Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 14, 2004, 3–16

Abdulla Aljouiee
ON THE BRAUER MONOID OF S3
(submitted by M. Arslanov)

ABSTRACT. In [HLS], the authors showed that the Brauer monoid of a finite Galois group can be written as a disjoint union of smaller pieces (groups). Each group can be computed following Stimets by defining a chain complex and checking its exactness. However, this method is not so encouraging because of the difficulty of dealing with such computations even with small groups. Unfortunately, this is the only known method so far. This paper is to apply Stimets’ method to some idempotent weak 2-cocycles defined on S3. In particular, the idempotent 2-cocycles whose associated graphs have two generators. Some nice results appear in the theory of noncommutative polynomials.

### 1. Preliminaries

Let KF be a finite Galois extension of fields and let G be its Galois group. A weak 2–cocycle is a function f : G × G K satisfying

1. fσ(τ,ν)f(σ,τν) = f(στ,ν)f(σ,τ)
2. f(σ, 1) = f(1,σ) = 1

for all σ,τ,ν G.

If we define an algebra Af associated with f to be the algebra generated as a K-vector space by the indeterminates {xσ : σ G} with the relations xσxτ = f(σ,τ)xστ and xσk = kσx σ for k K, x1 = 1, then Af is called a weak crossed product. Condition (i) above guarantees the associativity of Af.

In the classical theory f does not take the value 0, and in that case Af is a central simple F-algebra. The set H = {σ G : f(σ,σ1) = 0} is a subgroup of G called the inertial subgroup of f. We define an order “” on GH by σH τH if and only if f(σ,σ1τ) = 0. This order is lower subtractive, that is if σH τH then σH αH τH if and only if σ1αH σ1τH. In this order, H is the unique minimal element (root) and has the property f(H × H) K [HLS]. The weak crossed product Af can be written as

Af = σHKxσ σHKxσ = B J

where J = σHKxσ is the radical of Af and B = σHKxσ is a classical crossed product algebra for fH×H. In particular B is a central simple KH–algebra.

Given a weak 2-cocycle f, define a function e : G × G {0, 1} by e(σ,τ) = 0 if and only if f(σ,τ) = 0. Then e is a weak 2-cocycle called the idempotent weak 2-cocycle associated to f.

Two weak 2-cocycles f,g are called cohomologous (or equivalent) if there is a function α : G K such that

f(σ,τ) = α(σ)ασ(τ) α(στ) g(σ,τ) for all σ,τ G.

Any two cohomologous weak 2-cocycles have the same associated idempotent cocycle. Under the equivalence relation introduced above the set of classes of weak 2-cocycles from G × G to K forms a monoid denoted by M2(G,K). The subgroup of invertible elements of this monoid is the usual cohomology group H2(G,K).

Let e be an idempotent weak 2-cocycle. If f is a weak 2-cocycle associated to e then we can define a function g : G × G K by

g(σ,τ) = (f(σ,τ))1if f(σ,τ) = 0 0 otherwise.

Then g is a weak 2-cocycle associated to e. If [ ] denotes the equivalence class in the relation above, let Me2(G,K) = {[f] M2(G,K) [f][e] = [f] and there is a weak 2-cocycle g such that [f][g] = [e]}. Then Me2(G,K) is a group with identity [e] and M2(G,K) = eMe2(G,K) (disjoint) where the union is over all idempotent weak 2-cocycles, [H1]. In a similar way, we can define the group Mei(G,K) of ith dimension.

The set of all idempotent cocycles on G × G with inertial subgroup H is in 1 1 correspondence with all lower subtractive orders (graphs) on GH with unique root H. Whenever we refer to the graph for e, we mean the graph associated to the weak 2-cocycle e. If [f] Me2(G,K) then f and e have the same associated graph. The following results and definitions are from [S1] and [S2]. Each idempotent 2-cocycle e with trivial inertial subgroup H determines a ring Re = {xσ : σ G}Ie where Ie is the ideal generated by {xσxτ xστ σ στ}. The ring Re is called the derived ring of e. To define a graded Re–module, it is more convenient to use the notation gi for the elements of the group G and [g1 ,g2,,gk] for the free generators of the Re–module, where gi gi+1 according to the relation defined above.

