The partial Schur multiplier of a group

Journal of Algebra 392 (2013) 199–225

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Journal of Algebra www.elsevier.com/locate/jalgebra

The partial Schur multiplier of a group ✩ M. Dokuchaev a,∗ , B. Novikov b , H. Pinedo a a b

Instituto de Matemática e Estatística, Universidade de São Paulo, Rua do Matão, 1010, 05508-090 São Paulo, SP, Brazil Kharkov National University, Svobody sq., 4, 61077, Kharkov, Ukraine

a r t i c l e

i n f o

Article history: Received 20 June 2012 Available online xxxx Communicated by Volodymyr Mazorchuk MSC: primary 20C25 secondary 20M30, 20M50 Keywords: Partial projective representation Schur multiplier Factor sets

a b s t r a c t Refining the technique worked out in previous papers, we give a characterization, up to equivalence, of the factor sets σ of partial projective representation of a group G over an algebraically closed field K in terms of equalities satisfied by σ . This allows one to conclude that any component of the partial Schur multiplier pM (G ) is an epimorphic image of a direct power of K ∗ . Moreover, it is shown that any component of pM (G ) is an epimorphic image of the component pM G ×G (G ) of the equivalence classes of the totally defined partial factor sets. Examples with cyclic G are considered, in particular, the total component pM G ×G (G ) is determined when G is an arbitrary finite cyclic group. In this case pM G ×G (G ) is a direct power of K ∗ . © 2013 Elsevier Inc. All rights reserved.

1. Introduction This article can be considered as a continuation of [7] and [8] which gives a further development of the technique for the study of the partial Schur multiplier pM (G ) of a group G over an algebraically closed field K , permitting to obtain a deeper understanding of the structure of pM (G ). The concept of the partial Schur multiplier of a group appeared in [7] where partial projective group representations were introduced and studied. Partial group representations and the related notion of a partial group action were worked out in the theory of operator algebras, motivated by the desire to provide useful technical tools for the study of C ∗ -algebras generated by partial isometries (see [3]). Several important classes of C ∗ -algebras were characterized as crossed products by partial actions, permitting

✩ The first author was partially supported by FAPESP of Brazil and partially by CNPq of Brazil. The second and third authors were supported by FAPESP of Brazil. Corresponding author. E-mail addresses: [email protected] (M. Dokuchaev), [email protected] (B. Novikov), [email protected] (H. Pinedo).

*

0021-8693/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2013.07.002

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one to obtain relevant results on their K -theory, ideal structure and representations. Also partial representations were used to study amenability of Fell bundles in [10], while in [13] the development of a machinery was concluded based on a fruitful interaction between partial representations and partial actions. This interaction was previously used in [12] to define and investigate the Cuntz–Krieger algebras with infinite number of states, and then in [14] to study their KMS state structure. The algebraic study of partial representations and partial actions began in [5] and [4], stimulating further algebraic investigations. One of them was the idea to introduce and study partial projective representations of groups. The possibility to define them on one hand, and, on the other, the concept of a twisted partial group action already being in use, suggested that some cohomological theory should fit these notions. It is well-known that the (usual) Schur multiplier M (G ) of a group is isomorphic to the second cohomology group H 2 (G , C∗ ) where G acts trivially on C∗ . Hence the theory of partial projective group representations, which naturally comes together with the concept of the partial Schur multiplier of a group, is a reasonable initial testing laboratory for the development of such cohomological ideas. The interaction between projective partial group representations and the so-called K -valued twisted partial actions of groups (which are a simplified version of general twisted partial actions from [9] and [6]) obtained in [7] and [8] brings to the projective case equilibrium of notions, first enjoyed in the C ∗ -setting and also confirmed in the ring theoretic case. The notion of a partial projective representation of a group G is intimately related to that of a projective representation of the semigroup E (G ), defined by R. Exel to deal with partial actions and partial representations (see the semigroup S (G ) in [11]). Later it was shown by J. Kellendonk and M.V. Lawson [15] that E (G ) is isomorphic to the Birget–Rhodes expansion of G (see [1]), which gives an additional convenience in its use, heavily employed here and in [8]. In fact, an alternative definition of a partial projective representation of G can be given using E (G ) (see [7, Proposition 1]), and this way projective representations of semigroups and their Schur multipliers, the theory elaborated in [16–18] (see also [19]), comes into the picture as an essential working tool. The usual cohomology of semigroups does not serve the projective semigroup representations, instead the more general 0-cohomology [18] fits them with its natural partial flavor. It was shown in [7] that pM (G ) is a semilattice of abelian groups pM X (G ), called components, where X runs over the domains of the partial projective representations of G. One of the surprising facts in [7] is that the domains X are exactly the T -invariant subsets of G × G where T is a semigroup of order 25 acting on G × G. The structure of T does not depend on G and T is a disjoint union of the symmetric group S 3 and an ideal which is a completely 0-simple semigroup. After obtaining this rough description of pM (G ) we felt that the total component pM G ×G (G ) should be of special interest and concentrating on its study should be beneficial for the other components. Note that pM G ×G (G ) contains the usual Schur multiplier M (G ) as a subgroup, but in general it is essentially bigger than M (G ). We fully confirm this feeling in the present paper showing, in particular, that any other component of pM (G ) is an epimorphic image of pM G ×G (G ). A description of the total component of pM (G ) was obtained in [8] by a technique which we are able to improve and obtain a characterization of any component of pM (G ). This includes two significant simplifications in the study of the factor sets of projective representations of E (G ) associated to the partial factor sets of G made in [8] and fully adopted now for other components: a passage from the semigroup E (G ) to its quotient E 3 (G ) and a reduction to the locally trivial factor sets of E 3 (G ). An interesting development is a description, up to equivalence, of partial factor sets σ of G by simple equalities satisfied by σ . Notice that in [8] the partial factor sets σ were characterized as k-valued twistings, i.e. functions σ : G × G → K which satisfy the 2-cocycle equality for triples (a, b, c ) determined by a partial action of G. The equalities which we obtained now to determine σ are more simple than the 2-cocycle equality and surprisingly give a better understanding of the structure of pM (G ) than one has for that of the usual Schur multiplier. The article is structured as follows. After the introduction we recall in Section 2 some necessary background. In Section 3 it is shown in Theorem 3.2 that an arbitrary factor set of E 3 (G ) is equivalent to a factor set which is locally trivial in some sense. In Section 4 an essential technical step is made describing the locally trivial factor sets of E 3 (G ) using functions ν : E 3 (G ) × E 3 (G ) → K defined by rather simple equalities (see Theorems 4.6 and 4.9). This is applied in the main Section 5

