Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 1, 223-231.

Generators of Order Two for S_{n} and its Two Double Covers

John Brinkman

Department of Mathematics, Liverpool Hope University College Hope Park, Liverpool L16 8ND, England

1. Introduction

This paper considers the minimum number of involutions, i(G), required to generate both of the double covers G of the symmetric group. In particular, explicit generators, of order two, for each of the groups are also introduced. These generators may, for example, be useful for implementation in Magma-Cayley or GAP.

As a starting point we observe that if i(G) = 1 then G is cyclic and if i(G) = 2 then G is dihedral. Hence for any group of order greater than two that is not isomorphic to a dihedral group we have i(G) ≥ 3.

It is also well-known that the symmetric group S_{n}, n ≥ 3, may be generated by the two cycles (1, 2) and (1, 2, 3, …, n). But as we may write (1, 2, 3 …, n) as the product of, multiplying from the left,

(1, n − 1)(2, n − 2) … (r, n − r) and (1, n)(2, n − 1) … (t, n + 1 − t)

where r is the integer part of (n − 1)/2 and t the integer part of n/2, it is clear that S_{n}, n ≥ 4, may be generated by three involutions. Moreover, for n ≥ 4, i(S_{n}) = 3.

2. i(S̃_{n})

This double cover of S_{n}, which lifts a transposition of S_{n} to an element of order 4, will be denoted by S̃_{n}. So that S̃_{n} is the group with generators z, r_{1}, r_{2}, …, r_{n−1} and relations

z^{2} = 1,

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2000 Heldermann Verlag

224 J. Brinkman: Generators of Order Two for S_{n} and…

zr_{i} = r_{i}z, r_{i}^{2} = z for 1 ≤ i ≤ n − 1,
(r_{j}r_{j+1})^{3} = z for 1 ≤ j ≤ n − 2,
r_{k}r_{h} = zr_{h}r_{k} for | h − k |> 1 and 1 ≤ h, k ≤ n − 1.

For computations in S̃_{n} we will use a method first described by Conway and others at Cambridge (Atlas [2]). This method is outlined in a paper by David B. Wales [5]. In this the elements of S̃_{n} are products of the form ±[σ_{i}], where the σ_{i} are disjoint cycles in S_{n} and ±[σ_{i}] are the corresponding lifts in S̃_{n}.

Definition 2.1. For distinct elements a_{1}, …, a_{m} we define [a_{1}a_{2} … a_{m}] = a_{1}a_{2} … a_{m}a_{1}. We call ±[a_{1}a_{2} … a_{k}] signed cycles in S̃_{n}. Each is a lift of the cycle (a_{1}a_{2} … a_{k}) in S_{n}.

In fact each a_{i} corresponds to an element of a subgroup of a Clifford algebra which is isomorphic to S̃_{n}. But the following rules are sufficient to enable the calculation of products of disjoint signed cycles in S̃_{n} (these appear as 2.3 and 2.4 in [5]).

[a_{i}] = −1,
[a_{1}a_{2} … a_{m}] = (−1)^{m+1}[a_{2}a_{3} … a_{m}a_{1}],
[a_{1}a_{2} … a_{m−1}]a_{m} = (−1)^{m}a_{m}[a_{1}a_{2} … a_{m−1}].

In particular, these are used in [1] to prove the following proposition.

Proposition 2.2. Any product of k disjoint signed transpositions in S̃_{n} has order two if the integer part of (k + 1)/2 is a multiple of two, and order four otherwise.

Hence in S̃_{n} an involution is of the form ±π where π is a signed cycle consisting of k disjoint transpositions and k is congruent to 0 or 3 modulo 4. Also as the only other element of order two in S̃_{n} is −1, we have immediately that S̃_{4} and S̃_{5} may not be generated by involutions.

Also, as we will be making repeated use of this fact, it is convenient at this point to note that factoring out S̃_{n} by the subgroup Z = ⟨1, −1⟩ recovers S_{n}. That is S̃_{n}/Z ≅ S_{n}.