Define a graded Re–module M by M = nMn where

Mn = g1<<gnRe[g1,,gn]n 1,gi = 1 Re n = 0 n = 1 0 n 2

with differentials dn[g1,,gn] = xg1[g11g 2,,g11g n] + (1)i[g 1,, g ̂ i, , gn ]. We call the pair (M,d) the chain complex of e. This definition can be given in a similar manner if H = {1}. Suppose f is a function from Gn to a field K which satisfies

1. in case f(g1,g2,,gi1, 1,gi+1,,gn) = 0,

we have f(g1,g2,,gi1, 1, gi+1,,gn) = 1 for all g1,g2,,gn in G and all 1 i n,

2. f(1, 1,, 1,g, 1,, 1) = 1 for all g G, and
3. for each g1,g2,,gn+1 in G,

fg1 (g2,,gn+1) i evenf(g1,,gigi+1,,gn+1) = f(g1,,gn) i oddf(g1,,gigi+1,,gn+1)

if n is even, and

fg1 (g2,,gn+1)f(g1,,gn) i evenf(g1,,gigi+1,,gn+1) = i oddf(g1,,gigi+1,,gn+1)

if n is odd. We then call f a weak n-cocycle. The first condition is analogous to the standard degeneracy conditions used in homological algebra, but still allows for the possibility that certain cochains may take on non-invertible values.

The second condition ensures that the cochains (cocycles) have a sufficient amount of invertibility.

If f is a weak n-cocycle and there is another weak n-cocycle h and an invertible cochain β : Gn1 K such that f = β h, where

β(g1,,gn) = βg1 (g2,,gn)β(1)n (g1,,gn1) β(1)i (g1,,gigi+1,,gn),

then we say f is cohomologous to h and write f h. Let Mn(G,K) be the monoid of weak n-cocycles modulo the equivalence relation . Then Mn(G,K) is called the weak Galois cohomology monoid. The class of cocycles equivalent (cohomologous) to the identity are known as weak coboundaries.

Since the groups Mei(G,K) are components of the required monoid, we are interested in computing these groups. Stimets ([S1],[S2]) has shown that under a sharp condition (exactness of (M,d)) we have the following isomorphism Mei(G,K) Ext Rei(,K) for all i, where the groups ExtRei(,K) are relatively easier to deal with because they are well known and Re acts on K in the obvious way. (See [S1],[S2] for details). In this paper, we investigate exactness in some special cases. Exactness is always guaranteed at M0 ([S1]). It is quite difficult to check exactness in general but using the “contraction process” makes the situation somewhat easier. If g1 < < gn, call [g1 ,,gn] (n)–cell in Mn.

Proposition 1.1 (Contraction process) (Stimets). Let Cn Mn, Cn1 Mn1 be two cells such that Cn does not appear in the boundary of Mn+1 and in the boundary of an n-cell not equal to Cn, the cell Cn1 does not appear. But if Cn1 appears in the boundary of Cn, then the chain complex (M,d) obtained by removing Cn and Cn1 is exact at Mn if and only if (M,d) is exact at Mn. Moreover, (M,d) and (M,d) have the same homology groups.

This procedure allows us to cancel cells gradually until we reach a point after which we cannot proceed any further. Then we can investigate exactness at fewer modules.

### 2. Contracting to a Simpler Graph

In some cases, we can simplify the given graph to a smaller one. This section is devoted to demonstrating a class of graphs that can be contracted to a specific form of Z4 and showing that such graphs possess exact chain complex.

Lemma 2.1. In the ring R = {x,y}(x2 = y2), if α(x 1) = β(y 1), then α = h(x + 1), β = h(y + 1) for some h R.

