EXCEPTIONAL SUBGROUPS OF SL(2, F)

DARRYL MCCULLOUGH

Abstract. For A, B ∈ SL(2, ℂ), R. C. Churchill gave simple conditions in terms of tr(A), tr(B), and tr(AB) to determine whether the image of the subgroup ⟨A, B⟩ in PSL(2, ℂ) is isomorphic to A_{4}, S_{4}, or A_{5}. We prove that the same conditions work for SL(2, F) for any field F, except when the image is A_{5} and F has characteristic 3, 11, 19, or 29.

Introduction

Some work of R. C. Churchill [1] includes a characterization of twogenerator subgroups of SL(2, ℂ) that generate a subgroup of PSL(2, ℂ) isomorphic to either A_{4}, S_{4}, or A_{5}. It may be useful to have an extension of this characterization to SL(2, F) for other F. To this end, drawing considerably on Churchill’s ideas, we will prove that it extends to all fields, except for certain cases in characteristics p = 3, 11, 19, and 29.

To set up the statement of the main result, fix a field F of characteristic p, possibly 0. Replace F by its algebraic closure, if necessary, to assume that it is algebraically closed. Write \sqrt{2} for one of the nonzero solutions of x^{2} = 2, which exist only when p ≠ 2. Write μ_{1} for one of the solutions of x^{2}−x−1 = 0, and put μ_{2} = μ_{1}^{−1} = μ_{1}−1. Note that −μ_{2} is the other solution of x^{2} − x − 1 = 0, and that μ_{2} and −μ_{1} are the solutions of x^{2} + x − 1 = 0. Define T_{2} = \{0\}, T_{3} = \{±1\}, T_{4} = \{±\sqrt{2}\}, and T_{5} = \{±μ_{1}, ±μ_{2}\}. It is an elementary fact (see section 1) that for 2 ≤ n ≤ 5 and A ∈ SL(2, F) (A ≠ ±I), A has order n in PSL(2, F) if and only tr(A) ∈ T_{n}. In particular, if at least two of tr(A), tr(B), and tr(AB) are 0, then the subgroup that A and B generate in PSL(2, F) is dihedral or is a subgroup of C_{2} × C_{2}.

Our Main Theorem involves the trace of the commutator [A, B] of two elements of SL(2, F). It can be expressed in terms of tr(A), tr(B), and tr(AB) using the well-known Fricke trace identity

tr([A, B]) = tr^{2}(A) + tr^{2}(B) + tr^{2}(AB) − tr(A) tr(B) tr(AB) − 2 .

Finally, write ⟨A, B⟩ for the subgroup of SL(2, F) generated by A and B, and P: SL(2, F) → PSL(2, F) for the canonical projection. In the statement of the Main Theorem and throughout our work, two groups are considered to be equal whenever they are isomorphic.

Date: June 14, 2005.

2000 Mathematics Subject Classification. Primary 20G15; Secondary 20F29.

Key words and phrases. SL, projective, tetrahedral, octahedral, icosahedral, alternating, finite, trace, commutator.

Main Theorem. Let A, B ∈ SL(2, F), with no more than one of tr(A), tr(B), or tr(AB) equal to 0. Then

1) P (⟨A, B⟩) = A_{4} if and only if tr(A), tr(B), tr(AB) ∈ T_{2} ∪ T_{3} and tr([A, B]) = 0.

2) P (⟨A, B⟩) = S_{4} if and only if tr(A), tr(B), tr(AB) ∈ T_{2} ∪ T_{3} ∪ T_{4} and tr([A, B]) = 1.

3) Apart from the exceptional cases listed below, P (⟨A, B⟩) = A_{5} if and only if tr(A), tr(B), tr(AB) ∈ T_{2} ∪ T_{3} ∪ T_{5} and tr([A, B]) ∈ {−μ_{2}, 1, μ_{1}}. In each of the following exceptional cases, one can find examples of pairs A, B with tr(A), tr(B), tr(AB) ∈ T_{2} ∪T_{3} ∪T_{5}, where P (⟨A, B⟩) does equal A_{5}, and where it does not equal A_{5}:

(a) p ∈ {3, 11, 29} and tr([A, B]) ∈ {μ_{1}, −μ_{2}}.

(b) p = 19 and tr([A, B]) = 1.

