ON BLOCKS STABLY EQUIVALENT TO A QUANTUM COMPLETE INTERSECTION OF DIMENSION 9 IN CHARACTERISTIC 3 AND A

CASE OF THE ABELIAN DEFECT GROUP CONJECTURE

RADHA KESSAR

Abstract. Using a stable equivalence due to Rouquier, we prove that Broué’s abelian defect group conjecture holds for 3-blocks of defect 2 whose Brauer correspondent has a unique isomorphism class of simple modules. The proof makes use of the fact, also due to Rouquier, that a stable equivalence of Morita type between self-injective algebras induces an isomorphism between the connected components of the outer automorphism groups of the algebras.

1. Introduction

The abelian defect group conjecture of Broué [1, Question 6.2], [9, Conjecture, page 132] states that any p-block (p a prime number) of a finite group G whose defect groups are abelian is derived equivalent to its Brauer correspondent. The conjecture is known to hold when the defect groups in question are cyclic [13], Klein 4-groups [14], and when the concerned blocks are nilpotent [11]. The conjecture has been proved in several situations under additional global assumptions, that is, assumptions on G. In this paper, we prove the following result.

Theorem 1.1. Let k be an algebraically closed field of characteristic 3. Let G be a finite group, A a block of kG, P a defect group of A and C the Brauer correspondent of A in kN_{G}(P). Suppose that P ≅ C_{3} × C_{3} and that C is a non-nilpotent block with a unique isomorphism class of simple modules. Then A and C are Morita equivalent as k-algebras.

The above theorem is the first case where the abelian defect group conjecture is proved under a prescribed non-nilpotent local structure of wild representation type. We point out that we do not know if the equivalence constructed for the proof of Theorem 1.1 is splendid or if it can be lifted to a complete local ring in characteristic zero.

The proof of Theorem 1.1 fits in with the inductive approach to the abelian defect group conjecture outlined by Rouquier in [15] : the starting point is the gluing result of [15], which yields a stable equivalence of Morita type between the blocks A and C. Moreover, given the assumptions on C and Külshammer’s structure results on blocks with normal defect groups in [7], it is known that C is Morita equivalent to the quantum complete intersection algebra k⟨X, Y ⟩/⟨X^{3}, Y^{3}, XY + Y X⟩. Finally, by [12], it is known that A and C have isomorphic centers and that A also has a unique isomorphism class of simple modules. We obtain Theorem 1.1 as a consequence of the following result.

Theorem 1.2. Let k be an algebraically closed field of characteristic 3. Let

B = k⟨X, Y ⟩/⟨X^{3}, Y^{3}, XY + Y X⟩

and let A be a local, symmetric k-algebra. Suppose that dim_{k}(A) = 9, Z(A) ≅ Z(B) and that there is a stable equivalence of Morita type between A and B. Then A is isomorphic to B.

The proof of Theorem 1.2 has two main steps. First, we use arguments in the style of [8] to describe 9-dimensional local symmetric algebras whose center is isomorphic to that of B - there are infinitely many isomorphism classes of these. In the second step, we show that amongst the algebras obtained in step 1, the only ones which are stably equivalent (á la Morita) to B are isomorphic to B. The crucial ingredient here is the fact that a stable equivalence of Morita type between self-injective algebras preserves the connected component of the outer automorphism group of the algebras ([16]).

Remarks 1.3. (i) By work of Kiyota in [5], with the notation of Theorem 1.1, it is known that A has a unique isomorphism class of simple module if and only if C has a unique isomorphism class of simple modules. Thus Theorem 1.1 remains true if one replaces the hypothesis that C has a unique isomorphism class of simple modules by the hyothesis that either A or C has a unique isomorphism class of simple modules.

(ii) The proof of Theorem 1.2 is very computational; however it seems possible that the methods will find application in other situations, for instance, for p-blocks of defect 2 whose Brauer correspondent has one isomorphism class of simple modules for other primes p (see [2], [4]).

2. Background Results on algebras and their automorphism groups

Let k be an algebraically closed field. All k-algebras will be assumed to be finite dimensional vector spaces over k. Recall that a k-algebra C is symmetric if there exists a k-linear function s : C → k such that s(xy) = s(yx) for all x, y ∈ C and no non-zero left ideal of C is contained in the kernel of s. Also, recall that since k is algebraically closed, C is local if and only if dim_{k}(C/J(C)) = 1.

The following gathers some well known properties of local symmetric algebras; we refer to [6] for proofs.

Lemma 2.1. Suppose that C is a local symmetric k-algebra. Then (i) dim_{k}Soc(C) = 1. (ii) Soc(C) ⊆ Soc(Z(C)). (iii)Soc(C) ∩ [C, C] = 0. (iv) dim_{k}(C) = dim_{k}(Z(C)) + dim_{k}[C, C].

(v) Z(C) is local and Z(C) ∩ J(C) = J(Z(C)).

(vi) If n is the least natural number such that J^{n+1}(C) = 0, then Soc(C) = J^{n}(C).

We will also use the following results from [8].

Lemma 2.2. [8, Lemma E] Let C be a k-algebra and let I be a two-sided ideal of C. Let m, n be natural numbers with m ≤ n. Suppose that

I^{n} = k − span of {x_{i1} ⋯ x_{in} : i = 1, ⋯, d} + I^{n+1}

with elements x_{ij} ∈ I. Then,

I^{n+m} = k − span of {x_{j1} ⋯ x_{jm}x_{i1} ⋯ x_{in} : i, j = 1, ⋯, d} + I^{n+m+1}.

Lemma 2.3. [8, Lemma G] Let C be a local symmetric k-algebra. If n is a natural number such that dim_{k}J^{n}(C)/J^{n+1}(C) = 1, then J^{n−1}(C) ⊆ Z(C).

For a finite dimensional k-algebra C, let Aut(C) be the group of automorphisms of C, viewed as a (linear) algebraic group. Clearly, Aut(C) is a subgroup of the algebraic group GL(C) of the automorphisms of the k-vector space C. For any Aut(C)-invariant subspaces U, V of C with V ⊆ U, let f_{U,V} : Aut(C) → GL(U/V) denote the map defined by f_{U,V} (φ)(u + V) = φ(u) + V, u ∈ U. Any power J^{i}(C) of the radical of C is Aut(C)- stable, hence Aut(C) is contained in the parabolic subgroup of GL(C) of elements which stabilize the flag C ⊃ J(C) ⊃ J^{2}(C) ⊃ · · ·. For each i ≥ 0, let f_{i} denote the map f_{J^{i}(C),J^{i+1}(C)}.

Let Inn(C) denote the closed connected normal subgroup of Aut(C) consisting of inner automorphisms of C and let Out(C) be the quotient group Aut(C)/Inn(C). Let Aut^{0}(C) denote the connected component of the identity of Aut(C) and let Out^{0}(C) = Aut^{0}(C)/Inn(C) denote the connected component of Out^{0}(C).

The following easy results will come in handy for calculating automorphism groups. Recall that for a connected linear algebraic group G over k, the unipotent radical R_{u}(G) of G is the largest closed connected normal unipotent subgroup of G.

Proposition 2.4. Suppose that C is a local algebra. Then, (i) Ker(f_{1}) is a unipotent subgroup of Aut(C). (ii) Inn(A) ≤ Ker(f_{1}). (iii) There exists a bijective morphism of algebraic groups

f̂ : Out^{0}(C)/R_{u}(Out^{0}(C)) → f_{1}(Aut(C))^{0}/R_{u}(f_{1}(Aut(C))^{0}).

The proof of the above is a consequence of the following standard structural result.

Lemma 2.5. (i) Let G be a connected linear algebraic group G over k. Then R_{u}(G) contains every normal unipotent subgroup of G.

(ii) Let φ : G → G′ be an epimorphism of connected algebraic groups such that Ker(φ) is a unipotent subgroup of G. Then there is a bijective morphism

φ̂ : G/R_{u}(G) → G′/R_{u}(G′)

such that φ̂(gR_{u}(G)) = φ(g)R_{u}(G′) for all g ∈ G.

Proof. (i) This follows by [3, Theorem 30.4 (b)] and [3, Exercise 26.14].

(ii) Set K = Ker(φ). By (i), K ≤ R_{u}(G) and by [3, Corollary 21.3F,Theorem 30.4 (b), Exercise 26.14], φ(R_{u}(Ḡ)) = R_{u}(G′). The result follows.

□

Proof of Proposition 2.4. (i) Since C is local, Aut(C) = Ker(f_{0}). On the other hand, Ker(f_{1}) ≤ Ker(f_{i}), for all i ≥ 1. Hence, if φ ∈ Ker(f_{1}), then φ induces the identity on J^{i}(C)/J^{i+1}(C) for all i ≥ 0. This proves (i).

(ii) Let x be an invertible element of C. Since C is local, x = λ(1+y) for some 0 ≠ λ ∈ k and some y ∈ J. So x^{−1} = λ^{−1}(1 + y′), y′ ∈ J and conjugation by x induces the identity on J(C)/J^{2}(C).

(iii) By [3, Proposition 7.4B], f_{1}(Aut^{0}(C)) = f_{1}(Aut(C))^{0}. So, by (ii), the restriction of f_{1} to Aut^{0}(C) yields a surjective morphism

f̄_{1} : Out^{0}(C) → f_{1}(Aut(C))^{0}.

By (i), the kernel of f_{1} is a unipotent subgroup of Aut(C), hence the kernel of f̄_{1} is a unipotent subgroup of Out^{0}(C). So (iii) follows from Lemma 2.5 applied with φ equal to f̄_{1}.

□

For u, v elements of a ring, [u, v] denotes the commutator uv − vu of u and v.

Lemma 2.6. Let C be a k-algebra such that J^{3}(C) ⊆ Z(C). For u ∈ J(C), let φ_{u} : C → C be the automorphism defined by φ_{u}(v) = (1 − u)v(1 − u)^{−1}, v ∈ C. Then, for any u, v ∈ J(C), φ_{u}(v) = v + [v, u] + [vu, u].

