A Subgroup of SO(3, ℝ) generated by Rotations of orders 4 and 8

Geoffrey R. Robinson, Department of Mathematical Sciences, University of Aberdeen, Aberdeen, AB 24 3UE, Scotland e-mail: g.r.robinson@abdn.ac.uk

Introduction: In [2], Radin and Sadun determined the structure of certain subgroups of SO(3, ℝ) generated by a pair of rotations of finite order whose axes of rotation are at an angle which is a rational multiple of π. The essential case is when the axes of rotation are orthogonal, and the case that one of the rotations has order 4 is especially critical. With few explicit exceptions when the groups are finite, such groups are described as free products with amalgamation over common subgroups of pairs of groups which are either all cyclic, dihedral (including Klein 4-groups), or S_{4}, the symmetric group of degree 4.

In this paper, we will study the behaviour of these groups under reduction (mod p), which is available for all odd primes p. We will only consider the case that the two axes of rotation are orthogonal, and ( deviating from the notation of [2]), we will denote by G_{(m,n)} the group generated by

x_{m} = ( \begin{matrix}cos(\frac{2π}{m}) & sin(\frac{2π}{m}) & 0 \\ −sin(\frac{2π}{m}) & cos(\frac{2π}{m}) & 0 \\ 0 & 0 & 1 \end{matrix} ) ,

and

y_{n} = ( \begin{matrix}1 & 0 & 0 \\ 0 & cos(\frac{2π}{n}) & sin(\frac{2π}{n}) \\ 0 & −sin(\frac{2π}{n}) & cos(\frac{2π}{n}) \end{matrix} )

for positive integers m and n. We will be particularly interested in the group G_{(4,8)}.

We summarize here (some of) the results of [2] : as usual, for groups A and B with a common subgroup C, the notation A ∗_{C} B is the free product of A and B amalgamated over C ( if C is omitted, then the product is just the free product of A and B). We also use D_{n} to denote the dihedral group of order n ( that is, with 2n elements) and D_{2} to denote the Klein 4-group.

THEOREM( Radin and Sadun,[2]):

i) If m is odd, we have G_{(m,4)} = D_{m} ∗_{ℤ_{2}} ℤ_{4}. ii) If m is even, but not divisible by 4, then we have G_{(m,4)} ≅ D_{m} ∗_{D_{2}} D_{4}. iii) If 4|m, then G_{(m,4)} ≅ D_{m} ∗_{D_{4}} S_{4}. iv) If p and q are both odd, then G_{(p,q)} ≅ ℤ_{p} ∗ ℤ_{q}. v)If p is even and q is odd, then G_{(p,q)} ≅ ℤ_{p} ∗_{ℤ_{2}} D_{q}. vi) If p, q are both even, but q is not divisible by 4, then G_{p,q} ≅ D_{p} ∗_{D_{2}} D_{q}. vii) If p and q are both divisible by 4, then we have G_{(p,q)} ≅ G_{(lcm(p,q),4)}, so this group may also be expressed as a free product with amalgamation, as in part iii).

We remark here that while (with the noted exceptions) the groups involved are infinite, and the representations as given are not realised over a ring of algebraic integers, it is still possible to perform the process of reduction (mod p) for all odd primes p, (but it is not possible to perform reduction (mod 2), since it is easy to check that trace(x_{m}y_{n}) has negative valuation at all prime ideals containing the rational prime 2 (of the relevant ring of cyclotomic integers) ).

When nm is divisible by p, we will see that the reduction (mod p) of G_{(m,n)} depends only on the “smaller” pair of integers (m_{p′}, n_{p′}). As usual, for a positive integer n and a prime p, we have set n_{p} = p^{ν_{p}(n)} and n_{p′} = \frac{n}{n_{p}}.

We note that ⟨x_{4}, y_{4}⟩ ≅ S_{4}, and that (x_{4}y_{4})^{2} has order 3. Furthermore, x_{2} = x_{4}^{2} inverts y_{8} and y_{2} inverts x. Notice also that (using [2] ) ⟨x_{2}, y_{8}⟩ and ⟨x_{4}, y_{4}⟩ are maximal finite subgroups of G.

THEOREM 1: G has epimorphic images as follows for odd primes p :

i) PSL(2, p) whenever p ≡ ±1 (mod 16). ii) PGL(2, p) whenever p ≡ ±7 (mod 16). iii) PGL(2, p^{2}) whenever p ≡ ±3 (mod 8). Moreover, the epimorphisms may be chosen so that each has torsion free kernel and so that the intersection of the kernels of any infinite number of them is trivial. Also, G has no non-identity solvable normal subgroup and [G : G′] = 2.