Proposition 2.3. For n ≥ 3, S̃_{n} may be generated by a = ±[1, 2] and b = ±[1, 2, …, n].

Proof. As S_{n} is generated by (1, 2) and (1, 2, 3, …, n) it follows that a and b will generate at least one, up to parity of sign, of every type of signed cycle. Hence we only need show that we may also generate −1. But a^{2} = −1. □

Proposition 2.4. i(S̃_{6}) = 5 and i(S̃_{7}) = 3.

Proof. That i(S̃_{6}) = 5 is proved in Section 3. However, in order to give specific generators, we note that (1, 2)(3, 4)(5, 6), (1, 3)(2, 4)(5, 6), (1, 4)(2, 3)(5, 6), (1, 5)(2, 6)(3, 4) and (1, 6)(2, 3)(4, 5) generate S_{6}. Thus, as in Proposition 2.3, the corresponding signed cycles will generate S̃_{6}.

For S̃_{7} we only need note that

(1, 2)(3, 4)(5, 6), (1, 4)(3, 5)(2, 7) and (2, 3)(3, 7)(1, 6)

generate S_{7}. Hence the corresponding signed cycles generate S̃_{7}. □

Proposition 2.5. For 8 ≤ n ≤ 12, the following involutions generate S̃_{n}. When n = 8

a = [1, 2][3, 8][4, 7][5, 6] , b = [1, 3][4, 8][5, 7] and c = [3, 8][4, 7][5, 6].

For 9 ≤ n ≤ 12 a = [1, 2][3, 9][4, 8][5, 7], b = [1, 3][4, 9][5, 8][6, 7]

and c = \{ \begin{matrix}[3, 9][4, 8][5, 7] & when & n = 9, \\ [3, 10][4, 8][5, 7] & when & n = 10, \\ [3, 10][4, 11][5, 7] & when & n = 11, \\ [3, 10][4, 11][5, 12] & when & n = 12. \end{matrix}

Proof. For n = 8 and 9 we have ac = ±[1, 2] and ab = ±[1, 2, …, n] from which the result follows.

When n = 10, 11 or 12, (ac)^{3} = ±[1, 2] and ab = ±[1, 2, …, 9] so for each n we may generate the subgroup S̃_{9}, in particular the signed cycles d = [1, 3], e = [1, 4] and f = [1, 5].

It only remains to show that [1, 2, …, n] can also be generated for each n. But when n = 10 we have abcdc = ±[1, 2, …, 10], when n = 11, abcdec = ±[1, 2, …, 11] and when n = 12, abcdefc = ±[1, 2, …, 12]. □

Proposition 2.6. i(S̃_{13}) = 4.

Proof. Note that in S̃_{13} the only involutions are −1 and signed cycles of type 2^{3} and 2^{4}. Also we require at least 12 signed transpositions in our generators to ensure that all the numbers from 1 to 13 have some link. However we cannot use three signed cycles of type 2^{4}, the minimum needed, as they are all even and thus cannot generate any odd signed cycles. Thus i(S̃_{13}) > 3. That i(S̃_{13}) = 4 follows by noting that (1, 12)(2, 11)(3, 10)(4, 9), (5, 8)(6, 7)(1, 13)(2, 12), (3, 11)(4, 10)(5, 9)(6, 8) and (1, 2)(4, 11)(5, 6) will generate S_{13}. Thus the corresponding signed cycles generate S̃_{13}. □

Proposition 2.7. For 14 ≤ n ≤ 16, the following involutions generate S̃_{n}.

n = 14  a = [1, 2][3, 14][4, 13][5, 12][6, 11][7, 10][8, 9],
b = [1, 3][4, 14][5, 13][2, 9],
c = [6, 12][7, 11][8, 10][2, 9].
n = 15  a = [2, 15][3, 14][4, 13][5, 12][6, 11][7, 10][8, 9],
b = [1, 2][3, 15][4, 14][5, 13][6, 12][7, 11][8, 10],
c = [3, 14][4, 13][6, 11][7, 10].
n = 16  a = [2, 16][3, 15][4, 14][5, 13][6, 12][7, 11][8, 10],
b = [1, 2][3, 16][4, 15][5, 14][6, 13][7, 12][8, 11][9, 10],
c = [3, 16][4, 15][5, 14][6, 13][7, 12][8, 11][9, 10].