Proof. We claim that the set V = {yɛxyxyxyxi ɛ {0, 1}, i 0} forms a -basis for R. Clearly V generates R. We show the independence. Let M be a free module on T = {Y ɛXY Y Xi ɛ {0, 1}, i 0}. The module M can be viewed as a right R-module by defining the action:

(Y ɛXY 1XXY jXi) x = Y ɛXY 1XXY jXi+1 (Y ɛXY 1XXY jXi) y = Y ɛXY 1XXY j+1Xi1if i is odd Y ɛXY 1XXY j1Xi+3if i is even.

This action is well-defined and notice that

(Y ɛXY 1XXY jXi) y2 = (Y ɛXY 1XXY jXiy) y = Y ɛXY 1XXY j+1Xi1 yif i is odd Y ɛXY 1XXY j1Xi+3 yif i is even = (Y ɛXY 1XXY jXY j+1 y)Xi1 Y ɛXY 1XXY jXi+2 = Y ɛXY 1XXY jXi+2 = (Y ɛXY 1XXY jXi) x2.

So we can define a homomorphism ϕ : R End(M) by

ϕ(x) = right multiplication by x ϕ(y) = right multiplication by y

Now, for zi V , if aizi = 0, then ϕ( aizi)(1) = 0 = aiZiai = 0 for all i, and we showed the claim. Now, we define a total order on the basis elements as follows: To each yɛxy 1xxyjxi V , assign a degree (ɛ, 2, 2,, 2 j, 1, 1,, 1 i) and let deg m = (1 2) for all m , deg0 = 0. If u, v V , we define u v if the length of the corresponding tuple of v is greater than the length of the corresponding tuple of u. If they have the same length, then compare the first numbers on the right, if they are the same go to the next numbers and so on. Notice that

ɛ, 2, 2,, 2 j, 1, 1,, 1 i x = ɛ, 2, 2,, 2 j, 1, 1,, 1 i+1

and

ɛ, 2, 2,, 2 j, 1, 1,, 1 iy = ɛ, 2, 2,, 2 j1, 1, 1,, 1 i+3i is even ɛ, 2, 2,, 2 j+1, 1, 1,, 1 i1i is odd.

This shows that multiplying an element of V by x or y increases the degree. Since any monomial in V ends from the right with either y or xi, i > 0, we can write any element in V as a sum f + gy where f and g are sums of monomials of V ending with xi, i > 0. We say f is the component of 1 and g is the component of y. Remember that in this form any term in the sum g must end with x1. Let α = f1 + g1y, β = h1 + k1y where f1 ,g1h1 and k1 are sums of elements in V which end with xi, i > 0. Let f, g,h and k be of the largest degrees in f1,g1,h1 and k1 respectively. The equation α(x 1) = β(y 1) gives that

fx + gyx gy + lower terms in both components = kx2 h + hy ky + lower terms in both components. (2.1)

Let us denote by (g) for the degree of g, (g, 2, 1) for the degree of gyx and (k, 1, 1) for the degree of kx2. From (2.1), if we assume that g = 0, and h ends with x1, we get

Case 1. If (h) < (k) then (g) = (k) and (g, 2, 1) (k, 1, 1) (impossible).

Case 2. If (h) (k) then (g) = (h) and (g, 2, 1) max{(h), (k, 1, 1, )} (impossible).

Now assume g = 0 and h ends with xi, i > 1. This implies that g = k and fx = hy. Let =0th be the component of 1 in β where h ends with xi, i > 1, for all and ht = h. Equation(2.1) implies that there exists hj for some j such that gyx = hjy or gyx = hj. The latter is clearly impossible since hj ends with xi, i > 1. The first is also impossible since hjy = hjxiy = hjyxi iis even hjxyxi1iis odd  3
and in both cases hjy ends with xr, r > 1. So, gyx = hjy for all j.

Thus g must be 0 and either k = 0 or k = h. In the first case, =0th = β where h ends with xi, i > 1 for all . Equation (2.1) implies that fx = hy. But hy always ends with x2i. so f ends with x2i1 and f = uyx2i1 for some u. So h = ux2i uyx2i+1 .

