbb{Q}}$ through $g:\left(\alpha ,\beta \right)\to \left(g\left(\alpha \right),g\left(\beta \right)\right)$. The orbits are called suborbitals. From the suborbital $O\left(\alpha ,\beta \right)$ containing $\left(\alpha ,\beta \right)$ we can form the suborbital graph $G\left(\alpha ,\beta \right)$ whose vertices are the elements of $\stackrel{ˆ}{\mathbb{Q}}$ and edges are the pairs $\left(\gamma ,\delta \right)\in O\left(\alpha ,\beta \right)$, which we will denote by $\gamma \to \delta$ and represent them as hyperbolic geodesics in .

Since Γ acts transitively on $\stackrel{ˆ}{\mathbb{Q}}$, every suborbital $O\left(\alpha ,\beta \right)$ contains a pair $\left(\mathrm{\infty },\frac{u}{n}\right)$ for $\frac{u}{n}\in \stackrel{ˆ}{\mathbb{Q}}$, $n\ge 0$, $\left(u,n\right)=1$. In this case, we denote the suborbital graph by ${G}_{u,n}$ for short.

As Γ permutes the blocks transitively, all subgraphs corresponding to blocks are isomorphic. Therefore we will only consider the subgraph ${H}_{u,n}$ of ${G}_{u,n}$ whose vertices form the block $\left[\mathrm{\infty }\right]=\left[\frac{1}{0}\right]$, which is the set . The following two results were proved in .

Theorem 1 There is an edge $\frac{r}{s}\to \frac{x}{y}$ in ${G}_{u,n}$ if and only if either
1. 1.
$x\equiv ur\phantom{\rule{0.25em}{0ex}}\left(modn\right)$
, $y\equiv us\phantom{\rule{0.25em}{0ex}}\left(modn\right)$ and $ry-sx=n$ or

2. 2.
$x\equiv -ur\phantom{\rule{0.25em}{0ex}}\left(modn\right)$
, $y\equiv -us\phantom{\rule{0.25em}{0ex}}\left(modn\right)$ and $ry-sx=-n$.

Lemma 1 ${G}_{u,n}={G}_{v,m}$ if and only if $n=m$ and $u\equiv v\phantom{\rule{0.25em}{0ex}}\left(modn\right)$.

The suborbital graph $F:={G}_{1,1}$ is the familiar Farey graph with $\frac{a}{b}\to \frac{c}{d}$ if and only if $ad-bc=±1$.

As it is illustrated in Figure 1, the pattern is periodic of period 1. That is, if $x\to y$ is an edge, then $x+1\to y+1$ is an edge as well. Figure 1 Farey graph.

Lemma 2 No edges of F cross in .

Theorem 1 clearly gives the following.

Theorem 2 Let $\frac{r}{s}$ and $\frac{x}{y}$ be in $\left[\mathrm{\infty }\right]$. Then there is an edge $\frac{r}{s}\to \frac{x}{y}$ in ${H}_{u,n}$ if and only if
1. 1.
$x\equiv ur\phantom{\rule{0.25em}{0ex}}\left(modn\right)$
, $ry-sx=n$, or

2. 2.
$x\equiv -ur\phantom{\rule{0.25em}{0ex}}\left(modn\right)$
, $ry-sx=-n$.

Theorem 3 Let $\frac{r}{s}$ and $\frac{x}{y}$ be in $\left[\mathrm{\infty }\right]$. Then there is an edge $\frac{r}{s}\to \frac{x}{y}$ in ${H}_{u,n}$ if and only if
1. 1.
$u=0$
and $ry-sx=1$ or

2. 2.
$u=1$
and $ry-sx=n$ or

3. 3.
$u=n-1$
and $ry-sx=-n$.

Proof Let $\frac{r}{s}\to \frac{x}{y}$ be an edge in ${H}_{u,n}$. Since $\frac{r}{s}$ and $\frac{x}{y}$ are in $\left[\mathrm{\infty }\right]$, $x,r\equiv 1\phantom{\rule{0.25em}{0ex}}\left(modn\right)$. Therefore, according to Theorem 2, we have $1\equiv u\phantom{\rule{0.25em}{0ex}}\left(modn\right)$, $ry-sx=n$ or $1\equiv -u\phantom{\rule{0.25em}{0ex}}\left(modn\right)$, $ry-sx=-n$. The first implies that $u=0$ and $u=1$, which proves (1). The second assures that $u=0$, $n=1$ or $u=1$, $n=2$ or $u=n-1$, which gives (3).