In the exceptional cases, it is easy to tell by examination of tr(A), tr(B), and tr(AB) whether or not P (⟨A, B⟩) = A_{5}. The method requires information from the proof of the Main Theorem, so we will defer its explanation until the very end of this paper.

Our proof of the Main Theorem uses only elementary trace identities and calculations, with occasional assistance from the GAP package of computer algorithms [2]. Unfortunately, a large number of cases must be considered; we hope that our presentation has organized them in a way that minimizes the pain.

Throughout, we freely use the elementary trace identity tr(AB^{−1}) + tr(AB) = tr(A) tr(B) and its consequences (such as tr(A^{2}) = tr(A)^{2} − 2 and the Fricke trace identity).

1. Preliminaries

The following facts about elements A ∈ SL(2, F) (A ≠ ±I) are easily checked:

1) A^{2} = I if and only if p = 2 and tr(A) = 0.

2) A^{2} = −I if and only if tr(A) = 0.

3) A^{3} = ±I if and only if tr(A) = ∓1.

4) A^{4} = −I if and only if tr(A)^{2} = 2.

5) A^{5} = ±I if and only if tr(A) = ∓μ_{1} or tr(A) = ±μ_{2}. These show that P (A) has order n if and only if tr(A) ∈ T_{n}.

We will use the following, which is part of Proposition 8.1 of [1].

Proposition 1.1. In the following, g and h denote elements of a certain group, for which g ∉ ⟨h⟩ and h ∉ ⟨g⟩. Then

1) If g, h ∈ A_{4}, then ⟨g, h⟩ = A_{4} if and only if at least one of g and h has order 3.

2) If g, h ∈ S_{4}, then ⟨g, h⟩ = S_{4} if and only if at least one of g, h, and gh has order 4, and at most one has order 2.

3) If g, h ∈ A_{5}, then ⟨g, h⟩ = A_{5} if and only if at least one of g, h, and gh has order 5, and at most one has order 2.

It will be convenient to use Nielsen equivalence. The Nielsen group for pairs is

𝕌 = C_{2} ∗ C_{2} ∗ C_{2} = ⟨t, u, v | t^{2}, u^{2}, v^{2}⟩ .

If K is any group, 𝕌 acts on the set of pairs K × K as follows.

1. t(A, B) = (A^{−1}, B)
2. u(A, B) = (B, A)
3. v(A, B) = (A^{−1}, AB)

The orbits of the action of 𝕌 on K ×K are called Nielsen equivalence classes of pairs. Put more concretely, pairs are Nielsen equivalent when one can be obtained from the other by any sequence of operations of replacing an element by its inverse, interchanging the two elements of the pair, or preor post-multiplying one element by the other, or by the inverse of the other. Note that Nielsen equivalent pairs generate the same subgroup of K.

An F -triple is an element of F × F × F. The Nielsen group acts on the set of F -triples as follows:

1) t(α, β, γ) = (α, β, αβ − γ)

2) u(α, β, γ) = (β, α, γ)

3) v(α, β, γ) = (α, γ, β)

This action is induced from the action of 𝕌 on SL(2, F) × SL(2, F) via the function Tr defined by Tr(A, B) = (tr(A), tr(B), tr(AB)); that is, if Tr(A, B) = (α, β, γ), then t ∘ Tr = Tr ∘t, u ∘ Tr = Tr ∘u, and v ∘ Tr = Tr ∘v. For u and v this is obvious, and for t it is simply the identity that tr(A^{−1}B) = tr(A) tr(B)−tr(AB). We declare two F -triples to be equivalent if they lie in the same 𝕌-orbit. In particular, permuting the entries of a triple produces an equivalent triple.

From now on, we assume that K = SL(2, F). For this case, we further declare (A, B) to be equivalent to each of (−A, B), (A, −B), and (−A, −B). Via Tr, these induce the further equivalences of (α, β, γ) with each of (−α, β, −γ), (α, −β, −γ), and (−α, −β, γ). Both equivalence relations are denoted by the symbol ∼.

For notational convenience, we define the Fricke polynomial Q: F^{3} → F by

Q(α, β, γ) = α^{2} + β^{2} + γ^{2} − α β γ − 2 ,

so that tr([A, B]) = Q(tr(A), tr(B), tr(AB)).