Proof. Let u, v ∈ J(C) and suppose that u^{n} = 0. Then, (1 − u)^{−1} = 1 + u + ⋯ + u^{n−1}, and for any v ∈ J(C),

φ_{u}(v) = v + [v, u] + [vu, u] + ⋯ + [vu^{n−1}, u].

Since J^{3}(C) ⊆ Z(C), the result follows. □

3. Preliminary Calculations

For the rest of the paper, unless stated otherwise, k will denote an algebraically closed field of characteristic p and B = k⟨X, Y ⟩/⟨X^{3}, Y^{3}, XY + Y X⟩ be as in Theorem 1.2. Throughout this section A will denote a k-algebra satisfying the following:

Hypothesis 3.1. A is symmetric, local, dim_{k}(A) = 9, and Z(A) ≅ Z(B) as k-algebras.

We fix a symmetrizing form s : A → k on A. For a subset U of A denote by U^{⊥} the subset of A consisting of elements orthogonal to U under the bilinear form associated to s, that is U^{⊥} is the set of elements a′ ∈ A such that s(aa′) = 0 for all a ∈ U. For each i ≥ 0, let J_{i} = J^{i}(A), Z = Z(A), Z_{i} := Z(A) ∩ J^{i}(A).

Lemma 3.2. Suppose that A satisfies Hypothesis 3.1. Then, Z has a k-basis {1, z_{i}, 1 ≤ i ≤ 5} such that

z_{1}z_{2} = z_{2}z_{1} = z_{5}, z_{i}z_{j} = 0 for all i, j s.t. {i, j} ≠ {1, 2}

and {z_{i}, 1 ≤ i ≤ 5} is a k-basis of Z_{1}. Further, {z_{i}, 3 ≤ i ≤ 5} is a k-basis of Soc(Z(A)) and dim_{k}[A, A] = 3.

Proof. Since Z(A) ≅ Z(B), in order to prove the first statement, it suffices to prove an analogous statement for Z(B). Set

z′_{1} := X^{2}, z′_{2} := Y^{2}, z′_{3} := XY^{2}, z′_{4} := Y X^{2}, z′_{5} := X^{2}Y^{2} ∈ Z(B).

Then {1, z′_{i}, i ≤ i ≤ 5} is a basis of Z(B) and z′_{1}z′_{2} = z′_{2}z′_{1} = z′_{5} and all other pairwise products are 0. Setting z_{i} to be the image of z_{i} under some isomorphism Z(A) ≅ Z(B) yields the first statement. The elements z_{i}, 1 ≤ i ≤ 5 are all nilpotent and central in A, hence z_{i} ∈ Z_{1} for 1 ≤ i ≤ 5. This proves the second assertion. The assertion on Soc(Z(A)) is immediate from the multiplication of the z′_{i} and the last assertion follows by Lemma 2.1. □

Lemma 3.3. Suppose that A satisfies Hypothesis 3.1. For x ∈ J and i ∈ ℕ, denote by C_{J_{i}}(x) the subring of J_{2} of elements of J_{2} which commute with x.

(i) J_{2} ⊈ Z_{2}.

(ii) J_{3} ⊆ Z_{3}.

(iii) dim_{k}(J_{1}/(Z_{1} + J_{2})) = 2 and dim_{k}(J_{2}/Z_{2}) = 1.

(iv) There exists a basis

{x + Z_{1} + J_{2}, y + Z_{1} + J_{2}}

of J_{1}/(Z_{1} + J_{2}) such that

C_{J_{2}}(x) = C_{J_{2}}(y) = Z_{2}.

(v) Z_{1} is an ideal of A.

(vi) For any u, v ∈ J, uv + vu ∈ Z_{2}.

(vii) Z_{1}^{2} = Soc(A).

(viii) [A, A] ⊈ Z_{1} and Soc(Z(A)) = [A, A] ∩ Z_{1} ⊕ Soc(A).

(ix) J_{5} = 0, J_{4} = Z_{1}^{2} = Soc(A).

Proof. (i) Note that since A is local, any subspace of J_{1} which contains J_{2} is an ideal of A, and further since [A, A] ⊆ J_{2}, any such subspace contains [A, A].

Suppose if possible that J_{2} ⊆ Z_{2} ⊆ Z_{1}. Then Z_{1} and hence Z_{1}^{2} is an ideal of A. In particular, Z_{1}^{2} ∩ Soc(A) ≠ 0. By Lemma 2.1, Soc(A) is 1-dimensional and by Lemma 3.2, Z_{1}^{2} is 1-dimensional. Hence, Z_{1}^{2} = Soc(A), whence by Lemma 2.1, [A, A] ∩ Z_{1}^{2} = 0. On the other hand, since [A, A] ⊆ Z_{1}, and since za = az for any a ∈ A, z ∈ Z_{1},

[A, A]Z_{1} ⊆ [A, A] ∩ Z_{1}^{2}.

Thus, [A, A]Z_{1} = 0, that is [A, A] ⊆ Soc(Z(A)). But

dim_{K}([A, A]) = 3 = dim_{k}(Soc(Z(A))),

hence [A, A] = Soc(Z(A)). This is a contradiction as by Lemma 2.1 (ii), Soc(A) ⊆ Soc(Z(A)) whereas by Lemma 2.1 (iii) [A, A] ∩ Soc(A) = 0.

(ii) By (i) and Lemma 2.3, dim_{k}(J_{3}/J_{4}) ≥ 2, dim_{k}(J_{2}/J_{3}) ≥ 2 and dim_{k}(J_{1}/J_{2}) ≥ 2. So, dim_{k}(J_{4}) ≤ 2. By Lemma 2.1 (i),(vi), it follows that dim_{k}(J_{4}/J_{5}) ≤ 1. Hence by Lemma 2.3, J_{3} ⊆ Z.

(iii) Let c be the codimension of J_{2} + Z_{1} in J_{1}. Let A be a basis of a complement of J^{2}(A) + Z_{1} in J(A) and let A′ be a (possibly empty) basis of a complement of J^{2}(A) in J^{2}(A) + Z_{1} consisting of elements of Z_{1}. Then {1} ∪ A ∪ A′ is a generating set for A as an algebra. If the elements of A pairwise commute, then {1} ∪ A ∪ A′ is a set of pairwise commuting generators of A, an impossibility as A is not commutative. So c ≥ 2. On the other hand, the codimension of Z_{1} in J_{1} is 3, so c ≤ 3. Suppose if possible that c = 3. Then

dim_{k}(J_{2} + Z_{1}) = 5 = dim_{k}(Z_{1}),

hence J_{2} ⊆ Z_{1}, a contradiction to (i). Thus, c = 2. The second assertion is immediate from the first and (i).

(iv) Let A = {x, y} and A′ be as in (ii). Since {1} ∪ A ∪ A′ is a generating set for A as an algebra,

Z_{2} = C_{J_{2}}(x) ∩ C_{J_{2}}(y).

As dim_{k}Z_{1} = 5 and x ∈ C_{J_{1}}(x), dim_{k}C_{J_{1}}(x) ≥ 6. Similarly, dim_{k}C_{J_{1}}(y) ≥ 6. On the other hand, since dim_{k}J_{1} = 8, and dim_{k}Z_{1} = 5, at least one of dim_{k}C_{J_{1}}(x) and dim_{k}C_{J_{2}}(y) is at most 6. Suppose that dim_{k}C_{J_{1}}(x) = 6 and dim_{k}C_{J_{1}}(y) = 7. Then by the same argument applied to {x + y, y}, it follows that dim_{k}C_{J_{1}}(x + y) = 6. Hence replacing y with x + y, we may assume that

dim_{k}C_{J_{1}}(x) = 6 = dim_{k}C_{J_{1}}(y).

Thus, C_{J_{1}}(x) is the k-span of {x, Z_{1}} and C_{J_{2}}(y) is the k-span of {y, Z_{1}}.Consequently, C_{J_{2}}(x) ⊆ Z_{2} and C_{J_{2}}(y) ⊂ Z_{2}.

From now on let A = {x, y} ⊆ J_{1} be such that {x + Z_{1} + J_{2}, y + Z_{1} + J_{2}} is a basis of J_{1}/(Z_{1} + J_{2}) and C_{J_{2}}(x) = C_{J_{2}}(y) = Z. Let A′ be a basis of a complement of J_{2} in J_{2} + Z_{1} consisting of elements of Z_{1}.

(v) Since A is local, and since A′ ⊆ Z_{1}, it suffices to prove that xZ_{1} ⊆ Z_{2} and xZ_{1} ⊆ Z_{2}. So, let z ∈ Z_{1}. Then, xz ∈ C_{J_{2}}(x), so by the above xz ∈ Z_{2}. Similarly, yz ∈ Z_{2}.

(vi) By (ii), J_{3} ⊆ Z_{3} and by (v), Z_{1} is an ideal of A. Hence it suffices to prove that x^{2}, y^{2}, xy + yx ∈ Z_{2}. Since x^{2} ∈ C_{J_{2}}(x), x^{2} ∈ Z_{2}. Similarly, y^{2} ∈ Z_{2}. Also,

x(xy + yx) = x^{2}y + xyx = yx^{2} + xyx = (xy + yx)x

and similarly,

y(xy + yx) = (xy + yx)y,

so xy + yx ∈ Z_{2}.

(vii) By (v), Z_{1} is an ideal of A. So, 0 ≠ Z_{1}^{2} is an ideal of A, and hence Soc(A) ⊆ Z_{1}^{2}. But since Z_{1}^{2} is one dimensional, it follows that Soc(A) = Z_{1}^{2}.

(viii) Since Z_{1}([A, A] ∩ Z_{1}) ⊆ [A, A] ∩ Z_{1}^{2} and since by (vii), Z_{1}^{2} = Soc(A), by Lemma 2.1 (iii),

([A, A] ∩ Z_{1})Z_{1} = 0.