For the converse, it is enough to verify (3) only. For this, let $u=n-1$ and $ry-sx=-n$. Then $x\equiv -r\left(n-1\right)\equiv 0\phantom{\rule{0.25em}{0ex}}\left(modn\right)$. This, by Theorem 2, completes the proof. □

Theorem 4 ${\mathrm{\Gamma }}_{1}\left(n\right)$ permutes the vertices and the edges of ${H}_{u,n}$ transitively.

Proof
1. 1.

Let v and w be vertices in ${H}_{u,n}$. Then $w=g\left(v\right)$ for some $g\in \mathrm{\Gamma }$. Since $v\sim \mathrm{\infty }$, $g\left(v\right)\sim g\left(\mathrm{\infty }\right)$, that is, $w\sim g\left(\mathrm{\infty }\right)$. Therefore, $g\left(\mathrm{\infty }\right)$ lies in the block $\left[\mathrm{\infty }\right]$ and so g is in ${\mathrm{\Gamma }}_{1}\left(n\right)$.

2. 2.

The proof for edges is similar. □

Definition 1 Let ${H}_{u,n}$ and ${H}_{v,m}$ be two suborbital graphs. If the map ϕ is an injective function from the vertex set of ${H}_{u,n}$ to that of ${H}_{v,m}$ and sends the edges of ${H}_{u,n}$ to the edges of ${H}_{v,m}$, then ϕ is called a suborbital graph homomorphism (homomorphism for short) and it will be denoted by $\varphi :{H}_{u,n}\to {H}_{v,m}$.

Theorem 5
1. 1.

If $m\mid n$, then $\varphi \left(v\right)=\frac{nv}{m}$ is a homomorphism from ${H}_{u,n}$ to ${H}_{u,m}$.

2. 2.

Let $m\mid n$ and $m\ne n$; then the homomorphism in (1) is not an isomorphism.

3. 3.

Let $\varphi :{H}_{1,n}\to {H}_{n-1,n}$, given by $\varphi \left(a\right)=a$ for all vertices and $\varphi \left(a\to b\right)=b\to a$, be an isomorphism.

Proof
1. 1.

Let $\frac{r}{sn}\to \frac{x}{yn}$ be in ${H}_{u,n}$. To see $\frac{r}{sm}\to \frac{x}{ym}$ is in ${H}_{u,m}$ is an easy consequence of Theorem 2.

2. 2.

Conversely, suppose that $h:{H}_{u,n}\to {H}_{u,m}$, $h\left(v\right)=\frac{nv}{m}$ is an isomorphism. Then there exists a vertex v in ${H}_{u,n}$ such that $h\left(v\right)=\frac{m+1}{m}$. Therefore, $v=\frac{m+1}{n}$. But, since $m\mid n$ and $m\ne n$, $m+1\not\equiv 1\phantom{\rule{0.25em}{0ex}}\left(modn\right)$. That is, $\frac{m+1}{n}$ is not a vertex in ${H}_{u,n}$. This gives the proof.

3. 3.

Since the subgraphs ${H}_{1,n}$ and ${H}_{n-1,n}$ have same set of vertices, ϕ is well defined. Now suppose $\frac{r}{sn}\to \frac{x}{yn}$ is an edge in ${H}_{1,n}$. Then, by Theorem 3(2), $ry-sx=n$. So, $sx-ry=-n$. That is, using Theorem 3(3), $\frac{x}{yn}\to \frac{r}{sn}$ is an edge in ${H}_{n-1,n}$. □

Corollary 1 If $m\mid n$, then ${H}_{u,n}\to {H}_{u,m}$, $v\to \frac{nv}{m}$ is an isomorphism if and only if $m=n$.

Corollary 2 $\varphi :{H}_{u,n}\to F$, given by $v\to nv$, is a homomorphism.

Proof Since ${H}_{u,1}=F$, Theorem 5(1) gives the result. □

Corollary 3 No edges of ${H}_{u,n}$ cross in .