It is straightforward to check that if A, B ∈ SL(2, F) and (A, B) ∼ (A′, B′), then P (⟨A, B⟩) = P (⟨A′, B′⟩). Moreover, tr([A, B]) = tr([A′, B′]), and consequently

Q(tr(A), tr(B), tr(AB)) = Q(tr(A′), tr(B′), tr(A′B′)) .

Thus in the proof of the Main Theorem, we may replace (A, B) by any equivalent pair, or (tr(A), tr(B), tr(AB)) by any equivalent triple, while assuming that P (⟨A, B⟩) and tr([A, B]) are unchanged.

2. Proof of the “only if” direction of the Main Theorem

In the course of the proof, we will develop some additional information about generating pairs of exceptional groups. Since some of it could be of independent interest, we will organize most of the argument as a pair of lemmas and a corollary. The lemmas are certainly well-known; in particular Nielsen equivalence for generating pairs of A_{5} was first analyzed in [4].

Lemma 2.1. Let H be A_{4} or S_{4}. Then any two generating pairs of H are Nielsen equivalent. Consequently:

1) Any generating pair of A_{4} is Nielsen equivalent to a pair (g, h) for which g^{3} = h^{3} = (gh)^{2} = 1.

2) Any generating pair of S_{4} is Nielsen equivalent to a pair (g, h) for which g^{4} = h^{3} = (gh)^{2} = 1.

Proof. For A_{4}, we may assume using proposition 1.1(a) that g has order 3, and by conjugation (which can be achieved by Nielsen equivalence) and possibly replacing g by g^{−1}, we may assume that g = (123). If h has order 2, replace it by gh which has order 3. Conjugating h by powers of g and possibly replacing h by h^{−1}, we may assume that h = (431). That is, every generating pair is Nielsen equivalent to ((123), (431)), which obey the relations in (a).

For S_{4}, we may similarly assume using proposition 1.1(b) that g = (1234). Not both h and gh can have order 4, since S_{4}/A_{4} ≅ C_{2}, so we may assume that h has order 3. Conjugating h by powers of g and possibly replacing h by h^{−1}, we may assume that h = (321). That is, every generating pair is Nielsen equivalent to ((1234), (321)), which obey the relations in (b). □

Lemma 2.2. Let H be A_{5}. Then any generating pair of H is Nielsen equivalent to exactly one of the pairs ((12345), (124)), ((12345), (12354)), or ((12354), (125)).

Proof. By proposition 1.1, we may assume that g is a 5-cycle. By conjugation we may make it one of (12345) or (12354); assume for now that it is the first. Using lemma 8.2 of [1], we may assume that h is not a 5-cycle.

Suppose first that h has order 2. Conjugating h by g, we may assume that h is either (12)(34), (13)(24), or (14)(23). The latter case is impossible since then, gh has order 2 and ⟨g, h⟩ is dihedral. In the first two cases, gh and respectively g^{2}h are 3-cycles. Replacing the pair by (g, gh) or by (g, g^{2}h), we may assume that h is a 3-cycle. Conjugating h by g, and possibly replacing h by h^{−1}, we may assume that h is either (123) or (124). If h = (124), we have the first pair, while if h = (123), then (g, g^{3}hg^{3}) is the second pair.

For the case when g is conjugate to (12354), a very similar argument (e. g. just conjugate everything in the previous argument by (45)) shows that (g, h) is Nielsen equivalent to the second or third pair.

To see that these three pairs are not Nielsen equivalent, regard A_{5} as SL(2, 4). We have {μ_{1}, μ_{2}} = 𝔽_{4} − {0, 1}, and we choose notation so that the matrix corresponding to (12345) has trace μ_{1} and the one corresponding to (12354) has trace μ_{2}. Then,

1) If (A, B) ∈ SL(2, 4) × SL(2, 4) corresponds to ((12345), (124)), then tr([A, B]) = Q(μ_{1}, 1, 0) = μ_{1}^{2} + 1 = μ_{1}.

2) If (A, B) ∈ SL(2, 4)×SL(2, 4) corresponds to ((12345), (12354)), then tr([A, B]) = Q(μ_{1}, μ_{2}, 0) = μ_{1}^{2} + μ_{2}^{2} = 1.

3) If (A, B) ∈ SL(2, 4) × SL(2, 4) corresponds to ((12354), (125)), then tr([A, B]) = Q(μ_{2}, 1, 0) = μ_{2}^{2} + 1 = μ_{2}.