Hence [A, A] ∩ Z_{1} ⊆ Soc(Z(A)) and again by Lemma 2.1 (iii), [A, A] ∩ Z_{1} ∩ Soc(A) = 0. The result follows since by (iii), the co-dimension of [A, A] ∩ Z_{1} in [A, A] is at most 1.

(ix) Let x_{i} ∈ A ∪ A′, 1 ≤ i ≤ 5. If one of the x_{i}’s is in A′, then by (vi) and (vii)

x_{1}x_{2}x_{3}x_{4}x_{5} ∈ Z_{1}J_{3}J_{1} ⊆ Z_{1}^{2}J_{1} = 0.

So, in order to show that J_{5} = 0, we may assume that x_{i} ∈ A, 1 ≤ i ≤ 5. First suppose that for some i, 1 ≤ i ≤ 4, x_{i} = x_{i+1}, say x_{i} = x_{i+1} = x, By (v) and (vi), for some r, s ≥ 0 such that r + s = 5,

x_{1}x_{2}x_{3}x_{4}x_{5} = x^{r}y^{s} ∈ Z_{1}^{2}J_{1} = 0.

Suppose now that no two consecutive x_{i}’s are equal, so

x_{1}x_{2}x_{3}x_{4}x_{5} = xyxyx

or

x_{1}x_{2}x_{3}x_{4}x_{5} = yxyxy.

In the former case, by (ii) and (v),

x_{1}x_{2}x_{3}x_{4}x_{5} = x^{2}yxy

and we are back in the previous situation. The second case is similar.

Suppose if possible that J_{4} = 0. Then J_{3} = Soc(A) is 1-dimensional, and by Lemma 2.3, J_{2} ⊆ Z_{1}, a contradiction to (i). □

Lemma 3.4. Suppose that A satisfies Hypothesis 3.1 and let {x + Z_{1} + J_{2}, y + Z_{1} + J_{2}} be a basis of J_{1}/(Z_{1} + J_{2}). Then, {xy + Z_{2}} is a basis of J_{2}/Z_{2}, dim_{k}[A, A] + J_{3}/J_{3} = 1, {[x, y] + J_{3}} is a basis of ([A, A] + J_{3})/J_{3} and {[x, y], [x, xy], [y, xy]} is a basis of [A, A], Proof. By Lemma 3.3 (v) and (vi), Z_{1}J_{1} ⊆ Z_{2} and x^{2}, y^{2}, xy + yx ∈ Z_{2}, and by Lemma 3.3 (i), J_{2} ⊈ Z_{2}. Hence, xy ∉ Z_{2} and it follows by Lemma 3.3 (iii) that {xy + Z_{2}} is a basis of J_{2}/Z_{2}.

By Lemma 3.3 (ii), J_{3} ⊆ Z and by Lemma 3.3 (viii) [A, A] ⊈ Z. So [A, A] ⊈ J_{3}. On the other hand, [A, A] + J_{3}/J_{3} is spanned by [x, y]. So, {[x, y] + J_{3}} is a basis of [A, A] + J_{3}/J_{3} and dim_{k}([A, A]+J_{3}/J_{3}) = 1. It follows from this that {x, y, [x, y], Z_{1}} spans J_{1}, and hence that {[x, y], [x, xy], [y, xy]} spans [A, A]. Since dim_{k}[A, A] = 3, {[x, y], [x, xy], [y, xy]} is a basis of [A, A].

□

Proposition 3.5. Suppose that A satisfies Hypothesis 3.1. Then one of the following holds.

(1)  dim_{k}(J_{1}/J_{2}) = 3, dim_{k}(J_{2}/J_{3}) = 2, dim_{k}(J_{3}/J_{4}) = 2, dim_{k}(J_{4}) = 1
(2)  dim_{k}(J_{1}/J_{2}) = 2, dim_{k}(J_{2}/J_{3}) = 3, dim_{k}(J_{3}/J_{4}) = 2, dim_{k}(J_{4}) = 1

Moreover, Z_{1}J_{1} ⊆ J_{3}.

Proof. As in the proof of Lemma 3.3 (ii),

dim_{k}(J_{3}/J_{4}) ≥ 2, dim_{k}(J_{2}/J_{3}) ≥ 2, dim_{k}(J_{1}/J_{2}) ≥ 2.

By Lemma 3.3, J_{4} = Soc(A) is 1-dimensional. Thus, we have the following possibilities:

dim_{k}(J_{1}/J_{2}) = 3, dim_{k}(J_{2}/J_{3}) = 2, dim_{k}(J_{3}/J_{4}) = 2, dim_{k}(J_{4}) = 1
dim_{k}(J_{1}/J_{2}) = 2, dim_{k}(J_{2}/J_{3}) = 3, dim_{k}(J_{3}/J_{4}) = 2, dim_{k}(J_{4}) = 1,
dim_{k}(J_{1}/J_{2}) = 2, dim_{k}(J_{2}/J_{3}) = 2, dim_{k}(J_{3}/J_{4}) = 3, dim_{k}(J_{4}) = 1.

We show that the last is not possible. Suppose, if possible that J_{1}/J_{2} has a basis {x + J_{2}, y + J_{2}} and that dim_{k}(J_{2}/J_{3}) = 2. Then {xy + J_{3}, yx + J_{3}, x^{2} + J_{3}, y^{2} + J_{3}} is a spanning set for J_{2}/J_{3}. By Lemma 3.3 (vi), x^{2}, y^{2}, xy + yx ∈ Z_{2} and by Lemma 3.3(i), J_{2} ⊈ Z_{2}. So, {xy + yx + J_{3}, x^{2} + J_{3}, y^{2} + J_{3}} spans Z_{2}/J_{3}. Suppose first that x^{2}, y^{2} ∈ J_{3}. Since Z_{2} + J_{3} has codimension 1 in J_{2}, xy + yx ∉ J_{3}. So, replacing x with x′ = x + y, we may assume that x^{2} ∉ J_{3} and hence that {x^{2} + J_{3}} is a basis of Z_{2}/J_{3}. Thus, by Lemma 3.4, {xy + J_{3}, x^{2} + J_{3}} is a basis of J_{2}/J_{3}. But then, by Lemma 2.2, {x^{2}y + J_{4}, x^{3} + J_{4}} is a spanning set of J_{3}/J_{4}, a contradiction. This proves the first assertion.

If dim_{k}(J_{1}/J_{2}) = 2, then Z_{1} ⊆ J_{2}, and the second statement is trivial. Thus, we may assume that dim_{k}(J_{1}/J_{2}) = 3 and hence by the first assertion that dim_{k}(J_{2}/J_{3}) = 2. Let {x+J_{2}, y+J_{2}, z+J_{2}} be a basis of J_{1}/J_{2} with z ∈ Z and suppose if possible that Z_{1}J_{1} ⊈ J_{3}. Since by Lemma 3.3 (iii), Z_{2} has codimension 1 in J_{2}, {x + Z_{1} + J_{2}, y + Z_{1} + J_{2}} is a basis of J_{1}/(Z_{1} + J_{2}). By Lemma 3.3(viii), (ix), z^{2} ∈ Soc(Z(A)) = J_{4}, hence either xz ∉ J_{3} or yz ∉ J_{3}, say xz ∉ J_{3}. By Lemma 3.3(v), xz ∈ Z_{2} and by Lemma 3.4, {xy + Z_{2}} is a basis of J_{2}/Z_{2}. Thus, {xy + J_{3}, xz + J_{3}} is a basis of J_{2}/J_{3}. By Lemma 2.2, {x^{2}y + J_{4}, x^{2}z + J_{4}} spans J_{3}/J_{4}. But by Lemma 3.3(vi), x^{2} ∈ Z_{1} whence by Lemma 3.3(vii), (ix)

x^{2}z ∈ Z_{1}^{2} = Soc(A) = J_{4}.

So (dim_{k}J_{3}/J_{4}) ≤ 1, a contradiction. □

4. The case dim_{k}J_{1}/J_{2} = 3.

In this section, we will work under the following hypothesis.

Hypothesis 4.1. A is symmetric, local, dim_{k}(A) = 9, Z(A) ≅ Z(B) as k-algebras and dim_{k}(J_{1}/J_{2}) = 3.

With the above hypothesis, by Lemma 3.3 J_{1}/J_{2} has a basis of the form {x + J_{2}, y + J_{2}, z + J_{2}}, with z ∈ Z_{1}.

Lemma 4.2. Suppose A satisfies Hypothesis 4.1. There exists a basis {x+J_{2}, y+J_{2}, z+J_{2}} of J_{2}/J_{3} such that z ∈ Z, x^{2} ∉ J_{3}, and xy + yx = 0.

Proof. Let {x + J_{2}, y + J_{2}, z + J_{2}} be a basis of J_{2}/J_{3} with z ∈ Z. By Proposition 3.5, xz, zx ∈ J_{3}, so {x^{2} + J_{3}, y^{2} + J_{3}, xy + J_{3}, yx + J_{3}} spans J_{2}/J_{3}. By Lemma 3.3 (vi), x^{2}, y^{2}, xy + yx ∈ Z_{2} and by Lemma 3.3 (i), J_{2} ⊈ Z_{2}. Thus, {x^{2} + J_{3}, y^{2} + J_{3}, xy + yx + J_{3}} spans Z_{2}/J_{3}. If either x^{2} ∉ J_{3} or y^{2} ∉ J_{3}, then interchanging x and y if necessary, we have x^{2} ∉ J_{3}. If both x^{2}, y^{2} ∈ J_{3} then xy + yx ∉ J_{3}. So,

(x + y)^{2} = x^{2} + y^{2} + (xy + yx) ∉ J_{3},

and replacing x with x + y we may assume that x^{2} ∉ J_{3}. Thus, since J_{2} ⊈ Z_{2} and dim_{k}(J_{2}/J_{3}) = 2, {x^{2} + J_{3}} is a basis of Z_{2}/J_{3}.