REMARK: In various private communications since January 2006, J-P. Serre has provided us with a proof that G_{(4,8)} ≅ SO(3, ℤ[\frac{1}{\sqrt{2}}]), which may be viewed as a strengthening of the above theorem, and has also provided examples to illustrate that, even in the cases when G_{(m,n)} is infinite, it is not always the case that G_{(m,n)} is the full special orthogonal group over the ring generated by the traces of its elements.

SECTION 1.1: ON THE GROUPS G_{(m,n)}

We note that for positive integers m and n both greater than 2, the group G_{(m,n)} is an absolutely irreducible linear group. For, in that case, (after extending scalars to ℂ) all eigenspaces of x_{m} and y_{n} are 1-dimensional and x_{m} and y_{n} have no common eigenvector. On the other hand, if the representation were reducible (over ℂ), then ( since the matrices are orthogonal) there would be an invariant 1-dimensional subspace, and hence x_{m} and y_{n} would have a common eigenvector.

We note for future use that x_{2} inverts each y_{n} and centralizes each x_{n}, while y_{2} inverts each x_{n} and centralizes each y_{n}. Hence the Klein 4-group G_{(2,2)} = ⟨x_{2}, y_{2}⟩ normalizes each of the groups G_{(m,n)}.

When min(m, n) ≤ 2, it is clear that G_{(m,n)} is reducible, while G_{(m,n)} is a finite dihedral group when the minimum is 2 ( including the possibility of a Klein 4 group). We note that ( for a given choice of m and n), we have G_{(m,n)} ≤ G_{(rm,sn)} for all positive integers r and s. If a, b, c, d are positive integers with gcd(a, c) = gcd(b, d) = 1, then G_{(ac,bd)} = ⟨G_{(a,b)}, G_{(c,d)}⟩. In that case, we note in particular that G_{(a,b)} ⊲ G_{(ac,bd)} if {c, d} ⊆ {1, 2}.

SECTION 1.2: REDUCTION (mod p)

Now let us fix a choice of G_{(m,n)} and an odd prime p. Let d(m, n) = lcm(4, m, n), and let ω = e^{\frac{2πi}{d(m,n)}} . Let ρ be a prime ideal of ℤ[ω] containing p. Let R denote the localization ℤ[ω]_{ρ} and ρ^{∗} denote the unique maximal ideal of R. Let F denote R/ρ^{∗}. We note that cos(\frac{2π}{n}), cos(\frac{2π}{m}), sin(\frac{2π}{n}), sin(\frac{2π}{m}) all lie in R, since 2 is a unit of R. We let G_{(m,n),ρ} denote the image of G_{(m,n)} under the natural epimorphism from SO(3, R) to SO(3, F). We refer to G_{(m,n),ρ} ( somewhat loosely) as a reduction (mod p) of G_{(m,n)}. We note that the field F so obtained, and the images of x_{m} and y_{n} in SO(3, F), only depend on the integers m_{p′} and n_{p′}. This is because 1 − ζ ∈ ρ whenever ζ is a d(m, n)-th root of unity whose order is a power of p, so that cos(\frac{2π}{n}) ≡ cos(\frac{2π}{n_{p′}}) (mod ρ^{∗}) ( and similarly for sin).

We also note that the group G_{(m,n),ρ} is absolutely irreducible whenever min(m_{p′}, n_{p′}) ≥ 3, (and is reducible otherwise). For, if G_{(m,n),ρ} were reducible ( after extension of scalars to an algebraically closed field) then we claim that there would be a 1-dimensional invariant subspace. If not, there is a 2-dimensional invariant subspace, say W. However, the 1-dimensional orthogonal complement, W^{⊥}, of W (which may be contained in W) is also invariant, since G_{(m,n)} ⊆ SO(3, F). In that case, the images of x_{m} and y_{n} would have a common eigenvector, which they do not, as in the complex case.

We remark that G_{(m,n)} embeds in the direct product of any infinite collection of G_{(m,n),ρ}’s. In fact, since there are infinitely many primes p congruent to 1(mod d(m, n)), we could also embed G_{(m,n)} in the direct product of any infinite collection of SO(3, p)’s, as p ranges through such (distinct) primes.