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to characterize, up to equivalence, the partial factor sets of G in terms of functions G × G → K with properties coming from those of ν (see Theorems 5.4 and 5.5). This permits one to determine, up to equivalence, the partial factor sets of G by three equalities intimately related to the T -action on G × G (see Theorem 5.6). The most important consequences are summarized in Theorem 5.9 which says that, on one hand, any component pM X (G ) is an epimorphic image of pM G ×G (G ) and, on the other, it is an epimorphic image of ( K ∗ )s where s is the cardinality of the set consisting of the S 3 -orbits of length 2 or 6 of X . Finally in Section 6 we consider the case when G is a finite cyclic group. Examples with small G are given and the total component of pM (G ) is computed for arbitrary finite cyclic G. It turns out to be a direct power of K ∗ . It is interesting to notice that if G is cyclic of order 6 then there is a non-total component of pM (G ) isomorphic to the total one. 2. Some background In this section we recall for reader’s convenience some information from [7] and [8]. Let K be a field, G be a group and let M be a monoid, i.e. a semigroup with an identity element. A map ϕ : G → M respecting the identity elements is called a (unital) partial homomorphism if for all a, b ∈ G









ϕ a−1 ϕ (a)ϕ (b) = ϕ a−1 ϕ (ab), 







ϕ (a)ϕ (b)ϕ b−1 = ϕ (ab)ϕ b−1 . Our partial projective representations involve the concept of a K -cancellative monoid, the most important example being the multiplicative semigroup Matn K of all n × n-matrices over K . By a K -semigroup we mean a semigroup S with 0 and a map K × S → S such that α (β x) = (α β)x, α (xy ) = (α x) y = x(α y ), 1 K x = x for any α , β ∈ K , x, y ∈ S and the zero of K multiplied by any element of S is 0 ∈ S. A K -cancellative semigroup is a K -semigroup S such that for any α , β ∈ K and 0 = x ∈ S one has that α x = β x ⇒ α = β . We assume that a homomorphism of K -semigroups is a K -map and that of monoids respects 1. Examples of K -cancellative semigroups are given by K -algebras considered as multiplicative semigroups. More generally, any multiplicative K -subsemigroup in a K -algebra is evidently K -cancellative. Thus one may take any subset in a K -algebra A and generate a K -subsemigroup or a K -submonoid (we assume that A has an identity element 1 A ). In particular the K -submonoid generated by a single a ∈ A is {αan : α ∈ K , n  0} (here a0 = 1 A ). A concrete example is {(α , α ), (0, β): α , β ∈ K } ⊆ A = K × K , the K -submonoid generated by a = (0, 1), which is not a subalgebra. Taking a = (1, −1), we obtain the K -submonoid {(α , α ), (β, −β): α , β ∈ K } ⊆ K × K , which is actually a group with zero and, consequently, a completely 0-simple semigroup. More generally, one may take as an example any completely 0-simple semigroup S with structure group K ∗ . Indeed, by the Rees–Sushkevich Theorem (see [2, Theorem 3.5]) S is isomorphic to









S 0 K ∗ ; I ; Λ; P = (x; i , λ): x ∈ K ∗ , i ∈ I , λ ∈ Λ ∪ {0} with a regular sandwich matrix P . If we define K × S → S by α · (x; i , λ) = (α x; i , λ), then S becomes a K -cancellative semigroup. For a K -cancellative monoid M, we define the congruence λ by setting xλ y (x, y ∈ M ) exactly when x = α y for some α ∈ K ∗ , where K ∗ stands for the non-zero elements of K . Set Proj M = M /λ, and denote by ξ the natural homomorphism ξ : M → Proj M. Observe that if we take M = Matn K then Proj M = PMatn K , the semigroup of projective (n × n)-matrices over K . Definition 2.1. A partial projective representation of G into a K -cancellative monoid M is a map Γ : G → M such that ξ Γ : G → Proj M is a partial homomorphism. Given a partial projective representation Γ : G → M, Theorem 3 from [7] ensures that there is a (unique) partially defined map σ : G × G → K ∗ , called the factor set of Γ , such that