Since xy + yx ∈ Z_{2}, there exists an α ∈ k such that

xy + yx ≡ αx^{2} modJ_{3}

Set y′ = y − \frac{1}{2}αx. Then,

xy′ + y′x ≡ 0 modJ_{3}.

Since {x + J_{2}, y′ + J_{2}, z + J_{2}} is also a basis of J_{1}/J_{2}, by replacing y be y′, we may assume that xy + yx ∈ J_{3}.

Since {x^{2} +J_{3}} is a basis of Z_{2}/J_{3}, by Lemma 3.4, {xy +J_{3}, x^{2} +J_{3}} is a basis of J_{2}/J_{3}, hence by Lemma 2.2, {x^{2}y, x^{3}} is a basis of J_{3}/J_{4}. So,

xy + yx ≡ αx^{3} + βx^{2}y mod J_{4}

for some α, β ∈ k.

Set x′ = x − \frac{1}{2}βx^{2} and y′ = y − \frac{1}{2}αx^{2}. Then \{x′ + J_{2}, y′ + J_{2}, z + J_{2}\} is still a basis of J_{1}/J_{2}, z ∈ Z , x^{′2} ∉ J_{3} and

x′y′ + y′x′ ≡ (x − \frac{1}{2}βx^{2})(y − \frac{1}{2}αx^{2}) + (y − \frac{1}{2}αx^{2})(x − \frac{1}{2}βx^{2})
≡ xy + yx − αx^{3} − βx^{2}y
≡ 0 mod J_{4}

Hence replacing x with x′ and y with y′, we may assume that xy + yx ∈ J_{4}. Arguing as above, again {x^{2}y, x^{3}} is a basis of J_{3}/J_{4} and by applying Lemma 2.2 again, {x^{3}y, x^{4}} spans J_{4}. Since dim_{k}(J_{4}) = 1, either xy + yx = αx^{3}y or xy + yx = αx^{4} for some α ∈ k. Replacing x by x − \frac{1}{2}αx^{3} in the first case and y by y − \frac{1}{2}αx^{3} in the second case yields the result. □

Lemma 4.3. Suppose A satisfies Hypothesis 4.1 and let {x + J_{2}, y + J_{2}, z + J_{2}} be a basis J_{2}/J_{3} such that z ∈ Z, x^{2} ∉ J_{3} and xy + yx = 0. Then,

(i) For any u ∈ Z_{1}, uxy = xyu = 0. In particular,

x^{3}y = yx^{3} = xy^{3} = yx^{3} = 0.

(ii) {xy + J_{3}, x^{2} + J_{3}} is a basis of J_{2}/J_{3}, {x^{2}y + J_{4}, x^{3} + J_{4}} is a basis of J_{3}/J_{4} and {x^{4}} is a basis of J_{4}.

(iii) y^{2} ∉ J_{3}.

Proof. (i) Let u ∈ Z_{1}. Since xy = −yx, and since by Lemma 3.3 (v), uy ∈ Z, we have

uxy = −uyx = −x(uy) = −x(yu) = −uxy.

Thus, uxy = 0. The second assertion follows from the first as by Lemma 3.3 (vi), x^{2}, y^{2} ∈ Z_{2}.

(ii) The first assertion was proved in the course of the proof of Lemma 4.2. It follows from this and Lemma 2.2 that J_{3}/J_{4} is spanned by {x^{2}y + J_{4}, x^{3} + J_{4}} and J_{4} is spanned by {x^{3}y, x^{4}}. But by (i), x^{3}y = 0. Hence {x^{4}} is a basis of J_{4}.

(iii) By Lemma 3.4, {[x, y], [x, xy], [y, xy]} is a basis of [A, A]. Since xy = −yx, [y, [xy]] = 2xy^{2}. If y^{2} ∈ J_{3}, then

[y, [xy]] = 2xy^{2} ∈ J_{4} = Soc(A),

a contradiction. □

Lemma 4.4. Suppose A satisfies Hypothesis 4.1. Then there exists a basis {x + J_{2}, y + J_{2}, z + J_{2}} of J_{2}/J_{3} such that z ∈ Z, {x^{2} + J_{3}} is a basis of Z_{2}/J_{3}, xy + yx = 0 and such that the following holds.

(i) zx = xz = 0.

(ii) z^{2} = x^{4}.

(iii) zy = yz = 0.

(iv) y^{2} = x^{2} + αx^{3} + βx^{2}y for some α, β ∈ k.

Proof. Let {x + J_{2}, y + J_{2}, z + J_{2}} be a basis J_{2}/J_{3} such that z ∈ Z, {x^{2} + J_{3}} is a basis of Z_{2}/J_{3} and xy + yx = 0. Such a basis exists by Lemma 4.2.

By Lemma 3.3, y^{2} ∈ Z_{2}, and by Lemma 4.3, y^{2} ∉ J_{3}, so y^{2} ≅ λx^{2} mod J_{3}, for some 0 ≠ λ ∈ k. Replacing y by \sqrt{λ}^{−1}y does not affect the relation xy + yx = 0, hence we may assume that y^{2} ≅ x^{2} mod J_{3}. In particular, x^{2}y^{2} = x^{4}.

By Proposition 3.5, zx ∈ J_{3}. So, by Lemma 4.3, zx = αx^{3} +βx^{2}y+γx^{4} with α, β, γ ∈ k. By Lemma 4.3 (i), and by (i),

0 = zxy = αx^{3}y + βx^{2}y^{2} = βx^{2}y^{2} = βx^{4}.

So, β = 0. Since z′ := z − αx^{2} − γx^{3} is an element of Z and z′ ≡ z mod J_{2}, replacing z by z′, we may assume that zx = xz = 0 and (i) holds.

Since {x^{2}, z, x^{2}y, x^{3}, x^{4}} ⊆ Z_{1} and dim_{k}(Z_{1}) = 5, {x^{2}, z, x^{2}y, x^{3}, x^{4}} is a basis of Z_{1} by Lemma 4.3. Now {x^{2}y, x^{3}, x^{4}} ⊆ Soc(Z(A)) and dim_{k}(Soc(Z(A)) = 3, so z ∉ Soc(Z(A)). Since zx^{2} = 0, this means that z^{2} ≠ 0. On the other hand, by Lemma 3.3 (vii), z^{2} ∈ Soc(Z(A)) = Soc(A). Thus, z^{2} = δx^{4}, with 0 ≠ δ ∈ k. Replacing z with a constant multiple does not affect any of the already established properties. Hence, we may assume that z^{2} = x^{4} and (ii) holds.

We now show (iii). Again by Proposition 3.5, zy ∈ J_{3}, so by Lemma 4.3,

zy = αx^{3} + βx^{2}y + δx^{4},

with α, β, δ ∈ k. By Lemma 4.3,

0 = zyx = αx^{4},

whence α = 0. Since zJ ⊆ J_{3}, zJ_{3} = 0, and we have shown above that y^{2} ≅ x^{2} mod J_{3}. Hence by (i),

0 = zx^{2} = zy^{2} = βx^{2}y^{2} = βx^{4}.

But then β = 0 and zy = δx^{4}. Set y′ = y − δz. Then by (i)

y′x = yx = −xy = −xy′,

by (i), (ii) and Proposition 3.5,

y′^{2} ≡ y^{2} + δ^{2}z^{2} − 2δyz ≡ y^{2} ≡ x^{2}mod J_{3},

and by (ii),

y′z = yz − δz^{2} = 0.

Replacing y with y′ gives (ii).

Since y^{2} ≡ x^{2} modJ_{3}, y^{2} = x^{2} + αx^{3} + βx^{2}y + γx^{4} for some α, β, γ ∈ k.

Set y′ = y + \sqrt{γ}xy. Then y′x + xy′ = 0, y′z = zy′ = 0 and by Lemma 4.3,

y^{′2} = y^{2} − γx^{2}y^{2} = x^{2} + αx^{3} + βx^{2}y = x^{2} + αx^{3} + βx^{2}y′.

So, replacing y by y′ yields (iv). □

Proposition 4.5. Suppose A satisfies Hypothesis 4.1. Then there exists a basis {x+J_{2}, y+ J_{2}, z +J_{2}} of J_{1}/J_{2} such that {xy +J_{3}, yx+J_{3}} is a basis of of J_{2}/J_{3}, {xyx+J_{3}, yxy +J_{3}} is a basis of J_{3}/J_{4}, {xyxy} is a basis of J_{4} and such that the following relations hold.

(i) zx = xz = zy = yz = 0.

(ii) z^{2} = xyxy.

(iii) y^{2} = x^{2} = αxyx + βyxy, α, β ∈ k.

(iv) xyxy = yxyx,

(v) For any u_{i} ∈ {x, y, z}, 1 ≤ i ≤ 5, u_{1}u_{2}u_{3}u_{4}u_{5} = 0.

Moreover, the above is a complete set of generators and relations for the algebra A, that is A ≅ k⟨x, y, z⟩/I, where I is the ideal of k⟨x, y, z⟩, generated by:

{zx, xz, zy, yz, z^{2} − xyxy, y^{2} − x^{2}, y^{2} − αxyx + βyxy} ∪ {u_{1}u_{2}u_{3}u_{4}u_{5}, u_{i} ∈ {x, y, z}, 1 ≤ i ≤ 5}, α, β ∈ k.

Proof. Let {x + J_{2}, y + J_{2}, z + J_{2}} be a basis of J_{1}/J_{2} satisfying the conditions of Lemma 4.4. Let i be a primitive 4-th root of unity in k and set x′ = x + iy, y′ = x − iy, z′ = z. Then x^{′2} = y^{′2} ∈ J_{3} and x′z′ = z′x′ = z′y′ = y′z′ = 0. A monomial of two terms in x′, y′, z′ which involves any of x^{′2}, y^{′2} or z′ is in J_{3}, hence {x′y′ + J_{3}, y′x′ + J_{3}} spans J_{2}/J_{3} and is therefore a basis of J_{2}/J_{3}. Similarly, a monomial of three terms in x′, y′, z′ which involves any of x^{′2}, y^{′2} or z′ is in J_{4}, hence {x′y′x′ + J_{4}, y′x′y′ + J_{4}} is a basis of J_{3}/J_{4} and J_{4} is spanned by x′y′x′y′ = y′x′y′x′. Write

x^{′2} = y^{′2} = α′x′y′x′ + β′y′x′z′ + δx′y′x′y′.