Since tr([A, B])) is an invariant of the Nielsen equivalence class, no two of the three pairs are Nielsen equivalent. □

Corollary 2.3. Let A, B ∈ SL(2, F), and suppose that P (⟨A, B⟩) = A_{5}. Then Q(tr(A), tr(B), tr(AB)) ∈ {μ_{1}, 1, −μ_{2}}. Moreover,

(a) (A, B) is Nielsen equivalent to a pair for which P (A)^{5} = P (B)^{3} = P (AB)^{2} = P (I) if and only if tr([A, B]) ∈ {μ_{1}, −μ_{2}}.

(b) (A, B) is Nielsen equivalent to a pair for which P (A)^{5} = P (B)^{5} = P (AB)^{2} = P (I) and P (A) is not conjugate to P (B) if and only if tr([A, B]) = 1.

Proof. If we can change (A, B) by equivalence so that P (A)^{5} = P (B)^{3} = P (AB)^{2} = P (I), then the associated traces are (±μ_{1}, ±1, 0) or (±μ_{2}, ±1, 0), and we find that tr([A, B]) = μ_{1} and tr([A, B]) = −μ_{2} respectively. If not, then by lemma 2.2 we may assume that P (A)^{5} = P (B)^{5} = P (AB)^{2} = P (I) with P (A) and P (B) not conjugate in A_{5}. Since F is algebraically closed, elements of SL(2, F) having the same trace are conjugate provided that the trace is not ±2. If p ≠ 5, then μ_{1} and μ_{2} are not ±2, and it follows that one of tr(A) or tr(B) is ±μ_{1} and the other is ±μ_{2}. Therefore tr([A, B]) = μ_{1}^{2} + μ_{2}^{2} − 2 = μ_{1} − μ_{2} = 1. If p = 5, then ±μ_{1} = ±μ_{2} = ±2, and tr([A, B]) = 4 + 4 − 2 = 1 for all possibilities. □

We remark that μ_{1} = −μ_{2} if and only if p = 5 (in which case both equal 3), so tr([A, B]) fails to distinguish the two Nielsen classes in part (a) if and only if p = 5.

Now we can deduce the only if direction of the Main Theorem. If P (⟨A, B⟩) is one of A_{4}, S_{4}, or A_{5}, then by consideration of the orders of elements in these three groups, it is immediate that the traces of A and B lie in the specified sets.

If P (⟨A, B⟩) = A_{4}, then P ([A, B]) lies in the commutator subgroup of A_{4}, which consists of involutions, so tr([A, B]) = 0.

If P (⟨A, B⟩) = S_{4}, then by lemma 2.1 we may change (A, B) by Nielsen equivalence to assume that P (A)^{4} = P (B)^{3} = P (AB)^{2} = P (I), in which case tr([A, B]) = Q(±\sqrt{2}, ±1, 0) = 1.

If P (⟨A, B⟩) = A_{5}, then corollary 2.3 completes the proof.

3. Proof of the “if” direction of the Main Theorem

In this section, we write α, β, and γ for tr(A), tr(B), and tr(AB) respectively.

Assume first that the conditions in part 1) hold. Since no two of α, β, and γ can be 0, we may use equivalence of triples to assume that α = β = 1. The condition that Q(1, 1, γ) = tr([A, B]) = 0 then forces γ = 0 or γ = 1. If γ = 0, then P (A)^{3} = P (B)^{3} = P (AB)^{2} = I so P (⟨A, B⟩) = A_{4}. If γ = 1, then (1, 1, 1) ∼ (1, 1, 0) and again P (⟨A, B⟩) = A_{4}.

Assume now that the conditions in part 2) hold. We may assume using proposition 1.1 that α = \sqrt{2}. Since no two of α, β, and γ are 0, we may assume that β is \sqrt{2} or 1.

Suppose first that β = \sqrt{2}, so (α, β, γ) is of the form (\sqrt{2}, \sqrt{2}, γ). The five possible values (−\sqrt{2}, −1, 0, 1, \sqrt{2}) of γ lead to the corresponding values (4+2\sqrt{2}, 5, 2, 1, 4−2\sqrt{2}) for Q(\sqrt{2}, \sqrt{2}, γ). Putting these equal to 1 leads to immediate contradictions (in particular, Q = 5 implies that p = 2 and hence that α = β = 0) except in the case when γ = 1. So (α, β, γ) = (\sqrt{2}, \sqrt{2}, 1) ∼ (\sqrt{2}, 1, \sqrt{2}) ∼ (\sqrt{2}, 1, 0) which gives P (A)^{4} = P (B)^{3} = P (AB)^{2} = P (I), so P (⟨A, B⟩) = S_{4}.