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dom σ = (a, b)  Γ (a)Γ (b) = 0 , and for all (a, b) ∈ dom σ

        Γ a−1 Γ (a)Γ (b) = Γ a−1 Γ (ab)σ (a, b) and Γ (a)Γ (b)Γ b−1 = Γ (ab)Γ b−1 σ (a, b). It is convenient to assume that σ is totally defined by setting σ (a, b) = 0 for each (a, b) ∈ G × G with Γ (a)Γ (b) = 0, and keep the symbol dom σ for the set of pairs (a, b) ∈ G × G with Γ (a)Γ (b) = 0. Without loss of generality we shall assume that Γ (1) = 1, in particular σ (a, 1) = σ (1, b) = 1 K for all a, b ∈ G such that (a, 1), (1, b) ∈ dom σ . The factor sets σ of partial projective representations of G form an inverse semigroup denoted by pm(G ) (see [7, Corollary 5]), and its quotient semigroup by an appropriate congruence is the partial Schur multiplier pM (G ) of G (see [7, p. 259]). There is a monoid E (G ) defined by R. Exel in [11] which plays a crucial role for partial representations and partial actions. Let G be a group, then E (G ) is defined as the monoid whose generators are the symbols [a] (a ∈ G) with relations









a−1 [a][b] = a−1 [ab],

    [a][b] b−1 = [ab] b−1 , [a][1] = [a] (it follows that [1][a] = [a]). It is proved in [11] that E (G ) is an inverse semigroup, and we recall some properties of E (G ) which will be used freely in what follows. The elements ea = [a][a−1 ] are idempotents, such that

[b]ea = eba [b] and ∀a, b ∈ G , and each element x ∈ E (G ) can be written in the form x = ea1 . . . ean [b] (n  0), which is unique up to the order of the factors eai , where ai = a j for i = j and 1 = ai = b for all i (see [11]). The monoid E (G ) is isomorphic to the semigroup of all pairs ( A , a), where A is a finite subset of G such that 1, a ∈ A, and multiplication is defined by

( A , a)( B , b) = ( A ∪ aB , ab) (see [15]). The latter implies that E (G ) is isomorphic to the Birget–Rhodes expansion of G (see [15] and [22]). We shall refer to A as the support of x = ( A , a) and write supp(x) = A. Moreover, the cardinality | A | of A shall be called the volume of the element x = ( A , a) and denoted by vol(x), we also write e x to denote the product xx−1 . We shall deal with a quotient semigroup of E (G ) defined in [8]. For an integer k  0 denote by N k the ideal

  ( A , a) ∈ E (G )  | A |  k + 1  E (G ),



and by E k (G ) the factor semigroup E (G )/ N k . Evidently, N 0 (G ) = E (G ). Since E (G ) is an inverse semigroup, so too is each E k (G ). We shall work with E 3 (G ) and we write for simplicity E = E (G ) and E 3 (G ) = E 3 . Notice that N 3 (G ) = ∅ if |G |  3. In this case E 3 = {∅} ∪ E = E 0 , where E 0 is the semigroup obtained from E by joining a zero.1

1

One may also set E 3 = E if N 3 (G ) = ∅. The results in this article as well as those in [8] remain true under this assumption.

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Denote by m( E ) the semigroup of factor sets of projective representations of E and by M ( E ) the Schur multiplier of E, i.e. the quotient semigroup M ( E ) = m( E )/∼. The reader is referred to [7] for the details. In particular, we know that m( E ) (as well as M ( E )) is a commutative inverse semigroup, which is a semilattice of groups m I ( E ), called components, i.e.

m( E ) =





M(E) =

m I ( E ),

I ∈Y ( E )

M I ( E ),

I ∈Y ( E )

where I runs over the set Y ( E ) of all (two-sided) ideals of E (including I = ∅). The set Y ( E ) is considered as a semilattice with respect to set-theoretic union. For any I ∈ Y ( E ) and ρ ∈ m I ( E ) one has I = { X Y | X , Y ∈ E (G ), ρ ( X , Y ) = 0}. Moreover, it is easily seen that m0 ( E / I ) ∼ = m I ( E ).2 In addition, we know (see [7, p. 255]) that

m I ( S )m J ( S ) ⊆ m I ∪ J ( S ).

(1)

Basic facts from Semigroup Theory shall be used freely through the paper, and one may consult them in [2]. The most important for our purposes amongst the Green’s relations is J which is defined on any semigroup S as follows:

(x, y ) ∈ J

⇔

S 1 xS 1 = S 1 y S 1 .

Evidently, if S has an identity element then S 1 xS 1 = SxS for all x ∈ S. This relation is an equivalence, each class J of which becomes a semigroup J 0 after joining an extra zero (i.e. a zero not contained in S) and defining a new operation as follows:

x◦ y=

xy

if xy ∈ J ,

0

if xy ∈ / J,

x, y ∈ J . This semigroup J 0 turns out to be 0-simple if S is inverse. Given a semigroup S, a factor set

ρ ∈ m( S ) and a J -class J of S, one may restrict ρ to J obtaining a factor set ρ1 of J 0 as follows:

ρ1 (x, y ) =

ρ (x, y ) if xy ∈ J , if xy ∈ / J.