Replacing x′ by x′ − \frac{δ}{2}y′x′y′ and y′ by y′ − \frac{δ}{2}x′y′x′ yields

x^{′2} = y^{′2} = α′x′y′x′ + β′y′x′y′.

Since z^{′2} is a non-zero element of J_{4}, replacing z′ by δz′ for a suitable δ, we may assume that z^{′2} = x′y′x′y′. Thus, {x′, y′, z′} satisfy the relations (i) -(v). The final assertion follows easily from this and the fact that dim_{k}(A) = 9. □

For the sake of completeness we record the following without proof.

Proposition 4.6. For any α, β ∈ k, the algebra k⟨x, y, z⟩/I, where I is the ideal generated by {zx, xz, zy, yz, z^{2} − xyxy, y^{2} − x^{2}, y^{2} − αxyx + βyxy} ∪ {u_{1}u_{2}u_{3}u_{4}u_{5}, u_{i} ∈ {x, y}, 1 ≤ i ≤ 5} is a local symmetric k-algebra of dimension 9 and with center isomorphic to Z(B).

5. The case dim_{k}(J_{1}/J_{2}) = 2.

The following lemma divides the case that dim_{k}(J_{1}/J_{2}) = 2 in two subcases.

Lemma 5.1. Suppose that A is symmetric, local, dim_{k}(A) = 9, Z(A) ≅ Z(B) as kalgebras and dim_{k}(J_{1}/J_{2}) = 2. Then there exists a basis {x + J_{2}, y + J_{2}} of J_{2}/J_{3} such that either xy + yx ∈ J_{3} or y^{2} ∈ J_{3}.

Proof. By Proposition 3.5, dim_{k}(J_{2}/J_{3}) = 3, and by Lemma 3.3,

dim_{k}(Z_{2}/J_{3}) = 2, dim_{k}(J_{2}/Z_{2}) = 2.

Let {x+J_{2}, y+J_{2}} be a basis of J_{1}/J_{2}. Then, {xy+J_{3}, yx+J_{3}, x^{2}+J_{3}, y^{2}+J_{3}} is a spanning set for J_{2}/J_{3}. By Lemma 3.3, {xy + yx, x^{2}, y^{2}} ⊆ Z_{2} and J_{2} ⊈ Z_{2}, so {xy + yx + J_{3}, x^{2} + J_{3}, y^{2} + J_{3}} spans Z_{2}/J_{3} and some two element subset of {xy + yx + J_{3}, x^{2} + J_{3}, y^{2} + J_{3}} is a basis of Z_{2}/J_{3}.

Suppose first that {x^{2}+J_{3}, y^{2}+J_{3}} is a basis of Z_{2}/J_{3}. Let xy+yx ≡ λx^{2}+μy^{2} mod J_{3}, λ, μ ∈ k. If λ = μ = 0, then xy + yx ∈ J_{3}. So, suppose that λ ≠ 0. By replacing y with λ^{−1}y, we may assume that λ = 1. First consider the case that μ ≠ 1. Set

σ = −1 + \sqrt{1 − μ}, τ = −1 − \sqrt{1 − μ}, x′ = x + σy, y′ = x + τy.

Then since σ ≠ τ, {x′ + J_{2}, y′ + J_{2}} is a basis of J_{1}/J_{2} and

x′y′ + y′x′ ≡ 0 mod J_{3}.

So, replacing {x, y} with {x′, y′} proves the result. Now consider the case μ = 1 and set y′ = y − x. Then

y^{′2} = y^{2} + x^{2} − (yx + xy) ∈ J_{3}.

Replacing y by y′ yields the result.

Now suppose that {xy + yx + J_{3}, x^{2} + J_{3}} is a basis of Z_{2}/J_{3} and let

y^{2} ≡ αx^{2} + β(xy + yx) mod J_{3}

for α, β ∈ k. Set y′ = y − βx. Then,

y^{′2} = y^{2} + β^{2}x^{2} − β(xy + yx) ≡ (α + β^{2})x^{2} mod J_{3}.

So, replacing y by y′ we may assume that

y^{2} ≡ γx^{2} mod J_{3}, γ ∈ k

If γ = 0, the result is proved. Otherwise, replacing y by γ^{−1}y, we may assume that

y^{2} ≡ x^{2} mod J_{3}.

Now set x′ = x + y and y′ = x − y. Then {x′ + J_{2}, y′ + J_{2}} is a basis of J_{1}/J_{2} and

x′y′ + y′x′ ∈ J_{3}.

So, replacing {x + J_{2}, y + J_{2}} with {x′ + J_{2}, y′ + J_{2}} yields the result. □

In view of the above Lemma, for the rest of the section, we will work under one of the following two hypotheses.

Hypothesis 5.2. A is symmetric, local, dim_{k}(A) = 9, Z(A) ≅ Z(B) as k-algebras dim_{k}(J_{1}/J_{2}) = 2 and J_{1}/J_{2} has a basis {x + J_{2}, y + J_{2}} with xy + yx ∈ J_{3}.

Hypothesis 5.3. A is symmetric, local, dim_{k}(A) = 9, Z(A) ≅ Z(B) as k-algebras, dim_{k}(J_{1}/J_{2}) = 2 and J_{1}/J_{2} has a basis {x + J_{2}, y + J_{2}} such that y^{2} ∈ J_{3}.

Lemma 5.4. Suppose that A satisfies Hypothesis 5.2. Then x and y may be chosen such that xy + yx = 0.

Proof. Since xy + yx ∈ J_{3}, [x, [xy]] ≡ 2x^{2}y mod J_{4} and [y, [xy]] ≡ −2xy^{2} mod J_{4}. On the other hand, by Lemma 3.4, and the fact that [A, A] ∩ J_{4} = 0, {[x, [xy]] + J_{4}, [y, [xy]] + J_{4}} is a basis of J_{3}/J_{4}. Thus, {x^{2}y + J_{4}, xy^{2} + J_{4}} is a basis of J_{3}/J_{4}. So,

xy + yx ≡ λx^{2}y + μxy^{2} mod J_{4}.

Set x′ = x − \frac{λ}{2}x^{2} and y′ = y − \frac{μ}{2}y^{2}. Then \{x′, y′\} is a basis of J_{1}/J_{2} and by Lemma 3.3,

x′y′ + y′x′ ≡ (x − \frac{λ}{2}x^{2})(y − \frac{μ}{2}y^{2}) + (y − \frac{μ}{2}y^{2})(x − \frac{λ}{2}x^{2}) ≡ 0 mod J_{4}

So, replacing {x, y} with {x′, y′}, we may assume that xy + yx ≡ 0 mod J_{4}. By the first part of the argument, {x^{2}y + J_{4}, xy^{2} + J_{4}} is a basis of J_{3}/J_{4}. Hence, by Lemma 2.2 xy + yx = λx^{2}y^{2} or xy + yx = λx^{3}y for some λ ∈ k. In the first case, replacing y by y′ := y − \frac{λ}{2}xy^{2} and in the second case replacing y by y′ := y − \frac{λ}{2}x^{2}y yields the result.

□

Proposition 5.5. Suppose that A satisfies Hypothesis 5.2. Then there exists a basis {x + J_{2}, y + J_{2}} of J_{1}/J_{2} such that {x^{2} + J_{3}, y^{2} + J_{3}, xy + J_{3}} is a basis of J_{2}/J_{3}, {x^{2}y + J_{4}, xy^{2} + J_{4}} is a basis of J_{3}/J_{4}, {x^{2}y^{2}} is a basis of J_{4}, and J^{5} = {0} and such that the following relations hold.

(i) xy + yx = 0.

(ii) x^{3} = αxy^{2} +βx^{2}y^{2} and y^{3} = γx^{2}y+δx^{2}y^{2} for some α, β, γ, δ ∈ k, with α, γ ∈ {0, 1}. (iii) x^{3}y = xy^{3} = 0. (iv)For any u_{i} ∈ {x, y}, 1 ≤ i ≤ 5, u_{1}u_{2}u_{3}u_{4}u_{5} = 0. The above is a complete set of generators and relations for the algebra A, that is A ≅

k⟨x, y, ⟩/I, where I is the ideal of k⟨x, y⟩, generated by:

{xy+yx, x^{3} −αxy^{2} −βx^{2}y^{2}, y^{3} −γx^{2}y−δx^{2}y^{2}, x^{3}y, xy^{3}}∪{u_{1}u_{2}u_{3}u_{4}u_{5}, u_{i} ∈ {x, y}, 1 ≤ i ≤ 5}, α, γ ∈ {0, 1}, β, δ ∈ k.

Proof. By Lemma 5.4, there is a basis {x + J_{2}, y + J_{2}} of J_{1}/J_{2} with xy + yx = 0. Since x^{3}, y^{3} ∈ J_{3} ⊆ Z by Lemma 3.3, the relation xy + yx = 0 forces x^{3}y = −yx^{3} and xy^{3} = −y^{3}x. Thus, x^{3}y = xy^{3} = 0. As noted in the proof of Lemma 5.4, {x^{2}y + J_{4}, xy^{2} + J_{4}} is a basis of J_{3}/J_{4} and either {x^{3}y} or {x^{2}y^{2}} spans J_{4}. But, x^{3}y = 0. Thus, x^{2}y^{2} is a basis of J_{4}. Write

x^{3} = λx^{2}y + αxy^{2} + βx^{2}y^{2}.

Multiplying on the right with y yields

0 = λx^{2}y^{2}

hence λ = 0. Similarly,

y^{3} = γx^{2}y + δx^{2}y^{2}.