REMARK: For the group G_{(4,8)} and an odd prime p, we don’t need to pass to such a large degree extension as in the previous section to perform reduction (mod p). For example, when p ≡ ±1 (mod 8), we know that 2 is a quadratic residue in ℤ/pℤ, and it follows that \sqrt{2} exists (and is a unit) in the (complete) padic integers ℤ_{p}. Hence we may embed the group G_{(4,8)} in SO(3, ℤ_{p}) and obtain a natural homomorphism to SO(3, p). When p ≡ ±3 (mod 8), we can view G_{(4,8)} as a subgroup of SO(3, ℤ_{p}[\sqrt{2}]), and obtain a natural homomorphism to SO(3, p^{2}). In the latter case, since trace(y_{8}) = 1 + \sqrt{2}, the resulting image can’t be realized (even up to equivalence) as a subgroup of GL(3, p), as the trace of the representation doesn’t take values in GF(p). For any odd prime p, we denote the resulting image by G_{(4,8),p}.

SECTION 2: THE GROUP G_{(4,8)}

We recall from the Theorem of Radin and Sadun that G_{(4,4n)} = G_{(4n,4n)} for each positive integer n. In fact, a direct calculation shows that (y_{4}x_{4})y_{4n}(y_{4}x_{4})^{−1} = x_{4n}, so that x_{4n} and y_{4n} are conjugate in G_{(4n,4n}). In particular, we have G_{(4,8)} = G_{(8,8)}.

We remind the reader that when F is a finite field of odd characteristic, PGL(2, F) ≅ SO(3, F). To see this, let GL(2, F) act by conjugation on 2 × 2 matrices of trace 0 ( with entries in F), and observe that the non-degenerate symmetric bilinear form ⟨a, b⟩ = trace(ab) is invariant under this action. This illustrates that the derived group of SO(3, F) has index 2, and is generated by the elements whose eigenvalues λ all satisfy λ^{\frac{|F |−1}{2}} = 1 if λ ∈ F and λ^{\frac{|F |+1}{2}} = 1 if λ ∉ F. For [PGL(2, F) : PSL(2, F)] = 2, and the elements of PGL(2, F)\PSL(2, F) are the images of those matrices in GL(2, F) whose determinants are non-squares in F.

We need to examine subgroups of GL(2, p) and GL(2, p^{2}) in some detail for odd primes p. It is well-known that a subgroup of GL(2, p) which has more that one Sylow p-subgroup contains SL(2, p). Similarly, it is easy to check that any subgroup of GL(2, p^{2}) which has more that one Sylow p-subgroup contains a subgroup isomorphic to SL(2, p). We prove:

LEMMA 2: Let p be a prime congruent to ±3 (mod 8). Let H be a subgroup of GL(2, p^{2}) which has more than one Sylow p-subgroup, and which contains an element t of order 8 with eigenvalues 1 and ω. Then H contains SL(2, p^{2}) and H maps onto PGL(2, p^{2}) under the natural projection.

PROOF: Since p ≡ ±3 (mod 8), we have p^{2} ≡ 9 (mod 16), so that \frac{p^{2}−1}{8} is odd and GF(p^{2}) contains no element of multiplicative order 16. Let L = O^{p′}(H) ≤ SL(2, p^{2}). It suffices to prove that p^{2}||L|, for distinct Sylow p-subgroups of SL(2, p^{2}) have trivial intersection and L has more than one Sylow p-subgroup. Hence, in that case, we see that L contains p^{2} + 1 Sylow p-subgroups. Since SL(2, p^{2}) is generated by its p^{2} + 1 Sylow p-subgroups, we then obtain L = SL(2, p^{2}). Also, by assumption, H contains an element (namely t, of order 8) whose determinant is not a square in the field of p^{2} elements, so H/(H ∩ Z(GL(2, p^{2}))) = PGL(2, p^{2}).

We remark that the hypotheses imply that H is absolutely irreducible as a linear group. Let P ∈ Syl_{p}(L), and let us note that O^{p}(C_{H}(u)) = Z(H) for each non-identity element u ∈ P. Now H = LN_{H}(P) and we may write t = sz for some s ∈ L and z ∈ N_{H}(P). Then det(z) = ω and we may suppose that z is a 2-element.

Since the 1-dimensional fixed-point space of P is z-invariant, we see that both eigenvalues of z are in GF(p^{2}). Hence both eigenvalues of z are 8-th roots of unity, and z itself has order 8. Let the eigenvalues of z be ω^{j} and ω^{9−j} where j ∈ {1, 3, 5, 7}. Then the eigenvalues of z^{4} are −1 and 1, since 9 − j is even.