Suppose now that β = 1, so (α, β, γ) is of the form (\sqrt{2}, 1, γ), where we may assume that γ ≠ \sqrt{2}. We find that 1 = Q(\sqrt{2}, 1, γ) = γ^{2} − \sqrt{2}γ + 1, so γ = 0 and again we conclude that P (⟨A, B⟩) = S_{4}.

Finally, assume that the conditions in part 3) hold. By proposition 1.1 we may assume that α ∈ {μ_{1}, μ_{2}}. If α = ±μ_{2}, then we change our notation so that μ_{1} becomes the other root −μ_{2} of x^{2} − x − 1. So α becomes ∓μ_{1}, and after possibly applying an equivalence to (α, β, γ), we may assume that α = μ_{1}. We will analyze the values of Q(μ_{1}, β, γ) for all possible cases.

Case I. (α, β, γ) = (μ_{1}, μ_{1}, γ)

The subcases where Q(μ_{1}, μ_{1}, γ) has each of the allowable values of μ_{1}, 1, −μ_{2} are indicated by the letters A, B, and C respectively. There are further subcases corresponding to the seven possible values (−μ_{1}, −μ_{2}, −1, 0, 1, μ_{2}, μ_{1}) for γ, and in our notation for cases we indicate them by the corresponding numerals 1 through 7. Thus, for example, the case when (α, β, γ) = (μ_{1}, μ_{1}, −μ_{2}) and Q(μ_{1}, μ_{1}, −μ_{2}) = μ_{1} is case IA2.

We calculate that Q(μ_{1}, μ_{1}, γ) = γ^{2} − μ_{1}γ − γ + 2μ_{1}. The corresponding values of Q(μ_{1}, μ_{1}, γ) are (5μ_{1} + 2, 2μ_{1} + 2, 3μ_{1} + 2, 2μ_{1}, μ_{1}, 2, μ_{1}). Thus in case IA2, we additionally know that (μ_{1}, μ_{1}, γ) = 2μ_{1} + 2, so we obtain the equality 2μ_{1} + 2 = μ_{1}. In our analysis of the various cases, we use the equality so obtained either to reach a contradiction showing that the subcase can never occur (as in Case IA4), or we show that the equality is an identity and hence always holds (as in Case IA5), or we show that it can only hold under certain conditions on p and μ_{1} (as in Case IB1). In analyzing the cases, one frequently uses the observation that if mμ_{1} = n, then n^{2} = m^{2}μ_{1}^{2} = m^{2}μ_{1} + m^{2} = mn + m^{2}.