Proof. For n = 16, bc = ±[1, 2] and ab = ±[1, 2, …, 16] from which the result follows. Now for n = 14, abc = ±[1, 2, …, 14] and for n = 15, ab = ±[1, 2, …, 15]. Thus in both cases we need only show that we may also generate ±[1, 2]. But when n = 15, (ab)^{4}((abc)^{2}bab)^{13}(ab)^{−4} = ±[2, 9] and (b. ± [2, 9])^{3} = ±[1, 2]. While for n = 14, d = ((ba)^{2}(ca)^{2}bc)^{15} = ±[1, 5][7, 13] and f = (abcedb)^{9} = ±[5, 6], from which we obtain (abc)^{4}f(abc)^{10} = ±[1, 2]. □

It is convenient at this point to introduce some notation.

Definition 2.8. We will denote the following product of δ + 1 signed transpositions [α, β][α + 1, β − 1][α + 2, β − 2] … [α + δ, β − δ] by T (α, α + δ, α + β).

As an example of this notation, in the previous proposition, we could express the generators for S̃_{16} as

a = T (2, 8, 18), b = [1, 2]T (3, 9, 19) and c = T (3, 9, 19).

Proposition 2.9. For n ≥ 17, the following involutions, which are dependent on the value of n modulo 8, generate S̃_{n}.

n ≡ 1  a = [1, 2]T (3, (n + 1)/2, n + 3),
b = [1, 3]T (4, (n + 3)/2, n + 4),
c = T (3, (n + 1)/2, n + 3).
n ≡ 2  a = [1, 2]T (3, n/2, n + 2),
b = [1, 3]T (4, (n + 2)/2, n + 3),
c = [3, n]T (4, n/2, n + 2).
n ≡ 3  a = [1, 2]T (3, (n − 1)/2, n + 1),
b = [1, 3]T (4, (n + 1)/2, n + 2),
c = [3, n][4, n − 1]T (5, (n − 1)/2, n + 1).
n ≡ 4  a = [1, 2]T (3, (n − 2)/2, n),
b = [1, 3]T (4, n/2, n + 1),
c = T (3, 5, n + 3)T (6, (n − 2)/2, n).
n ≡ 5  a = [1, 2]T (3, (n − 3)/2, n − 1),
b = [1, 3]T (4, (n − 1)/2, n),
c = T (3, 6, n + 3)T (7, (n − 3)/2, n − 1).
n ≡ 6  a = [1, 2]T (3, (n − 4)/2, n − 2),
b = [1, 3]T (4, (n − 2)/2, n − 1),
c = T (3, 7, n + 3)T (8, (n − 4)/2, n − 2).
n ≡ 7  a = [1, 2]T (3, (n − 5)/2, n − 3),
b = [1, 3]T (4, (n − 3)/2, n − 2),
c = T (3, 8, n + 3)T (9, (n − 5)/2, n − 3).
n ≡ 0  a = [1, 2]T (3, (n − 6/2, n − 4),
b = [1, 3]T (4, (n − 4)/2, n − 3),
c = T (3, 9, n + 3)T (10, (n − 6)/2, n − 4).

Note that when n = 24 we have (n − 6)/2 < 10. So we define T (10, (n − 6)/2, n − 4) to be 1.