0

We recall the next fact from [8] which determines the J -class of an arbitrary element ( A , a) in Exel’s semigroup. Proposition 2.2. J ( A ,a) = {( B , b) ∈ E | ∃c 1 , c 2 ∈ G , c 1 B ⊆ A , c 2 A ⊆ B }.

2

Remark 2.3. It is easy to see that the latter proposition implies







J ( A ,a) = ( B , b) ∈ E  ∃c ∈ G , c B = A . Therefore J ( A ,a) does not depend on a ∈ A, and for this reason in what follows we shall write J A instead of J ( A ,a) . We shall also say that the volume of J A is | A |.

2

We observe by the way that M ∅ ( E ) ∼ = H 2 ( E , K ∗ ).

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It is well-known that by joining an extra zero to the J -class J A , J 0A becomes a completely 0-simple semigroup. More precise, let A be a finite subset in G, with stabilizer St A = {b ∈ G | b A = A }. Let A / St A be a full set of representatives of right cosets of St A, which are contained in A. We have the next. Proposition 2.4. (See [8].) Let J A be some J -class of E. Then J 0A is a completely 0-simple inverse semigroup of the form

S 0 (St A ; A / St A , A / St A ; Id), where Id is the identity | A / St A | × | A / St A |-matrix.

2

Finally we recall the description of the second 0-cohomology group of a completely 0-simple semigroup. Proposition 2.5. (See [17].) Consider a completely 0-simple semigroup T of the form S 0 (G ; I , Λ; P ), then M 0 ( T ) = H 2 (G , K ∗ ), where K ∗ is considered as a trivial G-module. 2 So, in our case M 0 ( J 0A ) = H 2 (St A , K ∗ ) = M (St A ). We shall denote by C n the cyclic group of order n. 3. Reduction to locally trivial factor sets In all what follows, if otherwise is not stated, we assume that G is an arbitrary group and K is an algebraically closed field. Let I be a non-empty ideal of E 3 . We are interested in describing the component M I ( E 3 ) of the Schur multiplier M ( E 3 ), and for this we will first describe the component m I ( E 3 ) of m( E 3 ). To this end we shall follow the ideas in [8] where the local behavior of factor sets from m0 ( E 3 ) was described. So we are going to consider first certain restrictions of the factor sets from m I ( E 3 ). In doing this, identity factor sets of some subsemigroups will play again a crucial role. We recall (see [17] or [7]) that for any semigroup S and ρ ∈ m( S ), the 2-cocycle equality

ρ (x, y )ρ (xy , z) = ρ (x, yz)ρ ( y , z), x, y , z ∈ S, is always satisfied, and if the form

ρ ∈ m I ( S ) and ρ (x, yz)ρ ( y , z) = 0, we will write this identity in

∂ ρ (x, y , z) =

ρ (x, y )ρ (xy , z) = 1. ρ (x, yz)ρ ( y , z)

Recall that if S is a monoid, then by [7, Theorem 1] a map when ρ satisfies the 2-cocycle equality and, moreover,

ρ (x, y ) = 0 for all x, y ∈ S. Fix an arbitrary

⇔ 

ρ ∈ mI (E3) ∼ = m0 ( E 3 / I ) = m0 (

ρ : S × S → K belongs to m( S ) exactly

ρ (1, xy ) = 0 ( A ,a)∈ /I J A

(2)

∪ 0). Since | A |  3 by definition of E 3 ,

then the group St A is cyclic, and hence M 0 ( J 0A ) ∼ = C 1 (see [21, p. 523]). = M (St A ) ∼ = K ∗ /( K ∗ )| St A | ∼ Therefore the restriction of ρ to every J -class J A  I is equivalent to the identity factor set, i.e. to the function ε : J A × J A → K defined by

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ε(x, y ) =

1 if xy ∈ J A , 0

if xy ∈ / J A.

This means that for each A such that ( A , a) ∈ E 3 \ I there exists a map

∀x, y ∈ J A

xy ∈ J A

205

⇒

ρ (x, y ) =

α A : J A → K ∗ satisfying:

α A (x)α A ( y ) . α A (xy )

Since the intersection of two different J -classes is empty, one can let ( A , 1) run over a full set of representatives of J -classes not contained in I and join all such α A to obtain a map α : ( E 3 / I ) \ 0 → K ∗ . Setting

ρ  (x, y ) = ρ (x, y )

α (xy ) , α (x)α ( y )