Case IA1. Q(μ_{1}, μ_{1}, −μ_{1}) = μ_{1}. From 5μ_{1} + 2 = μ_{1} and the observation, we find that p = 2. This case is contained in case IA7 below. Case IA2. Q(μ_{1}, μ_{1}, −μ_{2}) = μ_{1}. From 2μ_{1} + 2 = μ_{1} and the observation, we find that this holds if and only if p = 5. But then, −μ_{2} = μ_{1}, so this case is contained in case IA7. Case IA3. Q(μ_{1}, μ_{1}, −1) = μ_{1}. We find that p = 2. This case is contained in case IA5 below. Case IA4. Q(μ_{1}, μ_{1}, 0) = μ_{1}. This gives the contradiction μ_{1} = 0. Case IA5. Q(μ_{1}, μ_{1}, 1) = μ_{1}. This holds for all p. Case IA6. Q(μ_{1}, μ_{1}, μ_{2}) = μ_{1}. This gives the contradiction that 0 = 1. Case IA7. Q(μ_{1}, μ_{1}, μ_{1}) = μ_{1}. This holds for all p. Case IB1. Q(μ_{1}, μ_{1}, −μ_{1}) = 1. This holds if and only if (p, μ_{1}) = (19, −4). For from 5μ_{1} +2 = 1 the observation gives −5+25 = 1 so p must be 19, and solving the linear equation gives μ_{1} = −4. And when p = 19, Q(−4, −4, 4) is indeed 1. Case IB2. Q(μ_{1}, μ_{1}, −μ_{2}) = 1. This gives the contradiction 0 = 1. Case IB3. Q(μ_{1}, μ_{1}, −1) = 1. This holds if and only if p = 5. But then, (μ_{1}, μ_{1}, −1) = (μ_{1}, −μ_{2}, −1) ∼ (μ_{1}, μ_{2}, 1), so this is contained in case IIB5. Case IB4. Q(μ_{1}, μ_{1}, 0) = 1. This holds if and only if p = 5. But then, (μ_{1}, μ_{1}, 0) = (μ_{1}, −μ_{2}, 0) ∼ (μ_{1}, μ_{2}, 0), so this is contained in case IIB4. Case IB5. Q(μ_{1}, μ_{1}, 1) = 1. This gives the contradiction that μ_{1} = 1. Case IB6. Q(μ_{1}, μ_{1}, μ_{2}) = 1. This gives the contradiction that 1 = 0. Case IB7. Q(μ_{1}, μ_{1}, μ_{1}) = 1. This gives the contradiction that μ_{1} = 1. Case IC1. Q(μ_{1}, μ_{1}, −μ_{1}) = −μ_{2}. This holds if and only if (p, μ_{1}) = (29, −5). Case IC2. Q(μ_{1}, μ_{1}, −μ_{2}) = −μ_{2}. This holds if and only if p = 5. But then, (μ_{1}, μ_{1}, −μ_{2}) = (μ_{1}, μ_{1}, μ_{1}), so this is contained in case IA7. Case IC3. Q(μ_{1}, μ_{1}, −1) = −μ_{2}. This holds if and only if (p, μ_{1}) = (11, −3). Case IC4. Q(μ_{1}, μ_{1}, 0) = −μ_{2}. This holds if and only if (p, μ_{1}) = (11, 4). Case IC5. Q(μ_{1}, μ_{1}, 1) = −μ_{2}. This holds if and only if p = 5, but then it is contained in case IA7. Case IC6. Q(μ_{1}, μ_{1}, μ_{2}) = −μ_{2}. Adding μ_{1} to both sides of 2 = −μ_{2} gives 2 + μ_{1} = 1 so μ_{1} = −1, a contradiction. Case IC7. Q(μ_{1}, μ_{1}, μ_{1}) = −μ_{2}. This holds if and only if p = 5, but then it is contained in case IA7.

Case II. (α, β, γ) = (μ_{1}, μ_{2}, γ)

Since γ = ±μ_{1} would give a triple equivalent to one of those in case I, we need only examine the subcases for which γ is one of (∗, −μ_{2}, −1, 0, 1, μ_{2}, ∗), where the ∗ means an ignored case. The corresponding values of Q(μ_{1}, μ_{2}, γ) are (∗, 2, 3, 1, 1, 2 − 2μ_{2}, ∗). We trust that the reader’s appetite to examine elementary cases is by now sated, so from now on we only give summary information. Cases IIA2, IIA4, IIA5, IIB2, IIB6, IIC2, IIC4, and IIC5. These lead to contradictions. Case IIB3. Q(μ_{1}, μ_{2}, −1) = 1. This holds if and only if p = 2, which is contained in case IIB5.

Cases IIA3, IIA6, IIC3, and IIC6. These hold if and only if p = 5. But then these cases are contained in cases IA5, IA7, IA5, and IA7 respectively. Case IIB4. Q(μ_{1}, μ_{2}, 0) = 1. This holds for all p. Case IIB5. Q(μ_{1}, μ_{2}, 1) = 1. This holds for all p. Case III. (α, β, γ) = (μ_{1}, 1, γ)

Since γ = ±μ_{1}, ±μ_{2} would fall under Case I or Case II, we need only examine γ one of (∗, ∗, −1, 0, 1, ∗, ∗), with corresponding values of Q(μ_{1}, 1, γ) equal to (∗, ∗, 2μ_{1} + 1, μ_{1}, 1, ∗, ∗). Cases IIIA3, IIIA5, IIIB4, IIIC5. These lead to contradictions. Case IIIB3. Q(μ_{1}, 1, −1) = 1. This holds if and only if p = 2, which is contained in case IIIB5. Case IIIC4. Q(μ_{1}, 1, 0) = −μ_{2}. This holds if and only if p = 5. But then it is contained in case IIIA4. Case IIIC3. Q(μ_{1}, 1, −1) = −μ_{2}. This holds if and only if p = 3. Case IIIA4. Q(μ_{1}, 1, 0) = μ_{1}. This holds for all p. Case IIIB5. Q(μ_{1}, 1, 1) = 1. This holds for all p.