Proof By Corollary 2 there is an isomorphism from ${H}_{u,n}$ to a subgraph of F. Also, by Lemma 2, no edges of F cross in . Therefore the result follows. □

## 4 Main calculations

In this final section, we state all conditions for ${H}_{u,n}$ to be connected and a forest.

Definition 2 For $m\in \mathbb{N}$, $m\ge 2$, let ${v}_{1},{v}_{2},\dots ,{v}_{m}$ be vertices of ${H}_{u,n}$. The configuration ${v}_{1}\to {v}_{2}\to \cdots \to {v}_{m}\to {v}_{1}$ (some arrows, not all, may be reversed) is called a circuit of length m.

If $m=3$, the circuit is said to be a triangle. If $m=2$, we call the self paired edge a 2-gon.

A graph is called a forest if it contains no circuits other than 2-gons.

As in examples $\mathrm{\infty }\to \frac{1}{2}\to \mathrm{\infty }$ is a 2-gon in ${H}_{1,2}$ and $\mathrm{\infty }\to 1\to 2\to \mathrm{\infty }$ is a triangle in ${H}_{1,1}$ and furthermore we will see below that $\mathrm{\infty }\to {v}_{1}\to {v}_{2}\to \cdots$ never becomes a circuit in ${H}_{1,n}$ for $n\ge 2$.

We now prove the connectedness of ${H}_{1,n}$ separately as follows.

Theorem 6 ${H}_{1,2}$ is connected.

Proof Since the situation, only for this case, coincides with the situation in , it is not necessary to give a proof. □

To understand subsequent proofs better, we start by giving the following example.

Example 1 The subgraph ${H}_{1,3}$ is not connected.

Solution 1 Since $\mathrm{\infty }\to \frac{1}{3}$ is an edge in ${H}_{1,3}$ and ${H}_{1,3}$ is periodic with period 1, we just need consider the strip $\frac{1}{3}\le Rez\le \frac{4}{3}$. It is clear that ∞ is adjacent to $\frac{1}{3}$ and $\frac{4}{3}$ in ${H}_{1,3}$, but to no intermediate vertices. We will show that no vertices of ${H}_{1,3}$ in the interval $\left[\frac{2}{3},1\right]$ are adjacent to vertices of ${H}_{1,3}$ outside this interval. Of course, there is some vertex of ${H}_{1,3}$, such as $\frac{19}{27}$, in $\left[\frac{2}{3},1\right]$.

As in Figure 2, suppose that the edge $\frac{a}{3b}\to \frac{c}{3d}$ in ${H}_{1,3}$ crosses $Rez=\frac{2}{3}$. Then Corollary 2 implies that $\frac{a}{b}\to \frac{c}{d}$ is an edge in F and furthermore $\frac{a}{b}<2<\frac{c}{d}$. This proves that the edges $\frac{a}{b}\to \frac{c}{d}$ and $\mathrm{\infty }\to 2$ cross in F, a contradiction. A similar argument shows that no edges of ${H}_{1,3}$ cross $Rez=1$. These conclude that ${H}_{1,3}$ is not connected. Figure 2 H 1 , 3 .

Note 1 The graphs ${H}_{1,3}$ and ${H}_{2,3}$ have at least two connected components.

Proof Example 1 and Theorem 5(3) give the result. □

We now give the following.

Theorem 7 ${H}_{1,n}$ is not connected if $n\ge 3$.

Proof Since ${H}_{1,n}$ is periodic with period 1, we can, again, work in the strip $\frac{1}{n}\le Rez\le \frac{n+1}{n}$. Note that ∞ is adjacent to $\frac{1}{n}$ and $\frac{n+1}{n}$ in ${H}_{1,n}$, but to no intermediate vertices. We will show that no vertices in ${H}_{1,n}$, between $\frac{2}{n}$ and 1, are adjacent to vertices outside this interval. We note that there are vertices of ${H}_{1,n}$ in $\left(\frac{2}{n},1\right)$ for $n\in \mathbb{N}$. Indeed as in Figure 3, if n is odd, take the vertex $\frac{6n+1}{9n}$ in $\left(\frac{2}{n},1\right)$ and if n is even, take the vertex $\frac{n+1}{2n}$ in $\left(\frac{2}{n},1\right)$. Figure 3 H 1 , n .