If α = 0, set y′ = y, and if α ≠ 0, set y′ = \sqrt{α^{−1}}y. If γ = 0, set x′ = x, and if γ ≠ 0, set x′ = \sqrt{γ^{−1}}x. Then replacing \{x, y\} with \{x′, y′\} gives α, γ ∈ \{0, 1\}. Thus (i)-(iv) are satisfied. The final assertion follows from the fact that any algebra satisfying (i)-(iii) has dimension at most 9. □

For A satisfying Hypothesis 5.3, we will require only partial structure results.

Lemma 5.6. Suppose that A satisfies hypothesis 5.3. Then, (i) {x^{2}, xy, yx} is a basis of J_{2}/J_{3}. (ii) yxy ∉ J_{4}. (iii) {xyxy} is a basis of J_{4}. (iv) {xyx + J_{4}, yxy + J_{4}} is a basis of J_{3}/J_{4}.

Proof. (i) is immediate from the first part of the proof of Lemma 5.1. By Lemma 3.4, {[xy, x], [xy, y]} is a basis of J_{3}/J_{4}. But [xy, y] = yxy − y^{2}x ≡ yxy mod J_{4}. This proves (ii). Suppose that xyxy = 0. Since yxy ∈ Z, this also means that yxyx = 0. Since y^{2} ∈ J_{3}, yyxy = 0 = yxyy. Thus yxy ∈ Soc(A) = J_{4}, contradicting (ii). Thus, xyxy ≠ 0, proving (iv). Since dim_{k}(J_{3}/J_{4}) = 2, by (ii), in order to prove (iv) it suffices to show that xyx + J_{4} ≠ λyxy + J_{4} for λ ∈ k. If this were the case then multiplying on both sides on the right with y would give xyxy = λyxyy = 0 and this would contradict (ii) as yxyx = xyxy. □

Lemma 5.7. Suppose that A satisfies Hypothesis 5.3. Then, x and y can be chosen such that

(i) x^{2}y ∈ J_{4}.

(ii) x^{3}y = 0.

(iii) x^{3} ≡ yxy mod J_{4}.

Proof. Let {x + J_{2}, y + J_{2}} be a basis of of J_{1}/J_{2} with y^{2} ∈ J_{3}. By Lemma 5.6, we may write

x^{2}y ≡ αxyx + βyxy mod J_{4}.

Then,

0 = x^{2}y^{2} = αxyxy.

Hence α = 0. Now

(x − βy)^{2}y ≡ x^{2}y − βyxy ≡ 0 mod J_{4}

so replacing x by x′ = x − βy we may assume that (i) holds. (ii) is immediate from (i).

By Lemma 5.6,

x^{3} ≡ αxyx + βyxy mod ∈ J_{4}.

Multiplying with y on the right yields α = 0. Now suppose if possible that β = 0. Then multiplying with x yields x^{4} = 0. By (ii)

x^{2}(xy + yx) = 2x^{3}y = 0.

Since {xy + yx, x^{2}, xyx, yxy, xyxy} is a basis of Z_{1} = J(Z) and since {xyx, yxy, xyxy} ⊆ Soc(Z(A), it follows that x^{2} ∈ Soc(Z(A)), and hence that dim_{k}(Soc(Z(A)) ≥ 4, a contradiction. So β ≠ 0. Replacing x by x′ := \sqrt{β^{−1}}x yields a basis satisfying (i), (ii), (iii).

□

6. Outer automorphism groups

In this section we will keep to the notation introduced for outer automorphism groups in Section 2.

Proposition 6.1. Suppose that A satisfies Hypothesis 4.1, and let α, β be as in Proposition 4.5.

(i) If either α or β is non-zero, then Out^{0}(A)/R_{u}(Out^{0}(A)) is contained in a onedimensional torus.

(ii) If α = β = 0, then Out^{0}(A)/R_{u}(Out^{0}(A)) is a two-dimensional torus.

Proof. Let {x + J_{2}, y + J_{2}, z + J_{2}} be a basis of J_{1}/J_{2} as in Proposition 4.5. Let B be the ordered basis {x, y, z, xy, yx, xyx, yxy, xyxy} of J_{1} and let φ : A → A be a k-linear map. For each u ∈ B, write φ(u) = ∑_{v∈B} λ_{v,u}v. Suppose that φ ∈ Aut(A). Then,

0 ≡ φ(x)^{2} ≡ λ_{x,x}λ_{y,x}(xy + yx) mod J_{3},
0 ≡ φ(y)^{2} ≡ λ_{x,y}λ_{y,y}(xy + yx) mod J_{3},
λ_{x,x}λ_{y,x} = λ_{x,y}λ_{y,y} = 0,

that is either λ_{x,y} = λ_{y,x} = 0 or λ_{x,x} = λ_{y,y} = 0. Identifying GL(J_{1}/(J_{2} + Z_{1}) with GL_{2}(k) through the ordered basis {x + J_{2} + Z_{1}, y + J_{2} + Z_{1}} it follows that the connected component of f_{J_{1},J_{2}+Z_{1}}(Aut(A))^{0} is contained in the subgroup of diagonal matrices. Hence, if φ ∈ Aut^{0}(A) then

(3)  λ_{x,y} = λ_{y,x} = 0.

We will assume from now on that this is the case.

Since φ(z) ∈ Z(A),

(4)  λ_{x,z} = λ_{y,z} = 0

So,

0 ≡ φ(z)φ(x) ≡ λ_{xy,z}λ_{x,x}xyx mod J_{4}, and 0 ≡ φ(z)φ(y) ≡ λ_{yx,z}λ_{y,y}yxy mod J_{4},

(5)  λ_{xy,z} = λ_{yx,z} = 0.

So,

λ_{x,x}^{2}λ_{y,y}^{2}xyxy = φ(x)φ(y)φ(x)φ(y) = φ(z)^{2} = λ_{z,z}^{2}xyxy,

yielding

λ_{z,z}^{2} = λ_{x,x}^{2}λ_{y,y}^{2}.

Thus, if φ ∈ Aut^{0}(A) then

(6)  λ_{z,z} = λ_{x,x}λ_{y,y}.

Identify GL(J_{1}/J_{2}) with GL_{3}(k) via the ordered basis {x+J_{1}, y+J_{1}, z+J_{1}} of J_{1}/J_{2}. By Equations 3 and 6, f_{1}(Aut^{0}(A)) is contained in the subgroup of lower triangular matrices (a_{ij}) satisfying a_{33} = a_{11}a_{22}, a_{2,1} = 0. So R_{u}(f_{1}(Aut^{0}(A)) is the subgroup of f_{1}(Aut^{0}(A)) consisting of matrices for which a_{ii} = 1, 1 ≤ i ≤ 3.

(i) Suppose first that β ≠ 0. Comparing the coefficient of yxy on both sides of the relation φ(x)^{2} = αφ(x)φ(y)φ(x) + βφ(y)φ(x)φ(y) yields

λ_{x,x}^{2}β = λ_{x,x}λ_{y,y}^{2}β,

hence, λ_{x,x} = λ_{y,y}^{2}. It follows from the above discussion that f_{1}(Aut^{0}(A))/R_{u}((Aut^{0}(A)) is isomorphic to a subgroup of the group of diagonal matrices of GL_{3}(k) satisfying a_{11} = a_{2,2} and a_{3,3} = a_{2,2}^{3}. So, the result follows by Proposition 2.4. Note that f_{1}(Aut^{0}(A)) = f_{1}(Aut(A))^{0} (see [3, Proposition 7.4B]).

Similarly, if α ≠ 0, then comparing the coefficient of xyx in the relation φ(y)^{2} = αφ(x)φ(y)φ(x) + βφ(y)φ(x)φ(y) yields that f_{1}(Aut^{0}(A))/R_{u}((Aut^{0}(A)) is isomorphic to a subgroup of the group of diagonal matrices of GL_{3}(k) satisfying a_{22} = a_{11} and a_{3,3} = a_{11}^{3}.

(ii) Suppose from now on that α = β = 0. Then,

0 = φ(x)^{2} = λ_{x,x}(λ_{xy,x} + λ_{yx,x})xyx + (2λ_{x,x}λ_{yxy,x} + λ_{xy,x}^{2} + λ_{yx,x}^{2})xyxy,

giving

(7)  λ_{yx,x} = −λ_{xy,x}, λ_{yxy,x} = −λ_{xy,x}^{2}λ_{x,x}^{−1}.

Similarly,

(8)  λ_{xy,y} = −λ_{yx,y}, λ_{xyx,x} = −λ_{yx,y}^{2}λ_{y,y}^{−1}.

Then,

0 = φ(z)φ(x) = (λ_{z,z}λ_{z,x} + λ_{x,x}λ_{yxy,z})yxyx,

and

0 = φ(z)φ(y) = (λ_{z,z}λ_{z,y} + λ_{y,y}λ_{xyx,z})yxyx,

which gives

(9)  λ_{yxy,z} = −λ_{z,z}λ_{z,x}λ_{x,x}^{−1}, λ_{xyx,z} = −λ_{z,z}λ_{z,y}λ_{y,y}^{−1}

Conversely, it is easy to check that any element of GL(J) satisfying the Equations 3, 4, 5, 6, 7, 8 and 9 and such that the image under φ of elements of B ∩ J_{2} is the multiplicative extension of the images of x, y and z of φ is in Aut^{0}(A). So, identifying as in (i), GL(J_{1}/J_{2}) with GL_{3}(k) via the ordered basis {x + J_{1}, y + J_{1}, z + J_{1}} of J_{1}/J_{2}, f_{1}(Aut^{0}(A)) is the subgroup of lower triangular matrices (a_{ij}) satisfying a_{33} = a_{11}a_{22}, a_{2,1} = 0, R_{u}(f_{1}(Aut^{0}(A))) is the subgroup of f_{1}(Aut^{0}(A)) consisting of matrices which satisfy additionally that a_{1,1} = a_{2,2} = a_{3,3} = 1 and f_{1}(Aut^{0}(A))/R_{u}(f_{1}(Aut^{0}(A))) is isomorphic to the subgroup of diagonal matrices satisfying a_{3,3} = a_{1,1}a_{2,2}. So, (ii) follows by Proposition 2.4. □

For the next result, note that the centre of an affine algebraic group is a closed subgroup of the group, and hence it makes sense to speak of the dimension of the centre as an algebraic variety.