for x, y ∈ E 3 / I , xy ∈ / I , we obtain a factor set ρ  , which is equivalent to ρ , and whose all restrictions to J -classes J A , with J A  I , are trivial in the sense that x, y , xy ∈ J A implies ρ  (x, y ) = 1. Thus up to equivalence we may assume that ρ | J A = 1 for all A in the above-mentioned sense. Denote by N 2 ⊂ E 3 / I , the set which consists of zero and all J -classes of volume 3 that are not contained in I . For | A | = 3 with ( A , a) ∈ / I , it follows by 1) and 2) of Lemma 4 from [8] that J A ∪ 0 is an ideal of E 3 . Consequently, N 2 is an ideal of E 3 / I . On the other hand the idempotents of E 3 / I are {( A , 1) ∈ E 3 / I } ∪ 0, so they form a commutative subsemigroup (subsemilattice) D of E 3 / I . The union D = N 2 ∪ D is a subsemigroup of E 3 / I . We recall the following: Lemma 3.1. (See [8, Lemma 5].) Let x ∈ N 2 , e ∈ D. If xe = 0 then xe = x; if ex = 0 then ex = x. We proceed with the next theorem. Theorem 3.2. An arbitrary factor set from m I ( E 3 ) is equivalent to a factor set ρ such that all restrictions of ρ on each J -class, which is not contained in I , as well as on D are trivial. Proof. Let ρ ∈ m0 ( E 3 / I ). As it was mentioned above, we can assume that ρ | J A = 1 for every J A with J A ∩ I = ∅. As the product of two different J -classes of volume 3 is zero, we get ρ | N 2 = 1. Furthermore, it was proved in [17] that the group M 0 ( S ) is trivial for any semilattice S. Hence, there is a map α : D \ 0 → K ∗ , such that

∀e , f ∈ D e f = 0

⇒

ρ (e, f ) =

α (e)α ( f ) . α (e f )

Taking e = f we have α (e ) = ρ (e , e ). But e belongs to some J -class outside I , so ρ (e , e ) = 1. Hence, ρ | D = 1. It remains to check that the restrictions of ρ on ( D \ N 2 ) × N 2 and N 2 × ( D \ N 2 ) are trivial. We consider the first case, the second one being similar. Let e ∈ D \ N 2 , x ∈ N 2 and ex ∈ / I . By Lemma 3.1, ex = x, so ρ (e , ex)ρ (e , x) = ρ 2 (e , x) = 0. Therefore

1 = ∂ ρ (e , e , x) = which gives

ρ (e, x) = 1. 2

ρ (e, e) 1 = , ρ (e, ex) ρ (e, x)

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4. The description of m0 ( E 3 / I ) According to the previous section it is enough to study only those factor sets ρ ∈ m I ( E 3 ), whose restrictions to D and to all J -classes which are not contained in I are trivial (i.e. take values 1 and 0 only). Such factor sets form a subsemigroup mI ⊂ m I ( E 3 ), which we are going to characterize. First we start with some technical but useful lemmas. Lemma 4.1. Let ρ ∈ mI , x, y ∈ E 3 , xy ∈ / I . Then, ρ (xe y , e x−1 y ) = 1. Proof. Observe that xe y e x−1 y = xy ∈ / I . We have xe y = xx−1 xy y −1 = xe x−1 y y −1 ∈ Ee x−1 y E and similarly e x−1 y ∈ Exe y E. So xe y and e x−1 y belong to the same J -class, and, consequently, ρ (xe y , e x−1 y ) = 1. 2 Lemma 4.2. If ρ ∈ mI , then ρ (x, y ) = ρ (x, e y )ρ (e x−1 , y ). Proof. Clearly xy ∈ I exactly when xe y ∈ I , and if this holds, then ρ (x, y ) = 0 = ρ (x, e y )ρ (e x−1 , y ). / I . Then by Lemma 4.1, ρ (x, e y )ρ (e x−1 , y ) = ρ (xe y , e x−1 y )ρ (x, e y )ρ (e x−1 , y ). Suppose xy ∈ By using the 2-cocycle identity several times we obtain

ρ (xe y , e x−1 y )ρ (x, e y )ρ (e x−1 , y ) = ρ (x, e y e x−1 y )ρ (e y , e x−1 y )ρ (e x−1 , y ) = ρ (x, e x−1 y )ρ (e y e x−1 , y )ρ (e y , e x−1 ) = ρ (x, e x−1 y )ρ (e x−1 e y , y )ρ (e x−1 , e y ) = ρ (x, e x−1 y )ρ (e x−1 , y )ρ (e y , y ) = ρ (x, y )ρ (x, e x−1 )ρ (e y , y ) = ρ (x, y ), as xJ e x−1 , e y J y and x, e x−1 , y , e y ∈ / I.

2

Lemma 4.3. If ρ ∈ mI , then ρ (x, e y ) = ρ (e xy , x), for x, y ∈ E 3 , xy ∈ / I. Proof. Since xy ∈ / I , we get xe y ∈ / I , e xy xe y = xe y e x−1 e y = xe x−1 e y = xe y ∈ / I , then we have

∂ ρ (e xy , x, e y ) =

ρ (e xy , x)ρ (e xy x, e y ) = 1. ρ (e xy , xe y )ρ (x, e y )

Write x = ( A , a), y = ( B , b). If x = 1 or y = 1, the lemma is true, so we assume | A |, | B |  2. If | A ∪ aB | = 3, then e xy x, e y , xe y , e xy ∈ D so ρ (e xy x, e y ) = ρ (e xy , xe y ) = 1, and hence ρ (x, e y ) = ρ (e xy , x). Now let | A ∪ aB | = 2. Then A = aB, and hence e xy J x, xJ e y . This implies ρ (e xy , x) = ρ (x, e y ) = 1. 2 From Lemma 4.2 and Lemma 4.3 we obtain the next: Corollary 4.4. If ρ ∈ mI and xy ∈ / I , then ρ (x, y ) = ρ (e xy , x)ρ (e x−1 , y ). We also give the following: Lemma 4.5. If ρ ∈ mI and yx ∈ / I , then ρ (e x , y −1 )ρ (e yx , y ) = 1.