We first analyze the six cases that hold for all p (the ten cases found in [1] for SL(2, ℂ) can quickly be reduced to these six by replacement of μ_{2} by −μ_{1} and/or applying equivalences). For each of the six cases, we will apply equivalences until the traces indicate a generating pair P (⟨A, B⟩) for which P (A)^{5} = P (B)^{3} = P (AB)^{2} = P (I). Case IA5. (μ_{1}, μ_{1}, 1) ∼ (μ_{1}, 1, μ_{1}) ∼ (μ_{1}, 1, 0). Case IA7. (μ_{1}, μ_{1}, μ_{1}) ∼ (μ_{1}, μ_{1}, 1), which is case IA5. Case IIB4. (μ_{1}, μ_{2}, 0). Using basic trace identities, we calculate tr(B^{2}) = tr^{2}(B) − 2 = μ^{2} − 2 = −μ_{1}, and tr(AB^{2}) = tr(AB) tr(B) − tr(A) = −μ_{1}. Since P (B) has order 5, we have P (⟨A, B⟩) = P (⟨A, −B^{2}⟩), and from our calculations, Tr(A, −B^{2}) = (μ_{1}, μ_{1}, μ_{1}), so case IA7 applies. Case IIB5. (μ_{1}, μ_{2}, 1) ∼ (μ_{1}, μ_{2}, 0), which is case IIB4. Case IIIA4. (μ_{1}, 1, 0) is immediate. Case IIIB5. (μ_{1}, 1, 1) ∼ (μ_{1}, 1, μ_{1} − 1) = (μ_{1}, 1, μ_{2}) ∼ (μ_{1}, μ_{2}, 1), which is case IIB5.

We now analyze the exceptional cases IB1, IC1, IC3, IC4, and IIIC3, to which all other exceptional cases have been reduced. Write G for ⟨A, B⟩. For each of p = 3, 11, 19, and 29, the six general cases just considered provide many examples with these values of p for which the conditions of the Main Theorem hold and P (G) = A_{5} (for example, Theorem 1 of [3] shows that for any triple (α, β, γ) that occurs in the six cases, there is a pair (A, B) in each of SL(2, 9), SL(2, 11), SL(2, 19), or SL(2, 29) for which Tr(A, B) = (α, β, γ), so we can just choose the triple to be (μ_{1}, μ_{1}, μ_{1}) from Case IA7 and achieve that P (G) = A_{5}). So it remains to produce examples for these values of p which satisfy the conditions of the Main Theorem but have P (G) ≠ A_{5}.

We begin with case IIIC3, for which we assume that p = 3. We have (μ_{2}, 1, 1) ∼ (μ_{2}, 1, μ_{2} − 1). Computing in GAP with u = Z(9), the standard GAP generator for the multiplicative group 𝔽_{9} − {0}, we find that μ_{1} can be u or u^{3} with corresponding values of μ_{2} either u^{7} or u^{5}. But then μ_{2} − 1 is either u^{7} − 1 = u^{2} or u^{5} − 1 = u^{6}. Since u^{2} = u^{3} + u^{−3} (or alternatively

because u^{2} ∈ T_{4}), a matrix of trace u^{2}, such as (\begin{matrix}u^{3} & 0 \\ 0 & u^{−3} \end{matrix}), has order 4

in PSL(2, 9). Since P (G) contains elements of order 4 as well as order 5, P (G) ≢ A_{5}. The case of u^{6} = u + u^{−1} is similar.

For the remaining cases, we have a fixed (odd) value of p and we use u to denote Z(p), the generator used by GAP for the multiplicative group of 𝔽_{p}. We use v for Z(p^{2})^{p−1} ∈ 𝔽_{p^{2}}, for which v^{(p+1)/2} = −1.