Proof. For n ≢ 1 we have (ac)^{3} = ±[1, 2] and when n ≡ 1, ac = ±[1, 2]. Hence we only need show that ±[1, 2, …, n] is also generated in each case. When n ≡ 1, we have directly that ab = ±[1, 2, …, n]. For the remaining values of n we note that when

n ≡ 2 ab = ±[1, 2, …, n − 1] so S̃_{n−1} is generated,
n ≡ 3 ab = ±[1, 2, …, n − 2] so S̃_{n−2} is generated,
n ≡ 4 ab = ±[1, 2, …, n − 3] so S̃_{n−3} is generated,
n ≡ 5 ab = ±[1, 2, …, n − 4] so S̃_{n−4} is generated,
n ≡ 6 ab = ±[1, 2, …, n − 5] so S̃_{n−5} is generated,
n ≡ 7 ab = ±[1, 2, …, n − 6] so S̃_{n−6} is generated.

In particular, as n ≥ 17, we may generate for each n the signed cycles d = [1, 9], e = [1, 8], f = [1, 7], g = [1, 6], h = [1, 5], j = [1, 4], k = [1, 3]. Thus we may obtain ±[1, 2, …, n] from abckc when n ≡ 2, abckjc when n ≡ 3, abchjkc when n ≡ 4, abcghjkc when n ≡ 5, abcfghjkc when n ≡ 6, abcefghjkc when n ≡ 7 and abcdefghjkc when n ≡ 0. □

The previous propositions give directly the following theorem.

Theorem 2.10. For n ≥ 7 and n ≠ 13 i(S̃_{n}) = 3.

3. i(Ŝ_{n})

We denote by Ŝ_{n} the double cover of S_{n} that lifts a transposition of S_{n} to an element of order 2. Hence Ŝ_{n} is the group with generators z, r_{1}, r_{2}, …, r_{n−1} and relations

z^{2} = 1,
zr_{i} = r_{i}z, r_{i}^{2} = 1 for 1 ≤ i ≤ n − 1,
(r_{j}r_{j+1})^{3} = 1 for 1 ≤ j ≤ n − 2,
r_{k}r_{h} = zr_{h}r_{k} for | h − k |> 1 and 1 ≤ h, k ≤ n − 1.

For computations in Ŝ_{n} we multiply each of the generators of the Clifford Algebra C(Ω), where Ω = {1, 2, …, n} ⋃{δ} (see [5]), by the complex number i to obtain an algebra over C generated by f_{1}, f_{2}, … f_{n}, f_{δ} where f_{j}^{2} = 1 and f_{j}f_{k} = −f_{k}f_{j} for j ≠ k. The subgroup of this complex algebra generated by

(f_{1} − f_{2})/\sqrt{2}, (f_{2} − f_{3})/\sqrt{2}, …, (f_{n} − f_{δ})/\sqrt{2}

is isomorphic to Ŝ_{n}.

By identifying D_{a_{j}} with (f_{j} − f_{δ})/\sqrt{2}, where a_{j} are distinct in Ω \ δ, the following two relations are readily verified.

D_{a_{j}}D_{a_{j}} = 1 and D_{a_{1}}D_{a_{2}} … D_{a_{m}}D_{a_{1}} = (−1)^{m+1}D_{a_{2}}D_{a_{3}} … D_{a_{m}}D_{a_{1}}D_{a_{2}}.

Now, misusing notation, as in [5], we may write these as

a_{j}a_{j} = 1 and a_{1}a_{2} … a_{m}a_{1} = (−1)^{m+1}a_{2}a_{3} … a_{m}a_{1}a_{2}.

We are now in a position to define a signed cycle in Ŝ_{n}.

Definition 3.1. For distinct elements a_{1}, …, a_{m} we define ⟨a_{1}a_{2} … a_{m}⟩ = a_{1}a_{2} … a_{m}a_{1}. We call ± ⟨a_{1}a_{2} … a_{k}⟩ signed cycles in Ŝ_{n}. Each is a lift of the cycle (a_{1}a_{2} … a_{k}) in S_{n}.

Using this definition with the above relations we obtain the following rules for multiplying signed cycles in Ŝ_{n}.

⟨a_{j}⟩ = 1,
⟨a_{1}a_{2} … a_{m}⟩ = (−1)^{m+1} ⟨a_{2}a_{3} … a_{m}a_{1}⟩ ,
⟨a_{1}a_{2} … a_{m−1}⟩ a_{m} = (−1)^{m}a_{m} ⟨a_{1}a_{2} … a_{m−1}⟩ .