2

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Proof. Since yx ∈ / I , e y−1 , ( yx)−1 , e x e y−1 ∈ / I . Then from ∂ ρ (e x , y −1 , y ) = 1 we obtain



 



ρ e x , y −1 ρ e x y −1 , y = 1. By Lemma 4.2

ρ (e x y −1 , y ) = ρ (e x y −1 , e y )ρ (e ye x , y ), so that 

1 = ρ e x , y −1

 



ρ e x y −1 , e y ρ (e yex , y ).

ρ (e x y −1 , e y ) = ρ (ee x y−1 y , e x y −1 ) = ρ (e x e y−1 , e x y −1 ), and on the

In view of Lemma 4.3 we obtain

other hand e ye x = ye x y −1 = e yx . Hence



1 = ρ e x , y −1

 



ρ e x e y−1 , e x y −1 ρ (e yx , y ).

Finally, the middle factor equals 1, because



supp(e x e y −1 ) = supp e x y −1



e x e y −1 J e x y −1 .

⇒

2

We proceed with the following theorem. Theorem 4.6. For every ρ ∈ mI , there exists a function ν : E 3 × E 3 → K , satisfying the following properties:

x−1 y ∈ I ,

(3)

ν (x, x) = 1 for all x ∈ E 3 \ I ,

(4)

ν (x, y ) = 0 

ν (x, y )ν y −1 x, y

⇔  −1

= 1,

if x−1 y ∈ / I;

(5)

and for all x, y ∈ E 3 ,

ν (x, y ) = ν (e x , y ),

(6)



 ρ (x, y ) = ν (xy , x)ν x−1 , y .

(7)

Proof. Define ν : E 3 × E 3  (x, y ) → ρ (e x , y ) ∈ K . Since e x y ∈ I ⇔ x−1 y ∈ I , condition (3) is satisfied as well as (4) and (6). By definition of ν and Lemma 4.5, ν (x, y )ν ( y −1 x, y −1 ) = ρ (e x , y )ρ (e y −1 x , y −1 ) = 1 which is (5). Finally (7) is clear if xy ∈ I . If xy ∈ / I , we use Corollary 4.4 to obtain ρ (e xy , x)ρ (e x−1 , y ) = ρ (x, y ). 2 Some useful properties of

ν (xy , x)ν (x−1 , y ) =

ν are given as follows.

Lemma 4.7. Let ν : E 3 × E 3 → K be a function, satisfying (3)–(6). (i) Without loss of generality we can suppose

ν (x, 1) = ν (1, x) = ν (e, f ) = 1,

(8)

for all x ∈ E 3 \ I and e , f ∈ D such that e f ∈ / I. / I . If x and y have the same support or vol( y ) = 3, then (ii) Suppose that x, y ∈ E 3 are such that x−1 y ∈

ν (x, y ) = 1.

(9)

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Proof. By (4) and (5), we have 1 = ν (1, x)ν (x−1 , x−1 ) = ν (1, x), for each x ∈ E 3 \ I , and by (5) we / I , by using again (5) we get 1 = obtain 1 = ν (x, 1)ν (x, 1) = ν 2 (x, 1). Consider e , f ∈ D with e f ∈ ν (e, f )ν (e f , f ) and 1 = ν (e f , f )ν (e f , f ) = ν 2 (e f , f ), so 1 = ν 2 (e, f ). Hence, if we define the function ν  : E 3 × E 3 → K by 

ν (x, y ) =



ν (x, y ) if ν (x, y ) = −1, 1 if ν (x, y ) = −1,

then ν  verifies (3)–(6) and ν  (x, 1) = ν  (1, x) = ν  (e , f ) = 1 for all x ∈ E 3 \ I and all e , f ∈ D with ef ∈ / I. / I . Assume first that x and y have the same support and Take arbitrary x, y ∈ E 3 \ I with x−1 y ∈ write x = ( A , a) and y = ( A , b). Then by (5), (4) and (6) we obtain



 

1 = ν ( A , 1), ( A , b)

 







ν b−1 A , b−1 , b−1 A , b−1 = ν ( A , 1), ( A , b) = ν (x, y ).

Assuming now vol( y ) = 3, we have y −1 e x = y −1 and thus 1 = ν (e x , y )ν ( y −1 e x , y −1 ) = ν (e x , y ) = ν (x, y ), using again (5), (4) and (6). 2 Due to the above fact in all what follows we shall assume that if we have a function E 3 → K , verifying (3)–(6), then ν also satisfies Lemma 4.7.

ν : E3 ×

Corollary 4.8. Let ν : E 3 × E 3 → K be a function, verifying (3)–(6), x, y ∈ E 3 and x−1 y ∈ / I. (i) If y = e ∈ D then ν (x, e ) = 1. (ii) If supp(x) ⊆ supp( y ) then ν (x, y ) = 1. Proof. (i) follows from (6) and (8), (ii) follows from (5) and Lemma 4.7(ii), noting that in this case supp( y −1 x) = supp( y −1 ). 2 The next theorem together with Theorem 4.6 characterize the elements of M I ( E 3 ). Theorem 4.9. Let ν : E 3 × E 3 → K be a function verifying (3)–(6), and define ρ : E 3 × E 3  (x, y ) → ν (xy , x)ν (x−1 , y ) ∈ K . Then ρ ∈ mI . Proof. We note first that if xy ∈ / I , then y −1 x−1 ∈ / I and y −1 x−1 x ∈ / I , hence ν (xy , x)ν (x−1 , y ) = 0 and ρ (x, y ) = 0. Moreover, by (3) xy ∈ I implies ρ (x, y ) = 0. We prove next that ρ is trivial in the J -classes. Let x = ( A , a), y = ( B , b) ∈ J C , such that xy ∈ J C .

