Disconnectedness of the subgraph for the group
© Akbaş et al.; licensee Springer 2013
Received: 6 December 2012
Accepted: 22 May 2013
Published: 5 June 2013
In this paper we show that the subgraph is disconnected and that for all integers m, we find all integers a and b such that and are square. It turns out that the set of numbers b comprises the Fibonacci numbers.
Dedicated to Professor Hari M Srivastava
where a, b, c, and d are rational integers, and let denote the group consisting of the cubes of the elements g of Γ, which is the group ; see .
Jones et al.  used the notion of the imprimitive action [3–5] for a Γ-invariant equivalence relation induced on by the congruence subgroup to obtain some suborbital graphs and examined their connectedness and forest properties.
In , a -invariant equivalence relation is introduced by using the subgroup to obtain suborbital graphs . There, the connectivity properties of all subgraphs other than are examined.
In this paper we show that the subgraph is disconnected and give some results, which seem important from the point of view of number theory.
Since , it is easily seen that the elements of are ones of the forms , , and where a, b, c, and in the third matrix. Furthermore, for each positive integer n, where is the stabilizer of ∞ generated by the element , and second inclusion is strict if .
Since the group is transitive on in , any reduced fraction in equals for some . Similar to that of , we get the following -invariant equivalence relation on by as if and only if , where and h is similar. Furthermore, the above equivalence relation is imprimitive, which means that it is different from the identity relation ( if and only if ) and the universal relation ( for all ).
From the above defined -invariant equivalence relation, we can verify that if and only if . The equivalence classes are called blocks, and a block containing the rational is denoted by .
Although the equivalence relations are resulting almost the same as in , the subgraph in  is easily shown to be connected, but here we will see that the subgraph is disconnected. So, using different subgroups changes the characters of the subgraphs.
The group acts on through . The orbits are called suborbital. From the suborbital containing , we can form the suborbital graph whose vertices are the elements of and the edges are the pairs , which we denote by and represent as hyperbolic geodesics in ℋ.
Since acts transitively on , every suborbital contains a pair for some in , , . In this case, we denote the suborbital graph by for short. From now on, we assume that .
As permutes the blocks transitively, all subgraphs corresponding to the blocks are isomorphic, as in . Therefore, we will only consider the subgraph of whose vertices form the block . The following two results were proved in .
Theorem 1 if and only if and .
if , then and , or
if , then , and , or
if , then , and .
We can easily get the following lemmas.
Lemma 1 is in if and only if is in .
Lemma 2 
No edges of cross in ℋ.
4 Disconnectedness of
Definition 1 For and , let be a finite sequence of vertices of . Then the configuration is called a finite path in . A subgraph ∧ of is called connected if every two vertices x and y of ∧ are connected by a finite path in . Otherwise, we call ∧ disconnected.
Now we give one of our main theorems.
Theorem 3 The graph is disconnected.
We prove Theorem 3 after giving some theorems, propositions and lemmas as follows.
It is clear that is in and the corresponding transformation is strictly increasing on . Furthermore, it is easily seen that is an edge in for all non-negative integers m. From this, we get, as an example, a finite path in as .
Lemma 3 Let T be as above, then the sequence is strictly monotone increasing and is an infinite path in in increasing order.
Proof The conclusion follows from Theorem 2 and from the fact that is strictly increasing on . □
Lemma 4 Let a and b be in ℕ and let , then .
Proof From we get . Then squaring gives the inequality . That is, . On the other hand, , then it is easily seen that . As , then taking square roots gives . This shows that . □
Proposition 1 Let T be as above and . Then is an edge in if and only if there exists a natural number u such that and .
Proof Let be an edge in . Then, by using Theorem 2 and Lemma 4, we get . Since , we have . This concludes that is an integer u.
Conversely, it is clear to see is in and that and . Therefore, by the definition of edges of , the configuration is an edge in . □
Theorem 4 The positive rational number is in if and only if there exists a natural number u such that and .
Proof From Proposition 1, the ‘if’ part is clear.
We know that is an edge in . From Lemma 2 it must be that . This gives the inequality . But, from , we arrive at the inequality , a contradiction. This concludes the proof of the theorem. □
Corollary 1 is an infinite path in , and all vertices of the path are smaller than , and the natural numbers x and y in the vertex are such that and are square. Furthermore, and , where T is as above.
Proof Lemma 3 and Theorem 4 conclude the proof. □
If we follow the way of the above, we arrive at the following two results without proofs.
Theorem 5 Let and . Then and is an edge in if and only if is a square and .
Corollary 2 is an infinite path in in decreasing order such that all vertices of the path are greater than , and the natural numbers x and y in the vertex are such that and are squares. Furthermore, and , where T is as above.
Theorem 6 Let k be a natural number and let the vertices in of be such that at least one is smaller and one is greater than . Then the path does not occur in .
Proof If the above situation occurs, since is not a vertex in , there exists such that and is an edge in .
Suppose first that . Then, multiplying by and using Theorem 5, we have for some . Therefore, from Theorem 2, we get . That is, is an edge in , a contradiction.
Suppose now that . Then, again multiplying by , we have . In this case, . But, in any case, , a contradiction.
Finally, let . Then must be an edge in , a contradiction. These complete the proof. □
Let . Then likewise we do before we give the following five results without a proof.
Lemma 5 Let . Then . Furthermore, is an edge in if and only if is a natural number and .
Lemma 6 Let . Then . Furthermore, is an edge in if and only if is a square and .
Lemma 7 is an infinite path in in decreasing order, and the limit of the sequence of vertices is .
Theorem 7 Let k be a natural number and let the vertices , in , of be such that at least one is smaller and one is greater than . Then does not occur in .
Proof We conclude the proof as in Theorem 6. □
Proof of Theorem 3 Theorems 6 and 7 conclude that the vertices of in are not connected to the vertex ∞. That is, the graph is disconnected. □
We finally give one of our main results as follows.
Theorem 8 For all natural numbers m, the natural numbers b that make the number square are .
And we conclude that the numbers make square.
Let us now show the only non-negative integers b such that is square.
From the first inequality, we conclude that . From the second inequality, we just have , a contradiction. Consequently, must be in the set . This completes the proof of the theorem. □
From Corollaries 1 and 2, we get the following without a proof.
Corollary 4 The non-negative integers b such that is square are .
Corollary 5 The non-negative integers b making square are .
From Corollaries 4 and 5, we conclude the following important corollary.
Corollary 6 Let and be the sequences and , respectively. Then the sequence is the Fibonacci sequence.
Proof Let us see that and for all n in ℕ by induction. Suppose that the assertion is true up to the natural number k. Let us show that . Since and , . The other is similar. □
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