Suppose that an edge crosses $Rez=\frac{2}{n}$, whence that it joins $v=\frac{a}{nb}$ to $w=\frac{c}{nd}$. By Corollary 2, nv and nw must be adjacent in F. As in Example 1, this is a contradiction. A similar argument shows that no edge crosses $Rez=1$, and since vertices between $\frac{2}{n}$ and 1 are not adjacent to ∞, it follows that ${H}_{1,n}$ is not connected.

Consequently, since there is no circuit like $\mathrm{\infty }\to \frac{1}{n}←{v}_{1}←\cdots ←\frac{n+1}{n}←\mathrm{\infty }$ in ${H}_{1,n}$, ${H}_{1,n}$ is not connected for $n\ge 3$. □

Theorem 8 ${H}_{u,n}$ is connected if and only if $n\le 2$.

Proof If $n=1,2$, it follows from ; otherwise, it follows from Theorem 7. □

Theorem 9 ${H}_{u,n}$ contains a triangle if and only if $n=1$.

Proof Let D be a triangle in ${H}_{u,n}$. From Theorem 3, $u=1$ or $u=n-1$. Using Theorem 5(3), we may only work in ${H}_{1,n}$. By Theorem 4, we may suppose that D has the form $\mathrm{\infty }\to {v}_{1}\to {v}_{2}\to \mathrm{\infty }$ or $\mathrm{\infty }\to {v}_{1}←{v}_{2}\to \mathrm{\infty }$. Let us do calculations only for the first triangle. We easily see that ${v}_{1}=\frac{x}{n}$ and ${v}_{2}=\frac{y}{n}$ for some $x,y\in \mathbb{Z}$. If $\frac{x}{n}<\frac{y}{n}$, then $x-y=-1$. Since $\frac{x}{n}$ and $\frac{y}{n}\in \left[\mathrm{\infty }\right]$, $x-y\equiv 0\phantom{\rule{0.25em}{0ex}}\left(modn\right)$. So, $n=1$. If $\frac{x}{n}>\frac{y}{n}$, then $x-y=1$. Therefore, again, $n=1$.

Conversely, if $n=1$, then $u=0$ or 1. But since ${H}_{0,1}={H}_{1,1}$, we have the triangle $\frac{1}{0}\to \frac{1}{1}\to \frac{2}{1}\to \frac{1}{0}$. □

Theorem 10 ${H}_{u,n}$ contains a 2-gon if and only if $n=1$ or 2.

Proof Suppose $\frac{x}{kn}\to \frac{y}{ln}\to \frac{x}{kn}$ is a 2-gon in ${H}_{u,n}$. Then, by Theorem 3, it is easily seen that $n=1$ or 2.

Conversely, if $n=1$ or 2, it is clear that $\frac{1}{0}\to \frac{1}{n}\to \frac{1}{0}$ is a 2-gon. □

We now give one of our main theorems.

Theorem 11 If $n\ge 2$, then ${H}_{1,n}$ and ${H}_{n-1,n}$ are forests.

Proof Let $n=2$. Assume that ${H}_{1,2}$ is not a forest. Therefore we suppose that there exists a circuit D, other than 2-gon, in ${H}_{1,2}$. By Theorem 4 and Theorem 3, we may assume that D has the form $\mathrm{\infty }\to {v}_{1}\to \cdots \to {v}_{k}\to \mathrm{\infty }$, where the vertices ${v}_{1},{v}_{2},\dots ,{v}_{k}$ are all different. Here, since the pattern for the subgraph ${H}_{u,n}$ is periodic with period 1, we may choose the vertices of D, apart from ∞, in the interval $\left[\frac{1}{2},\frac{3}{2}\right]$. By Theorem 3, ${v}_{1}=\frac{1}{2}$ or $\frac{3}{2}$. If ${v}_{1}=\frac{1}{2}$, then ${v}_{k}=\frac{2a+1}{2}\in \left[\frac{1}{2},\frac{3}{2}\right]$ and ${v}_{1}\ne {v}_{k}$ give that ${v}_{k}=\frac{3}{2}$. Since 1 is not a vertex in ${H}_{1,2}$, Corollary 2 implies that such a circuit D does not occur. Similarly, we can show that there is not a circuit D in the case where ${v}_{1}=\frac{3}{2}$. That is, ${H}_{1,2}$ is a forest.

