Question

Compute the center Z of SL2(Z/p)

Answer 1

We consider the discrete Heisenberg group G = Heis(3, Z) which is the group of matrices of the form:

, where a, b, c are from Z . We can consider also the group G as a set of all integer triples endowed with the group law: (a1, b1, c1)(a2, b2, c2) = (a1 + a2, b1 + b2, c1 + c2 + a1b2). (1)

It is clear that [(1, 0, 0),(0, 1, 0)] = (0, 0, 1) and the group G is generated by the elements (1, 0, 0) and (0, 0, 1) .

Let Aut(G) be the group of automorphisms of the group G . The starting point for this note was the group homomorphism Z → Aut(G) constructed in [3, § 5] and given by the formula

Rd((a, b, c))= (2)

where Rd is the automorphism of the group G induced by an element d ∈ Z . Formula (2) was obtained in [3] after some calculation of automorphisms which are analogs of ”loop rotations” in loop groups. But the fact that formula (2) defines a homomorphism from the group Z to the group Aut(G) is also an easy direct consequence of formulas (1) and (2). For the group G we have the following exact sequence of groups

, (3)

where 1 is the trivial group, the group C = {(0, 0, c) | c ∈ Z} ≃ Z is the center of G , and H = {(a, 0, 0) | a ∈ Z} ≃ Z , P = {(0, b, 0) | b ∈ Z} ≃ Z .

From exact sequence (3) we obtain the following sequence

, (4)

where the group of inner automorphisms Inn(G) ≃ G/C ≃ Z⊕Z , and the homomorphism ϑ is the homomorphism Aut(G) → Aut(Z ⊕ Z) , which is induced by the homomorphism λ from exact sequence (3). It is clear that Im(θ) ⊂ Ker (ϑ) . We claim that sequence (4) is exact in the term Aut(G) . Indeed, it is enough to prove that Ker (ϑ) ⊂ Im(θ) . Consider any ω ∈ Ker (ϑ) . We have ω((1, 0, 0)) = (1, 0, c1) and ω((0, 1, 0)) = (0, 1, c2) for some integer c1 and c2 . By direct calculations, we obtain (c2, −c1, 0)(1, 0, 0) = (1, 0, c1)(c2, −c1, 0) and (c2, −c1, 0)(0, 1, 0) = (0, 1, c2)(c2, −c1, 0) . Since elements (1, 0, 0) and (0, 1, 0) generate the group G , we obtain that the inner automorphism defined by the element (c2, −c1, 0) coincides with the automorphism ω . We claim also that the homomorphism ϑ from sequence (4) is surjective. Indeed, by formula (2) we have a homomorphism from the group Z to the group Aut(G) . It is easy to see that this homomorphism is a section of the homomorphism ϑ over the subgroup d∈Z of the group GL(2, Z) .

By formula (2), the action of the matrices from this subgroup on elements of the group G is given as:

. (5)

Symmetrically to the formula (5) we can write the following formula:

. (6)

By an easy direct calculation we have that formula (6) defines a correct automorphism of the group G . This automorphism depends on d ∈ Z and defines the homomorphism from the subgroup

d∈Z of the group GL(2, Z) to the group Aut(G) .

This homomorphism defines a section of the homomorphism ϑ over this subgroup. Besides, it is easy to see that the homomorphism from the subgroup of the group GL(2, Z) to the group Aut(G) given by the formula (7)

defines a section of the homomorphism ϑ over this subgroup.

By the classical result the group SL(2, Z) has a presentation: < ρ, τ | ρτ ρ = τ ρτ , (ρτ ρ)^ 4 = 1 > where the element ρ corresponds to the matrix A = and the element τ corresponds to the matrix B = . The group GL(2, Z) is generated by the elements of the group SL(2, Z) and the matrix D = . Therefore the group GL(2, Z) has a presentation: < ρ, τ, κ | ρτ ρ = τ ρτ , (ρτ ρ) 4 = 1, κτκ−1 = τ −1 , κρκ−1 = ρ −1 , κ 2 = 1 > , (8) where the element κ corresponds to the matrix D . Thus we obtained that the homomorphism ϑ is surjective.

We note that the group Aut(G) contains a distinguished subgroup Aut+ (G) = ϑ −1 (SL(2, Z)) of index 2 . Since for any ω ∈ Aut(G) we have ω((0, 0, 1)) = ω([(1, 0, 0),(0, 1, 0)]) = [ω(1, 0, 0), ω(0, 1, 0)] = (0, 0, det(ϑ(ω))), we obtain that the group Aut+(G) consists of elements ω of the group Aut(G) such that ω((0, 0, 1)) = (0, 0, 1) . In other words, the group Aut+ (G) consists of automoprhisms of the group G which act identically on the center of the group G .