As an example of the multiplication of signed cycles we recover our presentation for Ŝ_{n} as follows: Let r_{1} = ⟨1, 2⟩ , r_{2} = ⟨2, 3⟩ , …, r_{n−1} = ⟨n − 1, n⟩ and z = −1. Clearly z^{2} = 1 and

zr_{j} = − ⟨j, j + 1⟩ = r_{j}z, for 1 ≤ j ≤ n − 1,
r_{j}^{2} = ⟨j, j + 1⟩ ⟨j, j + 1⟩ = j, j + 1, j, j, j + 1, j = 1,
(r_{j}r_{j+1})^{3} = (⟨j, j + 1⟩ ⟨j + 1, j + 2⟩)^{3} = (− ⟨j + 1, j⟩ ⟨j + 1, j + 2⟩)^{3}
= (−j + 1, j, j + 2, j + 1)^{3} = −j + 1 ⟨j, j + 2⟩ ⟨j + 2, j⟩ j + 1
= j + 1 ⟨j + 2, j⟩ ⟨j + 2, j⟩ j + 1
= j + 1, j + 1 = 1 for 1 ≤ j ≤ n − 2

and

r_{j}r_{k} = ⟨j, j + 1⟩ ⟨k, k + 1⟩ = − ⟨k, k + 1⟩ ⟨j, j + 1⟩
= zr_{k}r_{j} for | j − k |> 1 and 1 ≤ j, k ≤ n − 1.

Note that, for n ≥ 4, Ŝ_{n} ≅ ˜S̸_{n} if n ≠ 6, see [3]. Also note that we will again be taking advantage of the fact that by factoring out Ŝ_{n} by the subgroup Z = ⟨1, −1⟩ we recover S_{n}. That is Ŝ_{n}/Z ≅ S_{n}.

Proposition 3.2. Any product of k disjoint signed transpositions in Ŝ_{n} has order two if the integer part of k/2 is a multiple of two, and order four otherwise.

Proof. Let S denote a product of k disjoint signed transpositions so that we have, for k ≥ 2,

S = ± ⟨a_{1}, b_{1}⟩ ⟨a_{2}, b_{2}⟩ … ⟨a_{k}, b_{k}⟩ ,

where a_{i} and b_{i} are distinct integers in the signed transpositions of Ŝ_{n}. Then a straight forward induction proof gives, for k > 1,

S^{2} = (−1)^{k−1}(−1)^{k−2} … (−1)^{2}(−1).

While for k = 1, ⟨a_{1}, b_{1}⟩ ⟨a_{1}, b_{1}⟩ = 1. □

So in Ŝ_{n} an involution is of the form ±π where π is a signed cycle consisting of k disjoint transpositions and k is congruent to 0 or 1 modulo 4.

Proposition 3.3. For n ≥ 4, Ŝ_{n} may be generated by a = ± ⟨1, 2⟩ and b = ± ⟨1, 2, …, n⟩ .

Proof. As in Proposition 2.3 we need only show that −1 is generated. Now in Ŝ_{n} we may generate either

g = + ⟨1, 2⟩ ⟨3, 4⟩ or h = − ⟨1, 2⟩ ⟨3, 4⟩ ,

but in either case g^{2} = h^{2} = −1. □

The investigation into the value of i(Ŝ_{n}) closely follows that of the previous section, except that here involutions are products of k signed transpositions for k ≡ 0 or 1 modulo 4. So in particular ± ⟨1, 2⟩ is an involution in this double cover.