Cancel fx with hy in (2.1) we get the term of the highest degree in the left hand side is f, but f = hy for any . Therefore f = h which is impossible. This forces h = k.

Hence the original equation takes the form f1 (x 1) = h(1 + y)(y 1) where h is the sum of either components. This implies that f1 (x 1) = h(y2 1) = h(x2 1) = h(x + 1)(x 1)α = h(x + 1) and β = h(y + 1).

Theorem 2.2. The chain complex of the idempotent weak 2-cocycle e whose graph is given by:

over 4 is exact.

Proof. We have M0 = {x,y}(x2 = y2) = R, M1 = R[1] R[2] R[3]. M2 = R[1, 2] R[3, 2], where x = x1, y = x3. We need to check the exactness of the following chain complex:

0M2d2M1d1M00.

As stated earlier, we always have exactness at M0. Now d2(α[1, 2] + β[3, 2]) = 0α(x1[1] [2] + [1]) + β(x3[3] [2] + [3]) = 0αx1 + α = 0, α β = 0 and βx3 + β = 0α = β = 0 and d2 is injective. To check exactness at M1, let z = a[1] + b[2] + c[3] ker d1, then z imd2 if and only if z = z + d 2(b[1, 2]) = (a + bx1 + b)[1] + c[3] imd2. Let s = a + bx1 + b. Now, z kerd 1s(x1 1) = c(x3 1). By the lemma above, the unique solution for such an equation is s = h(x1 + 1) and c = h(x3 + 1) for some h R. So z = h(x 1 + 1)[1] + h(x3 + 1)[3] = d2(h[1, 2] + h[3, 2]). Hence z imd2, so imd2 = kerd1.

Remark 2.3. We joint each weak 2-cocycle with a chain complex and a unique graph, so excising cells in the complex is equivalent to cancelling edges in the corresponding graph, and we are free to talk about one of these two contractions instead of the other.

Corollary 2.4. Let e be an idempotent weak 2-cocycle and let us denote by Γe its lower subtractive graph on a finite group G with a derived ring {x,y}(x2 = y2). If Γe can be contracted to a graph of the form:

with σ2 = τ2 = ν, then the chain complex of e is exact.

Proof. This is a direct consequence of Theorem 2.2 and Proposition 1.1.

Example 2.1. Consider the idempotent e which is given by its graph over the quaternion group Q = {±1,±i, ±j,±k}:

It is easy to check that Re = {x,y}(x2 = y2) where x = xi, y = xj.

d3[i,k,i] = xi[j,1] [k,i] + [i,i] [i,k]

d3[j,k,j] = xj[i,1] [k,j] + [j,j] [j,k]

The cells [i,k,i] and [j,k,j] can be excised with [i,k] and [k,j] respectively. We now only have the relations:

d2[i,1] = xi[i] [1] + [i]

d2[i,i] = xi[1] [i] + [i]

d2[j,k] = xj[i] [k] + [j]

d2[j,1] = xj[j] [1] + [j]

d2[j,j] = xj[1] [j] + [j]

d2[k,i] = xk[j] [i] + [k]

Notice that the cells [k,i], [j, j], [j, k] and [i, i] can be excised with [k], [j], [k] and [i] respectively. And then we are left with a graph of the form:

where i2 = j2 = 1. So, the chain complex of e is exact.

Corollary 2.5. In the previous example Mei(Q,K) Ext Rei(,K) where Q =
{±1,±i,±j,±k} is the Galois group of the extension KF for some base field F.

### 3. S3 with Two Generators

We will need some lemmas before we state and prove the main result in this work.

Lemma 3.1. There is no non-trivial solution for the equation α(x 1) = β(y 1) over the ring R = {x,y}(yx = xy2).