Proposition 6.2. Suppose that A satisfies Hypothesis 4.1 and that the elements α, β Proposition 4.5 are both equal to 0. Then

dim(Z(R_{u}(Out^{0}(A)))) = 3.

Proof. Let U = R_{u}(Out^{0}(A)), let {x+J_{2}, y+J_{2}, z+J_{2}} be a basis of J_{1}/J_{2} as in Proposition 4.5 and let B be the ordered basis {x, y, z, xy + yx, xy − yx, xyx, yxy, xyx, xyxy} of J (note the difference with the basis used in Proposition 6.1). Identify Aut^{0}(A) with a subgroup of GL_{8}(k) through the basis B. Let B_{8}(k) be the subgroup of GL_{8}(k) consisting of lower triangular matrices and let U_{8}(k) be the subgroup of B_{8}(k) of strictly unitriangular matrices. Let φ ∈ GL_{8}(k) be such that that the image under φ of elements of B ∩ J_{2} is the multiplicative and linear extension of the images of x, y and z under φ. By the proof of Proposition 6.1, φ ∈ Aut^{0}(A) if and only if

λ_{x,y} = λ_{y,x} = λ_{xy,z} = λ_{yx,z} = 0;
λ_{xy+yx,x} = λ_{xy+yx,y} = λ_{xy+yx,z} = λ_{xy−yx,z} = 0;
λ_{yxy,x} = −λ_{xy−yx,x}^{2}λ_{x,x}^{−1}, λ_{xyx,y} = −λ_{xy−yx,y}^{2}λ_{y,y}^{−1};
λ_{yxy,z} = −λ_{z,z}λ_{z,x}λ_{x,x}^{−1}, λ_{xyx,z} = −λ_{z,z}λ_{z,y}λ_{y,y}^{−1}.

In particular, Aut^{0}(A) is contained in B_{8}(k) and R_{u}(Aut^{0}(A)) = Aut^{0}(A)∩U_{8}(k). Thus R_{u}(Aut^{0}(A)) is the following closed subgroup of U_{8}(k):

\{( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ a & e & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ b & f & 0 & 0 & 1 & 0 & 0 & 0 \\ c & −f^{2} & −e & 0 & −2f & 1 & 0 & 0 \\ −b^{2} & g & −a & 0 & −2b & 0 & 1 & 0 \\ d & h & i & 2(g + ae + 2bf + c) & 0 & 0 & 0 & 1 \end{matrix} ) : a, b, c, d, ef, g, h, i ∈ k\}

We now calculate Inn(A). Let u ∈ J_{1}. So by Lemma 2.6, φ_{u}(v) = v + [v, u] + [vu, u] for all v ∈ V. By Lemma 3.3, Z ∩ J_{1} is an ideal of A, hence we may assume that u = ϵ_{1}x + ϵ_{2}y + ϵ_{3}(xy − yx), ϵ_{i} ∈ k, 1 ≤ i ≤ 3. By Lemma 3.3 one easily calculates

φ_{u}(x) = x + ϵ_{2}(xy − yx) + (ϵ_{1}ϵ_{2} − 2ϵ_{3})xyx − ϵ_{2}^{2}yxy;
φ_{u}(y) = y − ϵ_{1}(xy − yx) − ϵ_{1}^{2}xyx + (ϵ_{1}ϵ_{2} + 2ϵ_{3})yxy;
φ_{u}(z) = 1.

Thus Inn(A) consist of matrices above such that a = e = d = h = i = 0 and c + g = −2bf.

For each φ ∈ Aut^{0}(A), let φ^{0} := φ−Id. Then from the above description of Aut^{0}(A), we have that for all φ ∈ Aut^{0}(A), {xyx, yxy, xyxy} ⊂ Ker(φ^{0}), φ^{0}(xy − yx), φ^{0}(xy + yx) and φ^{0}(z) are contained in the k-span of {xyx, yxy, xyxy} and φ^{0}(x) and φ^{0}(y) are contained in the k-span of {z, xy − yx, xy + yx, xyx, yxy, xyxy}. Hence, φ^{∘}τ^{∘}ψ^{∘} = 0 for all φ, τ, ψ ∈ Aut^{0}(A). It follows that for all φ ∈ Aut^{0}(A), φ^{−1} = Id+φ^{∘}+φ^{∘}^{2} and for all φ, τ ∈ Aut^{0}(A)

φτφ^{−1}τ^{−1} = Id + φ^{∘}τ^{∘} − τ^{∘}φ^{∘}.

Further, since {z, xy − yx, xy + yx, xyx, yxy, xyxy} is contained in the kernel of φ^{∘}τ^{∘},

φ^{∘}τ^{∘}(u) − τ^{∘}φ^{∘}(u) = 0

for all u ∈ 𝔹 different from x, y. Let φ be the matrix displayed above and let τ be the matrix with a replaced by a′ etc. Then,

φ^{∘}τ^{∘}(x) − τ^{∘}φ^{∘}(x) = (ae′ − a′e + 2bf′ − 2b′f)xyx + (a′i − ai′)xyxy,
φ^{∘}τ^{∘}(y) − τ^{∘}φ^{∘}(y) = (ea′ − e′a + 2fb′ − 2f′b)yxy + (e′i − ei′)xyxy.

Thus, for any φ ∈ Aut^{0}(A), φτφ^{−1}τ^{−1} ∈ Inn(A) for all τ ∈ Aut^{0}(A) if and only if

a = e = i = 0.

Let Ẑ be the inverse image in R_{u}(Aut^{0}(A)) of Z(U). There is is an obvious injective morphism of varieties k^{6} → GL_{8}(k) whose image is Ẑ and Ẑ is a closed subgroup of Aut^{0}(A) and hence of GL_{8}(k). Thus, Ẑ has dimension 6 (see [3, Corollary 4.3]). Similarly Inn(A) has dimension 3. So, it follows that Z(U) has dimension 3 (see [3, Proposition 7.4B]).

□

Note that if A is as in Proposition 5.5 with α = β = γ = δ = 0, then A ≅ B.

Proposition 6.3. Suppose that A satisfies Hypothesis 5.2 and that α, β, γ, δ are as in Proposition 5.5.

(i) If any one of α, β, γ or δ is non-zero, then Out^{0}(A)/R_{u}(Out^{0}(A)) is contained in a 1-dimensional torus.

(ii) If α = β = γ = δ = 0, then Out^{0}(A)/R_{u}(Out^{0}(A)) is a two-dimensional torus.

Proof. Let {x + J_{2}, y + J_{2}} be a basis of J_{1}/J_{2} as in Proposition 5.5. Let B be the ordered basis {x, y, x^{2}, y^{2}, xy, xyx, yxy, xyxy} of J_{1}. Let φ ∈ Aut(A) and for u ∈ B write φ(u) = ∑_{v∈B} λ_{v,u}v. Then,

0 ≡ φ(x)φ(y) + φ(y)φ(x) ≡ λ_{x,x}λ_{x,y}x^{2} + λ_{y,x}λ_{y,y}y^{2} mod J_{3},

So,

λ_{x,x}λ_{y,x} = λ_{x,y}λ_{y,y} = 0,

that is either λ_{x,y} = λ_{y,x} = 0 or λ_{x,x} = λ_{y,y} = 0. Identifying GL(J_{1}/J_{2}) with GL_{2}(k) through the ordered basis {x + J_{2}, y + J_{2}} it follows that f_{1}(Aut(A))^{0} is contained in the subgroup of diagonal matrices. Hence, if φ ∈ Aut^{0}(A) then

(10)  λ_{x,y} = λ_{y,x} = 0.

We will assume from now on that this is the case.

(i) Now suppose that α ≠ 0. Comparing the coefficient of xy^{2} in the relation φ(x)^{3} = αφ(x)φ(y)^{2} + βφ(x)^{2}φ(y)^{2} gives

λ_{x,x}^{3}α = λ_{x,x}λ_{y,y}^{2}α,

and hence λ_{x,x}^{2} = λ_{y,y}^{2}. Thus, identifying f_{1}(Aut(A))^{0} = f_{1}(Aut^{0}(A)) with its image in GL_{2}(k) as above, f_{1}(Aut(A))^{0} is contained in the subgroup of diagonal matrices (a_{ij}) satisfying a_{1,1} = a_{2,2}. This proves (i) if α ≠ 0 and similarly if γ ≠ 0. Suppose now that α = γ = 0 and β ≠ 0. Comparing the coefficient of x^{2}y^{2} in the relation φ(x)^{3} = βφ(x)^{2}φ(y)^{2} gives

λ_{x,x}^{3}β + 3λ_{x,x}^{2}λ_{y^{2},x} = βλ_{x,x}^{2}λ_{y,y}^{2},

and hence (since k is of characteristic 3),

λ_{x,x} = λ_{y,y}^{2}.

Thus f_{1}(Aut(A))^{0} is contained in the subgroup of diagonal matrices (a_{ij}) satisfying a_{1,1} = a_{22}^{2}. This proves (i) if α = γ = 0, β ≠ 0. The proof for the case that α = γ = 0, δ ≠ 0 is similar.

(ii) Suppose that α = β = γ = δ = 0. The coefficient of x^{2}y in φ(x)φ(y) + φ(y)φ(x) is 2λ_{x,x}λ_{y^{2},y} = 0 and the coefficient of xy^{2} in φ(x)φ(y) + φ(y)φ(x) is 2λ_{y,y}λ_{x^{2},x}, hence

(11)  λ_{x,x^{2}} = λ_{y,y^{2}} = 0.