In particular, z^{4} is non-scalar, so that zC_{H}(P) has order 8 in N_{H}(P)/C_{H}(P). Thus P is not cyclic, as p ≡ ±3 ( mod 8), and the Lemma is proved.

THEOREM 3: Let G_{(m,n),ρ} be a reduction (mod p) constructed as before, where p is an odd prime. Suppose that min(n_{p′}, m_{p′}) ≥ 3. Then ( except for the case (m_{p′}, n_{p′}) = (4, 4)), the group G_{(m,n),ρ} has order divisible by p and is an absolutely irreducible and primitive linear group.

PROOF: If mn is odd, then G_{(m,n)} ⊲ G_{(2m,2n)}, while if m is odd but n is even, then G_{(m,n)} ⊲ G_{(2m,n)}. In each case, the factor group has order dividing 4.

Suppose then that we can establish that G_{(m′,n′),ρ} is (absolutely) primitive of order divisible by p, where m′ = lcm(m, 2) and n′ = lcm(n, 2). In that case, it would certainly be the case that G_{(m,n),ρ} would have order divisible by p, since p is odd. If G_{(m,n),ρ} were imprimitive ( even over an extension field), then it would have an Abelian normal subgroup with factor group isomorphic to a subgroup of S_{3}. In particular, G_{(m,n),ρ} would be solvable.

However, the Fitting subgroup F = F (G_{(m,n),ρ}) is normal in G_{(m′,n′),ρ}, so Z(F) is normal in G_{(m′,n′),ρ}, and then central, by primitivity. Thus Z(F) consists of scalars, by absolute irreducibility. However, each element of G_{(m′,n′),ρ} has an eigenvalue 1, so that Z(F) = 1 and then F = 1. But G_{(m,n),ρ} is solvable and non-trivial, a contradiction.

Hence it suffices to consider the case that m_{p′} and n_{p′} are both even. Suppose that m_{p′} and n_{p′} are both at least 6. Suppose further that G_{(m,n),ρ} has order prime to p. Then it “lifts” to an absolutely irreducible p′-subgroup, say X, of GL(3, ℂ) with Brauer character ( at a given element u) uniquely determined by the eigenvalues of the corresponding element of G_{(m,n),ρ}. In particular, X con-

tains (non-central) elements of orders m_{p′} and n_{p′} with eigenvalues 1, e^{\frac{2πi}{m_{p′}}} , e^{\frac{−2πi}{m_{p′}}}

and 1, e^{\frac{2πi}{n_{p′}}} , e^{\frac{−2πi}{n_{p′}}} . A Theorem of Blichfeldt [1] asserts that the element g of a finite primitive complex irreducible linear group Y is central in Y if there is an eigenvalue λ of g such that all eigenvalues of g are within \frac{π}{3} of λ on S^{1}. Hence X is imprimitive, and the associated complex representation of G_{(m,n),ρ} is equivalent to one induced from a 1-dimensional representation of a subgroup, H say, of index 3. The same is true (up to equivalence, and possibly over an extension field) for the given characteristic p representation of G_{(m,n),ρ}.

Let w_{m} be the image of x_{m} and z_{n} denote the image of y_{n} in G_{(m,n),ρ}. But no sum of eigenvalues of w_{m} is 0 in R/ρ^{∗} ( for example, if 1 + θ + θ^{−1} = 0, then θ^{3} = 1, whereas the eigenvalues of w_{m} are roots of unity of order at least 6). It follows by Mackey’s theorem that w_{m} is in core_{X}(H). A similar argument applies to z_{n}. But then w_{m} and z_{n} commute, and G_{(m,n),ρ} is Abelian, a contradiction, as it is an absolutely irreducible linear group. Hence p divides the order of G_{(m,n),ρ} (and G_{(m,n),ρ} is an absolutely primitive linear group).

We are left to consider the case m_{p′} = 4, n_{p′} ≥ 6. If the given representation of G_{(m,n),ρ} is equivalent (possibly over an extension field) to one induced from a 1-dimensional representation of a subgroup of index 3, say H, then z_{n} lies in K = core_{G_{(m,n),ρ}}(H). Since G_{(m,n),ρ}/K is isomorphic to a subgroup of S_{3} we see also that w_{m}^{2} lies in K. Thus [y_{n}, x_{2}] = y\frac{n}{2} lies in the kernel of the natural epimorphism from G_{(m,n)} to G_{(m,n),ρ}. This is a contradiction, since the p′-part of y_{n} has order at least 4, and the p′-part of z_{n} has the same order. Proof of Theorem 1: We will prove that G_{(4,8),p} ≅ PSL(2, p) whenever p ≡ ±1 (mod 16), that G_{4,8,p} ≅ PGL(2, p) whenever p ≡ ±7 (mod 16), and that G_{4,8,p} ≅ PGL(2, p^{2}) whenever p ≡ ±3 (mod 8).