Hence in Ŝ_{5}, Ŝ_{6} and Ŝ_{7} the only elements of order two are −1 and signed transpositions of the form ± ⟨a, b⟩. Thus we have immediately that i(Ŝ_{5}) = 4, i(Ŝ_{6}) = 5 (which, as Ŝ_{6} ≅ S̃_{6}, implies i(Ŝ_{6}) = 5) and i(Ŝ_{7}) = 6, generators being ⟨1, 2⟩ , …, ⟨1, n⟩. Also, as the only involutions in Ŝ_{8} are −1 and signed cycles of type 2^{1} and 2^{4}, it is readily verified (via GAP [4]) that i(Ŝ_{8}) > 3. But as

⟨1, 2⟩ , ⟨1, 6⟩ , ⟨1, 2⟩ ⟨3, 8⟩ ⟨4, 7⟩ ⟨5, 6⟩ and ⟨1, 6⟩ ⟨2, 3⟩ ⟨4, 8⟩ ⟨5, 7⟩

generate Ŝ_{8} we have i(Ŝ_{8}) = 4.

Proposition 3.4. For n ≥ 9, when n ≡ 1, 2 or 3 modulo 8, Ŝ_{n} may be generated by three involutions.

Proof. We only need apply the decomposition referred to in the introduction to see that we may express ± ⟨1, 2, …, n⟩ as the product ab where

a = ⟨1, n − 1⟩ ⟨2, n − 2⟩ … ⟨r, n − r⟩ , b = ⟨1, n⟩ ⟨2, n − 1⟩ … ⟨t, n + 1 − t⟩ .

Hence if we take a, b along with ⟨1, 2⟩ the result follows. □

Proposition 3.5. For 12 ≤ n ≤ 16, the following involutions generate Ŝ_{n}. When n = 16

a = ⟨1, 15⟩ ⟨2, 14⟩ ⟨3, 13⟩ ⟨6, 10⟩ ⟨7, 9⟩
b = ⟨4, 12⟩ ⟨5, 11⟩ ⟨7, 10⟩ ⟨2, 15⟩ ⟨3, 14⟩
c = ⟨4, 13⟩ ⟨5, 12⟩ ⟨6, 11⟩ ⟨1, 16⟩ ⟨8, 9⟩ .

For 12 ≤ n ≤ 15

a = ⟨1, 2⟩ ⟨3, 11⟩ ⟨4, 10⟩ ⟨5, 9⟩ ⟨6, 8⟩ , b = ⟨1, 3⟩ ⟨4, 11⟩ ⟨5, 10⟩ ⟨6, 9⟩ ⟨7, 8⟩

and c = \{ \begin{matrix}⟨3, 12⟩ ⟨4, 10⟩ ⟨5, 9⟩ ⟨6, 8⟩ & when & n = 12, \\ ⟨3, 12⟩ ⟨4, 13⟩ ⟨5, 9⟩ ⟨6, 8⟩ & when & n = 13, \\ ⟨3, 12⟩ ⟨4, 13⟩ ⟨5, 14⟩ ⟨6, 8⟩ & when & n = 14, \\ ⟨3, 12⟩ ⟨4, 13⟩ ⟨5, 14⟩ ⟨6, 15⟩ & when & n = 15. \end{matrix}

Proof. For n = 16 we have abc = ± ⟨1, 2, …, 16⟩ so we need only show that ⟨1, 2⟩ is also generated. But

(abc)^{6}((c(ab)^{4}cb)^{3}(b(abcb)^{3}b)^{3})^{3}(abc)^{−6} = ± ⟨1, 4⟩ and
((abc)^{2}b(abc)^{3})^{2} ± ⟨1, 4⟩ ((abc)^{2}b(abc)^{3})^{−2} = ± ⟨1, 2⟩ .

When n = 12, 13, 14 or 15, (ac)^{3} = ± ⟨1, 2⟩ and ab = ± ⟨1, 2, …, 11⟩ so for each n we may generate the subgroup Ŝ_{11}, in particular the signed cycles d = ⟨1, 3⟩, e = ⟨1, 4⟩, f = ⟨1, 5⟩ and g = ⟨1, 6⟩.