Assume that p = 11, for which u = 2, and one of cases IC3 or IC4 occurs. For case IC3, we have μ_{1} = −3 = u^{3}, (μ_{1}, μ_{1}, −1) = (u^{3}, u^{3}, u^{5}) ∼ (u^{3}, u^{5}, u^{3}) ∼ (u^{3}, u^{5}, u^{9}). Since u^{9} = v+v^{−1}, where v has order 12, a matrix of trace u^{9} has projective order 6, showing that P (G) ≠ A_{5}. For case IC4, μ_{1} = u^{2}, (μ_{1}, μ_{1}, 0) = (u^{2}, u^{2}, 0) ∼ (u^{2}, u^{2}, u^{4}). Since u^{4} = v^{5} + v^{−5}, where v^{5} has order 12, a matrix of trace u^{4} has projective order 6 and P (G) ≠ A_{5}.

Assume now that p = 19, for which u = 2, and case IB1 occurs. We have μ_{1} = −4 = u^{11}, and (u^{11}, u^{11}, −u^{11}) ∼ (u^{11}, u^{11}, u^{15}). Since u^{15} = u + u^{−1}, a matrix of trace u^{15} has projective order 9, so P (G) ≠ A_{5}.

Finally, we consider p = 29, for which u = 2 once again, and assume that case IC1 occurs. We have μ_{1} = −5 = u^{8}, (u^{8}, u^{8}, −u^{8}) ∼ (u^{8}, u^{8}, u^{24}), and u^{24} = v^{4} + v^{−4} so a matrix of trace u^{24} has order 15.

This completes the proof of the Main Theorem.

As we remarked after the statement of the Main Theorem, in the exceptional cases, it is easy to determine from tr(A), tr(B), and tr(AB) whether or not P (⟨A, B⟩) = A_{5}. In practice, in the cases when P (⟨A, B⟩) ≠ A_{5}, random applications of the equivalence moves of triples to (α, β, γ) will soon yield an entry that is not in T_{2} ∪ T_{3} ∪ T_{5}. But one can tell for certain, as we will illustrate for p = 3 (the other cases being similar). Suppose that tr(A), tr(B), tr(AB) ∈ T_{2} ∪ T_{3} ∪ T_{5} (and at least two are nonzero) and tr([A, B]) ∈ {μ_{1}, −μ_{2}}. Write Tr(A, B) = (α, β, γ). From proposition 1.1, we know that one of the entries is one of ±μ_{1} or ±μ_{2}. Applying equivalences and possibly changing our choice of root of x^{2} − x − 1, we may assume that α = μ_{1}. One of β or γ is nonzero, so applying more equivalences we may make (α, β, γ) one of (μ_{1}, μ_{1}, γ), (μ_{1}, μ_{2}, γ), or (μ_{1}, 1, γ). If γ ∉ T_{2} ∪ T_{3} ∪ T_{5}, then we already know that P (⟨A, B⟩) ≠ A_{5}. If γ ∈ T_{2} ∪ T_{3} ∪ T_{5}, then from the case-by-case analysis of the proof of the “only if” direction of the Main Theorem, (μ_{1}, β, γ) and the value of Q(α, β, γ) must now be either as in one of the six general cases IA5, IA7, IIB4, IIB5, IIIA4, or IIIB5, or else as in the special case IIIC3. The proof showed that in the six general cases, P (⟨A, B⟩) = A_{5}, while in case IIIC3, P (⟨A, B⟩) ≠ A_{5}.

References

1. R. C. Churchill, Two generator subgroups of SL(2, ℂ) and the hypergeometric, Riemann, and Lamé equations, J. Symbolic Comput. 28 (1999) 521–545.

2. GAP: Groups, Algorithms, and Programming, available at the St. Andrews GAP

website http://turnbull.mcs.st-and.ac.uk/ gap/ .

3. A. M. Macbeath, Generators of the linear fractional groups, Number Theory (Proc. Sympos. Pure Math., Vol. XII, Houston, Tex., 1967), Amer. Math. Soc., Providence, R.I. (1969) 14–32.

4. B. H. Neumann and H. Neumann, Zwei Klassencharakteristischer Untergruppen und ihre Factorgruppen, Math. Nachr. 4 (1951) 106–125.

Department of Mathematics, University of Oklahoma, Norman, Oklahoma 73019, USA

E-mail address: dmccullough@math.ou.edu
URL: www.math.ou.edu/ dmccullough/