Since xy = ( A ∪ aB , ab), xy J x, we see that | A | = | A ∪ aB | = | B |, so A = aB, which implies xy = ( A , ab). Note also that x−1 = (a−1 A , a−1 ) and y = (a−1 A , b), therefore supp(xy ) = supp(x) and supp(x−1 ) = supp( y ). Thus by (ii) of Lemma 4.7 we obtain ρ (x, y ) = 1. Now consider x, y ∈ D, verifying xy ∈ / I . If x, y ∈ N 2 , then vol(x) = vol( y ) = 3. But xy ∈ / I , in particular, xy = 0, so we have xJ y, and consequently, ρ (x, y ) = 1. If x, y ∈ D are such that xy ∈ D, then by (8) we obtain ρ (x, y ) = 1. If x = e ∈ D, y ∈ N 2 , then since ey ∈ / I , by Lemma 3.1 ey = y, and using (6) we get ρ (e , y ) = ν ( y , e)ν (e, y ) = ν (e y , e)ν (e, y ) = ν (e, y ). In view of the fact that y has volume 3, we obtain ν (e, y ) = 1 thanks to (9). Finally if x ∈ N 2 and y = f ∈ D, we have xf = x, and ρ (x, f ) = ν (x, x)ν (x−1 , f ) = ν (x−1 , f ) = 1 by (i) of Corollary 4.8. Since ρ (x, y ) = 0 exactly when xy ∈ I , we see that ρ satisfies (2). Thus it remains to prove that ρ verifies the 2-cocycle equality. Take arbitrary x, y , z ∈ E 3 . If xyz ∈ I then ρ (xy , z) = ρ (x, yz) = 0 and evidently ρ (x, y )ρ (xy , z) = ρ (x, yz)ρ ( y , z). Also this equality is clear if one of the elements x, y , z / I and x = 1, y = 1 and z = 1. In particular we have ρ (x, yz)ρ ( y , z) = 0. equals 1. So, we suppose xyz ∈

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209

We need to prove the equality

1 = ∂ ρ (x, y , z) =

ρ (x, y )ρ (xy , z) ν (e xy , x)ν (e x−1 , y )ν (e xyz , xy )ν (e y−1 x−1 , z) = . ρ (x, yz)ρ ( y , z) ν (e xyz , x)ν (e x−1 , yz)ν (e yz , y )ν (e y−1 , z)

Write x = ( A , a), y = ( B , b) and z = (C , c ). Then the elements of E 3 involved above are:

e xy = ( A ∪ aB , 1),





e x−1 = a−1 A , 1 ,

e xyz = ( A ∪ aB ∪ abC , 1),





e y −1 x−1 = b−1 B ∪ b−1 a−1 A , 1 , e yz = ( B ∪ bC , 1),

xy = ( A ∪ aB , ab),

yz = ( B ∪ bC , bc ),





e y −1 = b −1 B , 1 .

Write W = A ∪ aB ∪ abC . Since xyz = 0, | W |  3. We consider next a series of cases and some subcases. Case 1. Suppose vol(xy ) = 2. Then xy = ( A ∪ aB , ab) = ( A , ab) = (aB , ab) whence e xy = e x , e y −1 x−1 = e y −1 and e x−1 = e y . Consequently, by (ii) of Corollary 4.8 we have that ν (e xy , x) = ν (e x−1 , y ) = ν (e x−1 , yz) = 1. Hence

∂ ρ (x, y , z) =

ν (e xyz , xy ) . ν (e xyz , x)ν (e yz , y )

Case 1.1. Suppose either a = 1 or b = 1. Then either x or y is an idempotent, and either xy = y or xy = x, respectively. Hence ∂ ρ (x, y , z) = 1, using (8). Case 1.2. Let a = 1 = b. From A = aB we have ab = 1, hence xy = e x , y = x−1 . Therefore

∂ ρ (x, y , z) =

1

ν (e xyz , x)ν (e yz , y )

= 1,

since by (5) and (6)

ν (e yz , y ) = ν ( yz, y ) =

1

ν (xyz, x)

=

1

ν (e xyz , x)

Hence, for the rest of the proof we can suppose vol(xy ) = 3. Then ν (e xy , x). This implies

∂ ρ (x, y , z) =

.

ν (e xyz , xy ) = 1 and ν (e xyz , x) =

ν (e x−1 , y )ν (e y−1 x−1 , z) . ν (e x−1 , yz)ν (e yz , y )ν (e y−1 , z)

Case 2. vol( yz) = 2. Similarly to Case 1, from yz = ( B ∪ bC , bc ) = ( B , bc ) = (bC , bc ) we have e yz = e y , e y −1 = e z , whence

∂ ρ (x, y , z) = in view of (ii) of Corollary 4.8.

