POWER SUBGROUPS OF HECKE GROUPS H ( √ n )

Results in discrete group theory are applied to some Hecke groups to determine the group theoretical structure of power subgroups.


Introduction.
Hecke groups H(λ) have been introduced by E. Hecke (see [2]).They are subgroups of PSL(2, R) generated by R(z) = −1/z and T (z) = z + λ.Hecke asked the question, "For what values of λ these groups are discrete?"In answering this question he proved that is a fundamental region for H(λ) if and only if λ ≥ 2 and real or λ = λ q = 2cos(π /q), q ∈ N, q ≥ 3. Therefore, H(λ) is discrete only for these values of λ.The most important and interesting Hecke group is the modular group H(λ 3 ) = PSL(2, Z).Next two interesting Hecke groups are obtained for q = 4 and q = 6.As λ 4 = √ 2 and λ 6 = √ 3, H( √ 2) and H( √ 3) denote the Hecke groups corresponding to λ 4 and λ 6 , respectively.One of the main reasons for H( √ 2) and H( √ 3) to be two of the most important Hecke groups is that apart from modular group, they are the only Hecke groups H(λ q ) whose elements can be completely described.Here we deal with the cases H( √ n), n square-free integer.H( √ n) consists of the set of all matrices of the following types: (i) Those of type (i) are called even while those of type (ii) are called odd.Even elements form a subgroup of index 2 called the even subgroup [1].
Let S = RT so that S(z) = −1/(z + λ).In the cases H( √ n), n = 2, 3, S is an element of order q = 2n.Thus R 2 = S q = I and RS = T is parabolic.It is known that H( √ n) is isomorphic to the free product C 2 * C q .Therefore H( √ n) has the signature (0; 2,q,∞), [1].In the case n > 3 square-free integer, S is an element of infinite order and H( √ n) is isomorphic to the free product C 2 * Z, [6].From the definition, one can easily deduce that and that 2), it is easy to deduce that Here (m, k) denotes the greatest common divisor of m and k.
2. Structure of power subgroups.We now discuss the group theoretical structure of these subgroups.First we have the following theorem.
√ n) is isomorphic to the free product of infinite cyclic group Z and two finite cyclic groups of order n.Also (2.1) The elements of H 2 ( √ n) are characterized by the property that the sums of the exponents of R and S are both even.
(ii) Let n > 3 square-free integer.The normal subgroup H 2 ( √ n) is the free product of three infinite cyclic groups.Also (2. 2) The elements of H 2 ( √ n) can be characterized by the requirement that the sums of the exponents of R and S are both even.
Proof.We use the Reidemeister-Schreier process to find a presentation of H 2 ( √ n), [5].We add the relation X 2 = 1 to the presentation of H( √ n).This gives a presentation of H( √ n)/H 2 ( √ n) the order of which is the index.We have (2.4) Clearly the elements of H 2 ( √ n) satisfy the requirements of the theorem, that is, the sums of the exponents of R and S are both even for each element.Note that we have S −1 = S 3 , S −1 = S 5 for n = 2, n = 3, respectively.Using the Reidemeister rewriting process, we get the relations (2.5) Therefore there are no nontrivial relations and H 2 ( √ n) is the free product of three infinite cyclic groups generated by x 1 ,x 2 , and x 3 .As each of R, S, and T goes to elements of order 2, they have the following permutation representations: By the permutation method (see [4,7]), the signature of H 2 ( √ 2) is (g;2, 2, ∞, ∞) = (g;2 (2) , ∞ (2) ) and the signature of H 2 ( √ 3) is (g;3 (2) , ∞ (2) ).Since the signature of all the Hecke groups H( √ n), n > 3 square-free integer, is (0; 2, ∞;1), we find the signature of H 2 ( √ n), n > 3 square-free integer, as (g; ∞ (2) ;2).Now by the Riemann-Hurwitz formula, we have g = 0 in all cases.Hence H 2 ( √ n), n > 3 square-free integer, is isomorphic to the free product of three Z's and H 2 ( √ 2) is isomorphic to the free product of Z and two finite cyclic groups of order 2 and H 2 ( √ 3) is isomorphic to the free product of Z and two finite cyclic groups of order 3.

Theorem 2.2. Let m be a positive odd integer. Then
Proof.Teh proof is clear as the quotient is trivial.
Theorem 2.3.Let m be a positive integer such that m ≡ 2 mod 4. Then H m ( √ 2) is the free product of the infinite cyclic group Z and m finite cyclic groups of order two.
Proof.It is easy to show that the quotient group is isomorphic to the dihedral group D m of order 2m.The permutation representations of R, S, and T are (2.7) Therefore, H 3 ( √ n) is generated by x 1 = R, x 2 = S 3 , x 3 = SRS −1 , and x 4 = S 2 RS −2 .Using the Reidemeister rewriting process, we get the relations (2.13) The permutation representations of R, S, and T are given in the statement of the theorem.
Theorem 2.9.Let m be a positive odd integer and n > 3 is a square-free integer.Then

.16)
By the permutation method, we find the signature of H m ( √ n) as (0; 2 (m) , ∞;1).Therefore, H m ( √ n) is isomorphic to the free product of m cyclic groups of order 2 and an infinite cyclic group.
Let m be a positive even integer and n > 3 is a square-free integer.Then we have

H
√ n /H m √ n = R, S; R 2 = S m = (RS) m = I , (2.17) that is, the factor group is the group whose signature (2,m,m).If m = 2, we have already seen that H 2 ( √ n) Z * Z * Z which is a normal subgroup of genus 0, then H( √ n)/H 2 ( √ n) is a group of automorphisms of a sphere with two boundary components and two punctures.If m = 4, we have a normal subgroup acting on the Euclidean plane.Because, in this case the factor group (2, 4, 4) is a group of infinite order and 1/4 + 1/4 = 1/2.If m ≥ 6 and even, the factor group (2,m,m) is a group of infinite order and 1/m+1/m = 2/m < 1/2.Therefore, in this case we have a normal subgroup acting on the hyperbolic 2-space (i.e., upper half plane).
Now we choose {I, R, S, RS} as a Schreier transversal for H 2 ( √ n).Then we can form all possible products