Since G_{(4,8),p} acts (absolutely) irreducibly in characteristic p, it has no nontrivial normal p-subgroup. By Theorem 3, G_{(4,8),p} has order divisible by p (and has more than one Sylow p-subgroup, since it is irreducible).

In the case that p ≡ ±1 (mod 8), we know that G_{(4,8),p} is a subgroup of SO(3, p)(≅ PGL(2, p)). Since G_{(4,8),p} has more than one Sylow p-subgroup, it contains ( a normal subgroup of index at most 2 isomorphic to) PSL(2, p).

The eigenvalues (other than 1) of the images of x_{4} and y_{8} are, respectively, primitive 4-th and 8-th roots of unity. If 8|p − 1, then the image will contain an element of PGL(2, p)\PSL(2, p) exactly when a primitive 8-th root of unity is not a square in GF(p), that is, when p ≡ 9 (mod 16). If 8|p + 1, then the image contains an element of PGL(2, p)\PSL(2, p) only when a primitive 8-th root of unity ζ ∈ GF(p^{2}) does not satisfy ζ^{\frac{p+1}{2}} = 1, that is, precisely when p ≡ 7 (mod 16).

In the case that p ≡ ±3 (mod 8), we know that G_{(4,8),p} is a subgroup of SO(3, p^{2}) ≅ PGL(2, p^{2}) which has more than one Sylow p-subgroup. The image of y_{8} in SO(3, p^{2}) is also the image in SO(3, p^{2}) under the natural homomorphism from GL(2, p^{2}) of an element of order 8 which has an eigenvalue 1. By Lemma 2, we see that G_{(4,8),p} ≅ PGL(2, p^{2}) in these cases.

For each odd prime p, the epimorphic image G_{(4,8),p} lies between a certain non-Abelian simple group PSL(2, q) and its automorphism group, so it has no non-trivial solvable normal subgroup. Hence any solvable normal subgroup of G_{(4,8)} is in the kernel of every reduction (mod p), so is trivial.

The principal ideal (p) remains prime in ℤ_{p}[\sqrt{2}] when p ≡ ±3 (mod 8) (and (p) is certainly prime in ℤ_{p} when p ≡ ±1 (mod 8). By the usual argument, if g ≠ I is an element of finite prime order with g ≡ I_{3} (mod p^{k}) (entrywise), but g ≢ I_{3} (mod p^{k+1}), then g^{p^{s}} ≡ I_{3} (mod p^{k+s}) but g^{p^{s}} ≢ I_{3} (mod p^{k+s+1}), for each positive integer s. If g has order prime to p, then we can choose arbitrarily large values of s for which g^{p^{s}} = g, a contradiction, since g ≠ I. If g^{p} = I, then we evidently have a contradiction.

Since G = G_{(4,8)} is generated by two conjugate elements of order 8, namely x_{8} and y_{8}, we know that G/G′ is generated by y_{8}G′. But x_{2} = x_{4}^{2} inverts y_{8}, so that [x_{2}, y_{8}] = y_{8}^{2} ∈ G′. Hence [G : G′] ≤ 2. However, we know that G_{(4,8)} has a homomorphic image of order 2, since (for example), it has PGL(2, 9) as a homomorphic image, which itself in turn has a factor group of order 2.

REMARK: With a little more effort, it is possible to prove in general (using a well-known theorem of L.E. Dickson) that G_{(m,n),ρ} contains PSL(2, F) where F is the field generated by the traces of its elements, whenever (4, 4) ≠ min(m_{p′}, n_{p′}) ≥ 3.

ACKNOWLEDGEMENTS: The author thanks Jean-Pierre Serre for many helpful communications about this work, which played a significant role in shaping this revised version and, in particular, for making us aware of the work of Radin and Sadun. It is also a pleasure to acknowledge helpful discussions related to this work with J. Kedra and C.Maclachlan ( both of the University of Aberdeen).

References

[1] Blichfeldt,H., Finite collineation groups, University of Chicago Press, (1917).

[2] Radin,C. and Sadun,L., On 2-generator subgroups of SO(3), Trans. AMS,351,11,4469-4480, (1999).