Proof. First, we show that the set {xiyji,j 0} forms a basis for R over . Let M be the free –module generated by {XiY ji,j 0}. Define the following action to make M a right R–module. Xi Y jx = Xi+1Y 2j, Xi Y jy = XiY j+1 for all i,j. This action is well-defined as we have seen in Lemma 2.1 and we can define a homomorphism ϕ : R End(M) by x multiplication from the right by x, y multiplication from the right by y. So, for any combination t = aijxiyj = 0, we have ϕ(t)(1) = 0 or aijXiY j = 0 and hence aij = 0 for all i,j. Let xsyt be denoted by (s,t).

We define a degree function on the basis elements in R by (s,t) < (s,t) if s < s. If s = s, then (s,t) < (s,t) if t < t. Notice that multiplying by x,y from the right gives (s,t)x(s + 1, 2t), (s, t)y(s,t + 1). Let α = (s1,t1) + (s2,t2)+ lower degrees, where (s1,t1) > (s2,t2), β = (s,t)+ lower degrees. α(x 1) = β(y 1) gives that (s1 + 1, 2t1) (s1,t1) + (s2 + 1, 2t2)+ lower degrees = (s,t + 1) (s,t)+ lower degrees (s1 + 1, 2t1) = (s,t + 1), that is, s = s 1 + 1 and t = 2t 1 1. We claim that (s,t) > max{(s 1,t1), (s2 + 1, 2t2)}. Clearly s > s 1 since s = s 1 + 1. s2 s1s2 + 1 s, if s2 + 1 < s, we are done. So, let s2 + 1 = s, thus, s1 = s2. This implies t1 > t2 (*). Assume that t 2t 2, so 2t1 1 2t2 or t2 t1 which contradicts (*).

Lemma 3.2. In the ring R = {x,y}(xyx = y2), α(x 1) = β(y 1) implies that α = k0(xn1 + xn2 + + 1)(1 + xy) and β = k0(1 xn + (xn1 + xn2 + + 1)y), for any k0 R and n .

Proof. Using an idea similar to what is in Lemma 3.1, we can show that S = {yɛxi1yxi2yyxi, ɛ {0, 1}, ij , + for j = , i {0, 1,}} forms a -basis for R. Assign to each element yɛxi1yxi2yyxi a degree (ɛ,i1,i2,,i) if i > 0 and (ɛ,i1,i2,,i1) for yɛxi2yyxi1y. If f ends from the right with xi, i > 0, denote by (f) for degree of f, (f + k) for fxk, (f, 1) for fyx. Notice also that we can express any element in R as f0 + g0y where f0 and g0 are sums of monomials from S ending with xi, i > 0. Call f0 the component of 1 and g0 the component of y.

Let f,g,h and k be of the largest degree in α,β in both components. So

 (f + gy + ⋯)(x − 1) = (h + ky + ⋯)(y − 1) (3.1)

fx + gyx gy+ lower terms = kxyx h + hy ky+ lower terms

1. (g) = (h k)
2. max((f + 1), (g, 1)) = max((k + 1, 1), (h)).

We discuss the following cases:

Case 1. If g = 0 then h = k and (f + 1) = (k + 1, 1) (impossible).

Case 2. labelcs2.6 If g = 0 and (h) < (k) then (g) = (k) and from (ii) (f + 1) = (k + 1, 1) (impossible).

Case 3. If g = 0 and (h) > (k) then (g, 1) = (k + 1, 1)g = mkx and h = rkx for some m, r . Substituting in the equation (3.1) yields m = 1 = r and so, g = kx = h. Let g , k,h be of the next largest degree (g) < (g), (k ) < (k), (h ) < (h), then we get

fx + kxyx + gyx kxy gy + lower terms = kxy + kx + hy + kxyx ky + kxyx + lower terms

(i) (g ) = (h k)

(ii) max((f + 1), (g, 1)) = max((k + 1), (k + 1, 1)).

Again if g = 0 then since h = k = tfx, t and equation (3.1) gives fx+ lower terms = tfx2+ lower terms f = 0 and hence h = k = f = g = g = h = 0.