Now let $n\ge 3$. If ${H}_{1,n}$ is not a forest, then, as we will see now by Theorem 3, D must be of the form $\mathrm{\infty }\to {v}_{1}←\cdots ←{v}_{k}←\mathrm{\infty }$ or $\mathrm{\infty }\to {v}_{1}\to \cdots \to {v}_{k}←\mathrm{\infty }$. As above, we choose the vertices in the finite interval $\left[\frac{1}{n},\frac{n+1}{n}\right]$. By Theorem 3, ${v}_{1}=\frac{1}{n}$ or $\frac{n+1}{n}$. If ${v}_{1}=\frac{1}{n}$, then, as above, ${v}_{k}$ must be $\frac{n+1}{n}$. In this case D has the form $\mathrm{\infty }\to \frac{1}{n}←\cdots ←\frac{n+1}{n}←\mathrm{\infty }$. As 1 is not a vertex in ${H}_{1,n}$, Corollary 2 implies that such a circuit D does not occur. Similarly, we can show that if ${v}_{1}=\frac{n+1}{n}$, then there does not exist a circuit D like $\mathrm{\infty }\to \frac{n+1}{n}\to \cdots \to \frac{1}{n}←\mathrm{\infty }$. Therefore ${H}_{1,n}$ is a forest. Using Theorem 5(3), we see that the subgraph ${H}_{n-1,n}$ is a forest as well. Therefore the proof is completed. □

Theorem 12 If ${H}_{u,n}$ contains a triangle, then ${\mathrm{\Gamma }}_{1}\left(n\right)$ contains an elliptic element of order 3.

Proof If ${H}_{u,n}$ contains a triangle, then by Theorem 9, $n=1$. So, ${\mathrm{\Gamma }}_{1}\left(1\right)=\mathrm{\Gamma }$ and $\left(\begin{array}{cc}1& -1\\ 3& -2\end{array}\right)\in {\mathrm{\Gamma }}_{1}\left(1\right)$ is an elliptic element of order 3. □

Remark 1 In general, the converse of Theorem 12 is not true. For example, the element $\left(\begin{array}{cc}1& -1\\ 3& -2\end{array}\right)\in {\mathrm{\Gamma }}_{1}\left(3\right)$ is an elliptic element of order 3, but ${H}_{1,3}$ does not contain a triangle. And also, by Theorem 5(3), ${H}_{2,3}$ does not contain a triangle either.

Remark 2 In , it is shown that the elliptic elements in ${\mathrm{\Gamma }}_{0}\left(n\right)$ correspond to circuits in the subgraph ${F}_{u,n}$ of the same order and vice versa. Here, in the case of ${\mathrm{\Gamma }}_{1}\left(n\right)$, owing to Theorem 12 triangles in the subgraph ${H}_{u,n}$ correspond to elliptic elements in ${\mathrm{\Gamma }}_{1}\left(n\right)$ of order 3. But the converse is not true as shown in Remark 1.

Theorem 13 ${H}_{u,n}$ contains a 2-gon if and only if ${\mathrm{\Gamma }}_{1}\left(n\right)$ contains an elliptic element of order  2.

Proof If ${H}_{u,n}$ contains a 2-gon, then by Theorem 10, $n=1$ or 2. So, $\left(\begin{array}{cc}1& -1\\ 2& -1\end{array}\right)$ is an elliptic element of order 2 in both ${\mathrm{\Gamma }}_{1}\left(1\right)$ and ${\mathrm{\Gamma }}_{1}\left(2\right)$.

Conversely, assume that ${\mathrm{\Gamma }}_{1}\left(n\right)$ contains an elliptic element of order 2. Then there is an element of ${\mathrm{\Gamma }}_{1}\left(n\right)$ of the form $\left(\begin{array}{cc}1+an& b\\ cn& 1+dn\end{array}\right)$ such that $2+\left(a+d\right)n=0$. From this we get $n=1$ or 2. Hence, the proof now follows from Theorem 10. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Authors’ Affiliations

(1)
Department of Mathematics, Recep Tayyip Erdoğan University, Rize, 53100, Turkey
(2)
Department of Mathematics, Karadeniz Technical University, Trabzon, 61080, Turkey

## References

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