Now the coefficient of x^{2}y^{2} in φ(x)φ(y) + φ(y)φ(x) is

2λ_{x,x}λ_{xy^{2},y} − 2λ_{xy,x}λ_{xy,y} + 2λ_{y,y}λ_{xy^{2},y} + 2λ_{y^{2},x}λ_{x^{2},y},

hence

(12)  λ_{xy^{2},y} = (λ_{xy,x}λ_{xy,y} − λ_{x,x}λ_{xy^{2},y} − λ_{x^{2},y}λ_{y^{2},x})λ_{y,y}^{−1}

Conversely, any element of GL(J) satisfying the Equations 10, 11 and 12, and such that the image under φ of elements of B ∩ J_{2} is the obvious multiplicative extension of the images of x, y and z of φ is in Aut^{0}(A). So, identifying as in (i), GL(J_{1}/J_{2}) with GL_{3}(k) via the ordered basis {x + J_{1}, y + J_{1}} of J_{1}/J_{2}, f_{1}(Aut^{0}(A)) is the subgroup of diagonal matrices. □

The next result gives the dimension of Z(R_{u}(Out^{0}(B))).

Proposition 6.4. Suppose that A satisfies Hypothesis 5.2 and that the elements α, β, γ, δ of Proposition 5.5 are all equal to 0. Then dim(Z(R_{u}(Out^{0}(A)))) = 2.

Proof. Let {x+J_{2}, y+J_{2}} be a basis of J_{1}/J_{2} as in Proposition 5.5. Let U = R_{u}(Out^{0}(A)) and let B be the ordered basis {x, y, x^{2}, y^{2}, xy, x^{2}y, xy^{2}, x^{2}y^{2}} of J and identify Aut^{0}(A) with a subgroup of GL_{8}(k) through the basis B. As in the Proposition 6.4, let B_{8}(k) be the subgroup of GL_{8}(k) consisting of lower triangular matrices and let U_{8}(k) be the subgroup of B_{8}(k) of strictly unitriangular matrices. By Equations, 10, 11 and 12, Aut^{0}(A) is contained in B_{8}(k) and hence by Lemma 2.5(i), R_{u}(Aut^{0}(A)) = Aut^{0}(A)∩U_{8}(k). Further, R_{u}(Aut^{0}(A)) is the following subgroup of U_{8}(k):

\{( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & e & 1 & 0 & 0 & 0 & 0 & 0 \\ a & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ b & f & 0 & 0 & 1 & 0 & 0 & 0 \\ i & g & 0 & −e & f & 1 & 0 & 0 \\ c & −bf − ae − i & −a & 0 & b & 0 & 1 & 0 \\ d & h & −c − b^{2} & −g − f^{2} & 0 & 0 & 0 & 1 \end{matrix} ) : a, b, c, d, ef, g, h, i ∈ k\}

We now calculate Inn(A). Let u ∈ J_{1}. By Lemma 2.6, φ_{u}(v) = v + [v, u] + [vu, u] for all v ∈ V. By Lemma 3.3, Z ∩ J_{1} is an ideal of A, hence we may assume that u = ϵ_{1}x + ϵ_{2}y + ϵ_{3}xy, ϵ_{i} ∈ k, 1 ≤ i ≤ 3. One easily calculates

φ_{u}(x) = x + 2ϵ_{2}xy + (2ϵ_{3} − 2ϵ_{1}ϵ_{2})x^{2}y + 2ϵ_{2}^{2}xy^{2};
φ_{u}(y) = y − 2ϵ_{1}xy + 2ϵ_{1}^{2}x^{2}y − 2(ϵ_{1}ϵ_{2} + ϵ_{3})xy^{2}.

Thus Inn(A) consist of matrices above such that a = e = d = h = 0 and c = −b^{2}, g = −f^{2}.

For each φ ∈ Aut^{0}(A), let φ^{0} := φ − Id. Then from the above description of Aut^{0}(A), we have that for all φ ∈ Aut^{0}(A), {x^{2}y, xy^{2}, x^{2}y^{2}} ⊂ Ker(φ^{0}), φ^{0}(xy), φ^{0}(x^{2}) and φ^{0}(y^{2}) are contained in the k-span of {x^{y}, xy^{2}, x^{2}y^{2}} and φ^{0}(x) and φ^{0}(y) are contained in the k-span of {x^{2}, y^{2}, xy, x^{2}y, xy^{2}, x^{2}y^{2}}. Hence, φ^{∘}τ^{∘}ψ^{∘} = 0 for all φ, τ, ψ ∈ Aut^{0}(A). It follows that for all φ ∈ Aut^{0}(A), φ^{−1} = Id + φ^{∘} + φ^{∘}^{2} and for all φ, τ ∈ Aut^{0}(A)

φτφ^{−1}τ^{−1} = Id + φ^{∘}τ^{∘} − τ^{∘}φ^{∘}.

Further, since {z, xy − yx, xy + yx, xyx, yxy, xyxy} is contained in the kernel of φ^{∘}τ^{∘},

φ^{∘}τ^{∘}(u) − τ^{∘}φ^{∘}(u) = 0

for all u ∈ 𝔹 different from x, y. Let φ be the matrix displayed above and let τ be the matrix with a replaced by a′ etc. Then,

φ^{∘}τ^{∘}(x) − τ^{∘}φ^{∘}(x) = (ae′ − a′e + b′f − bf′)x^{2}y + (a(g′ + f^{′2}) − a′(g + f^{2}))x^{2}y^{2},
φ^{∘}τ^{∘}(y) − τ^{∘}φ^{∘}(y) = (ea′ − e′a + fb′ − f′b)xy^{2} + (e(c′ + b^{′2}) − e′(c + b^{2}))x^{2}y^{2}.

Thus, for any φ ∈ Aut^{0}(A), φτφ^{−1}τ^{−1} ∈ Inn(A) for all τ ∈ Aut^{0}(A) if and only if

a = e = 0 and c = −b^{2}, g = −f^{2}.

Arguing as for Proposition 6.2, the inverse image in R_{u}(Aut^{0}(A)) of Z(U) has dimension 5, Inn(A) has dimension 3, and hence Z(U) has dimension 2, as claimed. □

Proposition 6.5. Suppose that A satisfies Hypothesis 5.3. Then Out^{0}(A)/R_{u}(Out^{0}(A)) is contained in a one-dimensional torus.

Proof. Let {x+J_{2}, y +J_{2}} be a basis of J_{1}/J_{2} satisfying the properties of Lemmas 5.6 and 5.7. Let B be the ordered basis {x, y, x^{2}, xy, yx, xyx, yxy, xyxy} of J_{1} and let φ ∈ Aut(A) be as before. The fact that φ(y)^{2} ≡ 0mod J_{3} implies that λ_{x,y} = 0 and we will assume that this is the case. Now consider the equation

φ(x)^{3} ≡ φ(y)φ(x)φ(y) mod J_{4}

λ_{x,x}^{3}x^{3} + λ_{x,x}^{2}λ_{y,x}xyx + 2λ_{x,x}^{2}λ_{y,x}x^{2}y + λ_{x,x}λ_{y,x}^{2}yxy ≡ λ_{y,y}^{2}y(λ_{x,x}x + λ_{y,x}y)y mod J_{4},

λ_{x,x}^{2}λ_{y,x}xyx + (λ_{x,x}^{3} + λ_{x,x}λ_{y,x}^{2})yxy ≡ λ_{y,y}^{2}λ_{x,x}yxy mod J_{4},
λ_{x,x}^{2}λ_{y,x} = 0, λ_{x,x}^{3} + λ_{x,x}λ_{y,x}^{2} = λ_{y,y}^{2}λ_{x,x},
λ_{y,x} = 0, λ_{x,x}^{3} = λ_{y,y}^{2}λ_{x,x},
λ_{y,x} = 0, λ_{y,y} = ±λ_{x,x}.

The result is follows from these calculations and Proposition 2.4. □

7. Proof of Theorem 1.2 and Theorem 1.1

Proof of Theorem 1.2. There is a stable equivalence of Morita type between A and B, by [16, Théorème 4.15], Out^{0}(A) and Out^{0}(B) are isomorphic as algebraic groups. In particular, Out^{0}(A)/R_{u}(Out^{0}(A)) ≅ Out^{0}(B)/R_{u}(Out^{0}(B)) and R_{u}(Out^{0}(A)) ≅ R_{u}(Out^{0}(B). By Proposition 6.3, and the remark preceding it, Out^{0}(B)/R_{u}(Out^{0}(B)) is a 2-dimensional torus, and by Proposition 6.4, dim(Z(R_{u}(Out^{0}(A)))) = 2. Hence, it follows from Propositions 6.1, 6.2, 6.3, and 6.5 that if A satisfies one of the hypotheses 4.1, 5.2 or 5.3, then A ≅ B. The result follows as by Proposition 3.5 and Lemma 5.1, A does satisfy one of the hypothesis 4.1, 5.2 or 5.3. □

Proof of Theorem 1.1. By the structure theory of blocks with normal defect groups [7], C is Morita equivalent to a twisted group algebra k_{α}P ⋊E, where E is a p′-subgroup of the automorphism group of P and α is an element of H^{2}(E, k^{×}). Since A is not nilpotent and C has up to isomorphism only one simple module, it is well known (see, for instance, [2, Theorem 1.1]), that E is a Klein-4-group and that C is Morita equivalent to the algebra B = k⟨X, Y ⟩/⟨X^{3}, Y^{3}, XY + Y X⟩.

By Rouquier’s work [16, 6.3] (see also [10, Theorem A.2]) there is a stable equivalence of Morita type between A and C. By [12] (see also [4]), the blocks A and C are perfectly isometric. Hence by [1], the centers of A and C are isomorphic as k-algebras and up to isomorphism A has one simple module. Consequently, the dimension of a basic algebra of A is equal to 9. The result follows from Theorem 1.2 applied to B and a basic algebra of A. □

Acknowledgments. I would like to thank Meinolf Geck, Markus Linckelmann and Nicole Snashall for helpful discussions on various aspects of this paper.

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Institute of Mathematics, University of Aberdeen, Fraser Noble Building, King’s College, Aberdeen AB24 3UE, U.K.