If g = 0, then (h ) (k)(g) = (k) and hence (g) = (k + 1)g = sk = tkx, s, t . As above we find s = t = 1. Repeat this step to get gi = kix, ki+1 = kix where (ki+1) > (ki) and equation (3.1) becomes

 (F +k0(xn+xn−1+⋯+x)y)(x−1) = (−k 0xn+H′+k 0(xn−1+⋯+1)y)(y−1) (3.2)

where F is the sum of components of 1 in α, H the sum of components of 1 in β that have degrees less than h. Let A := (xn1 + + 1), so equation (3.2) reads

(F + k0Axy)(x 1) = (k0xn + H + k 0Ay)(y 1) Fx F + k0Axyx k0Axy = Hy H + k 0Axyx k0xny k 0Ay + k0xn Fx F = Hy H k 0y + k0xn (since xn A = Ax 1) k0 = H F(x 1) = k0(xn 1) F(x 1) = k0(xn1 + + 1)(x 1) = k 0A(x 1) F = k0A.

Therefore

 α(x − 1) = (k0A + k0Axy)(x − 1) = k0A(1 + xy)(x − 1) (3.3)

and

β(y 1) = (k0xn + k 0 + k0Ay)(y 1) = k0(xn + 1 + Ay)(y 1) = k0(xny + xn + y 1 + Axyx Ay) = k0(Axy y + xn + y 1 + Axyx) (xny Ay = Axy y) = k0((xn 1) + Axy(x 1)) = k0(A(x 1) + Axy(x 1)) = k0A(1 + xy)(x 1) = α(x 1).

Lemma 3.3. In the ring R = {x,y}(xyx = yxy), the solutions of α(x 1) = β(y 1) are α = h(xy y + 1) and β = h(yx x + 1) for some h R.

Proof. Since xyx = yxy in R, any element in R can be written in a unique way as a combination of elements of the form xi1 yj1xi2yj2xiyj where i1,j 0, i2 , ,i1 2, j1 , j1 1, i 2 if j > 0 and i 1 if j = 0. The form yxy is always replaced by xyx. This set of monomials forms a -basis for R by applying the same trick in Lemma 3.1. Assign to each such element xi1 yj1xi2yj2xiyj a degree (i1,j2,i2,j2,,i,j) and notice that

(i1,j1,i2,j2,,i,j)x = (i1,j1,,i + 1, 0) if j = 0 (i1,j1,,i,j, 1, 0)if j = 0

(n1, m1,,mt1,nt,mt)y = = (n1,m1,,nt,mt + 1) if mt = 0 (n1,m1,,mt1,nt, 1) if mt = 0 and nt > 1  (n1,m1,,nt1 + 1, 1,mt1, 0)if mt = 0 and nt = 1

Obviously, these two terms can not be equal except if mt = 0 and nt = 1. If j = 0 then the equality implies (hxi1yxi)x = (hxi11yi+1x)y for some h R. If j = 0 then the equality gives j = mt1 = 1 and i = nt1 + 1. So, (h xi1y)x = (hxi1yx)y for some h R. But in the first case we note that hxi1yxi = hxi11yixy, so in all cases if (i1,j1,i2,j2,,i,j)x = (n1,m1,,mt1,nt,mt)y then there is h R such that hxy = (i1,j1,i2,j2,,i,j), hyx = (n1,m1,,mt1,nt,mt). Now, let α = i=1rα i, β = i=1sβ i, so by a suitable rearrangement, α(x 1) = β(y 1) implies that α1x = β1yα1 = h1xy and β1 = h1yx and hence h1xyxh1xy+α2xα2+α3xα3+ = h1yxyh1yx+β2yβ2+β3yβ3+α2x = h1yx or α2 = h1y. Similarly, we find β2 = h1x, α3 = h1, β3 = h1, α4 = h2xy, β4 = h2yx,α = i=1qh i(xy y + 1) and β = i=1qh i(yx x + 1) for some integer q. Take h = i=1qh i.