Hence we need only show that ⟨1, 2, …, n⟩ can also be generated for each n. But when n = 12 we have abcdc = ± ⟨1, 2, …, 12⟩. When n = 13, abcdec = ± ⟨1, 2, …, 13⟩. When n = 14, abcdefc = ± ⟨1, 2, …, 14⟩ and when n = 15, abcdefgc = ± ⟨1, 2, …, 15⟩. □

We again make use of the notation T (α, α + δ, α + β), as in the previous section, but here this represents the δ + 1 signed transpositions

⟨α, β⟩ ⟨α + 1, β − 1⟩ ⟨α + 2, β − 2⟩ … ⟨α + δ, β − δ⟩ .

Proposition 3.6. For n ≥ 17, the following involutions, which are dependent on the value of n modulo 8, generate Ŝ_{n}.

n ≡ 0  a = ⟨1, 2⟩ T (3, (n − 4)/2, n − 2),
b = ⟨1, 3⟩ T (4, (n − 2)/2, n − 1),
c = T (3, 7, n + 3)T (8, (n − 4)/2, n − 2).
n ≡ 4  a = ⟨1, 2⟩ T (3, n/2, n + 2),
b = ⟨1, 3⟩ T (4, (n + 2)/2, n + 3),
c = ⟨3, n⟩ T (4, n/2, n + 2).
n ≡ 5  a = ⟨1, 2⟩ T (3, (n − 1)/2, n + 1),
b = ⟨1, 3⟩ T (4, (n + 1)/2, n + 2),
c = ⟨3, n⟩ ⟨4, n − 1⟩ T (5, (n − 1)/2, n + 1).
n ≡ 6  a = ⟨1, 2⟩ T (3, (n − 2)/2, n),
b = ⟨1, 3⟩ T (4, n/2, n + 1),
c = T (3, 5, n + 3)T (6, (n − 2)/2, n).
n ≡ 7  a = ⟨1, 2⟩ T (3, (n − 3)/2, n − 1),
b = ⟨1, 3⟩ T (4, (n − 1)/2, n),
c = T (3, 6, n + 3)T (7, (n − 3)/2, n − 1).

Proof. For each n here we have (ac)^{3} = ± ⟨1, 2⟩, so we only need show that ± ⟨1, 2, …, n⟩ is also generated in each case. Now for

n ≡ 4 ab = ± ⟨1, 2, …, n − 1⟩ so Ŝ_{n−1} is generated,
n ≡ 5 ab = ± ⟨1, 2, …, n − 2⟩ so Ŝ_{n−2} is generated,
n ≡ 6 ab = ± ⟨1, 2, …, n − 3⟩ so Ŝ_{n−3} is generated,
n ≡ 7 ab = ± ⟨1, 2, …, n − 4⟩ so Ŝ_{n−4} is generated,
n ≡ 0 ab = ± ⟨1, 2, …, n − 5⟩ so Ŝ_{n−5} is generated.

In particular, as n ≥ 17, we may generate for each n the signed cycles d = ⟨1, 7⟩, e = ⟨1, 6⟩, f = ⟨1, 5⟩, g = ⟨1, 4⟩ and h = ⟨1, 3⟩. Thus we may obtain ± ⟨1, 2, …, n⟩ from abchc when n ≡ 4, abcghc when n ≡ 5, abcfghc when n ≡ 6, abcefghc when n ≡ 7 and abcdefghc when n ≡ 0. □

These last three propositions, along with the fact that (1, 2), (1, 3) and (1, 4) generate S_{4}, give directly the following theorem.

Theorem 3.7. For n = 4 and n ≥ 9, i(Ŝ_{n}) = 3.

Acknowledgements. This paper is an extension of a portion of a thesis which the author undertook while being supported by the SERC and supervised by Dr. J. F. Humphreys, to whom the author is greatly indebted, at Liverpool University.

References

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[4] Schönert, M. et.al.: GAP – Groups, Algorithms, and Programming. Lehrstuhl D für Mathematik, Rheinisch-Westfälische Technische Hochschule Aachen, Germany, fifth edition 1995.

[5] Wales, D. B.: Some projective representations of S_{n}. J. Algebra 61 (1979), 37–57.

Received November 4, 1998