Definition 3.1. Let e be a weak 2-cocycle and let Γe be its graph. Then, we call the elements of G of level 1 generators of Γe. That is the elements lie right above the root of Γe are called generators.

Note that generators of the graph certainly generate the group itself.

Theorem 3.4. Let e be an idempotent 2-cocycle whose graph has two generators over S3 = {σ,τσ2τ = τσ,σ3 = τ2 = 1}. Then e admits an exact chain complex.

Proof. Using a computer program, it can be shown that there are nine distinct lower subtractive graphs with two generators over S3. Five of them are trees which admit exact graded modules (see [S2]). The remaining four are:

By excising cells, we end up with Mj i = 0 for all j > 2, i = 1, 2, 3, 4. Let 0 M2id2iM1id1iM0i 0 be the chain complex of ei, i = 1, 2, 3, 4, which we get after excision. We always have exactness at M0i. At M2i, it is easy to show that d2i is injective for all i. To show exactness at M13, let z = a[σ2τ] + b[σ2] + c[σ] + d[στ] ker d 13, so z imd23 if and only if z = z + d 23(c[σ2,σ] + d[σ2,στ]) = α[σ2τ] + β[σ2] imd 23, for some α,β R3. But z ker d13z ker d 13α(x 1) = β(y 1) by Lemma 3.1 α = β = 0. Thus, z imd23 if and only if 0 imd23 which is always true. A similar idea can be applied to show exactness at M12. For M14, let u = p[σ2τ] + q[σ2] + r[τ] + s[στ] ker d 14, so u imd24 if and only if u = γ[σ2τ] δ[τ] imd 24 for some γ,δ Re4. Since u ker d 14 by Lemma 3.3 γ = h(xy y + 1), δ = h(yx x + 1) for some h Re4u imd24 if and only if u = h(xy y + 1)[σ2τ] h(yx x + 1)[τ] imd 24, but clearly d24{h(1 y)[σ2τ,σ2] + h[σ2,στ] h[τ,στ]} = u. For M11, let z = a[σ2τ] + b[σ2] + c[σ] + d[στ] ker d 11 z imd 21 if and only if z = z + d 21(c[σ2,σ] + d[σ2,στ]) = α(σ2τ) β[σ2] imd 21 for some α,β Re1. Since z ker d11 z ker d 11 and α(x 1) = β(x 1) in the ring {x,y}(xyx = y2) = R e1. This implies that α = k0(xn1 + + 1)(1 + xy), β = k0(1 xn + (xn1 + + 1)y) (by Lemma 3.2). So z imd21 if and only if z = k 0A(1 + xy)[σ2τ] k 0(1 xn + Ay)[σ2] imd 21. Notice that 1 xn = A Ax. So z imd 21 if and only if z = k 0A((1 + xy)[σ2τ] (1 x + y)[σ2]) imd 21. But z = d 21(x[σ2,στ] + [σ2τ,σ] [σ2,σ]).

Corollary 3.5. For any idempotent weak 2-cocycle e over S3 with two generators, Mei(S 3,K) ExtRei(,K).

### References

[A]   A. Aljouiee, On Weak Crossed Products, Frobenius Algebras and Weak Bruhat Ordering, J. Algebra, to appear.

[H1]   D. Haile, On Crossed Product Algebras Arising from Weak Cocycles, J. Algebra, 74, 1982, 270-279.

[H2]   D. Haile, The Brauer Monoid of a Field, J. Algebra, 81, 1983, 521-539.

[HLS]   D. Haile, R. Larson, M. Sweedler, Almost Invertible Cohomology Theory and the Classification of Idempotent Cohomology Classes and Algebras by Partially Ordered Sets with a Galois Group Action, Amer. J. Math., 105, 1983, 689-814.

[S1]   R. Stimets, Weak Galois Cohomology and Group Extension, Communications in Algebra, 2000.

[S2]   R. Stimets, Relative Weak Cohomology and Extension, J. Algebra, to appear.

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