ν (e x−1 , y )ν (e y−1 x−1 , z) , ν (e x−1 , yz)

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Case 2.1. Let b = 1. Then y is an idempotent and yz = z. Then we obtain

∂ ρ (x, y , z) = because by (5),

ν (e y−1 x−1 , z) ν ( y −1 x−1 , z) = 1, = ν (e x−1 , yz) ν (x−1 , z)

ν ( y −1 x−1 , z) and ν (x−1 , z) in our case have the same inverse.

Case 2.2. Let c = 1. Then z2 = z, yz = y and ∂ ρ (x, y , z) = 1. Case 2.3. Let b = 1 = c. Then bc = 1, y = z−1 and

∂ ρ (x, y , z) = ν (e x−1 , y )ν (e y −1 x−1 , z) = 1, thanks to (5) and (8). Case 3. vol(xy ) = vol( yz) = 3. Then e xy = e xyz and

∂ ρ (x, y , z) =

ν (e x−1 , yz) = 1 by (9). Hence

ν (e x−1 , y )ν (e y−1 x−1 , z) . ν (e yz , y )ν (e y−1 , z)

Since W = A ∪ aB = aB ∪ abC , we have











y −1 x−1 = b−1 B ∪ b−1 a−1 A , b−1 a−1 = b−1 a−1 W , b−1 a−1



and





y −1 yz = e y −1 z = b−1 B ∪ C , c = b−1 a−1 W , c .

ν ( y −1 x−1 , y −1 ) = ν (e y−1 x−1 , y −1 ) = ν (e y−1 yz , y −1 ) = ν ( y −1 yz, y −1 ), and using (5) we conclude ν (e x−1 , y ) = ν (x−1 , y ) = ν ( yz, y ) = ν (e yz , y ). It follows that

On the other hand, since vol( z−1 y −1 ) = 3, we have









ν z−1 y −1 x−1 , z−1 = ν z−1 y −1 , z−1 , and therefore by (5),

ν ( y −1 x−1 , z) = ν ( y −1 , z). Then ∂ ρ (x, y , z) = 1 follows by (6). 2

5. Partial factor sets For reader’s convenience we recall from [7] the definition of the semigroup T . Define the following transformations on G × G:







 −1

g : (x, y ) → xy , y −1 , h : (x, y ) → y −1 , x

t : (x, y ) → (x, 1). These transformations satisfy the relations:

,

(10) (11) (12)

M. Dokuchaev et al. / Journal of Algebra 392 (2013) 199–225

g 2 = h 2 = 1,

( gh)3 = 1,

t2 = t,

gt = t ,

211

t ght = thgh,

tht = 0,

(13)

where 0 is the map (x, y ) → (1, 1). Consider the abstract semigroup T generated by the elements g , h, t with the relations (13). The maps (10)–(12) determine an action of T on G × G. We recall a rough description of the structure of the semigroups pm(G ) and pM (G ). Denote by C (G ) the semilattice of all non-empty T -subsets of G × G with respect to the set-theoretic inclusion and intersection.3 By [7, Corollary 7] the elements of C (G ) are the domains of the partial factor sets. Theorem 5.1. (See [7, Theorem 5].) The semigroups pm(G ) and pM (G ) are semilattices of abelian groups

pm(G ) =





pM (G ) =

pm X (G ),

X ∈C ( G )

pM X (G ),

X ∈C ( G )

where pm X (G ) = {σ ∈ pm(G ) | dom σ = X } and pM X (G ) = pm X (G )/ ∼.

2

We also recall that the operation in the semigroup pm(G ) is given by pointwise multiplication of partial factor sets, and that the structural homomorphisms in the semilattices of abelian groups in Theorem 5.1 are determined by restriction. It follows that the multiplication formula in pM (G ) can be expressed as follows: for any X , Y ∈ C (G ), σ ∈ pm X (G ) and σ  ∈ pm Y (G ) one has

 

cls(σ ) · cls







 





σ = ψ XX∩Y cls(σ ) · ψ XY ∩Y cls σ  = cls σ | X ∩Y · σ  | X ∩Y ,

where ψ XX∩Y : pM X (G ) → pM X ∩Y (G ) and ψ XY ∩Y : pM Y (G ) → pM X ∩Y (G ) are structural homomorphisms, and cls(σ ) ∈ pM X (G ) and cls(σ  ) ∈ pM Y (G ) stand for the equivalence classes in pM (G ) of σ and σ  , respectively. It is possible to obtain every partial factor set of G from a factor set of E 3 (G ), in fact we recall the following result. Proposition 5.2. (See [8].) A map σ : G × G → K belongs to pm(G ) if and only if there is a factor set ρ ∈ m( E 3 ) such that:

∀x, y ∈ G (x, y ) ∈ dom σ ∀(x, y ) ∈ dom σ

σ (x, y ) =

⇔



 [x], [ y ] ∈ dom ρ ;

ρ ([x], [ y ])ρ ([x−1 ], [x][ y ]) . ρ ([x−1 ], [xy ])

(14)

2

(15)

If σ ∈ pm(G ) has domain X , then taking ρ as in Proposition 5.2, let I be the ideal of E 3 which determines the domain of ρ , i.e. ρ ∈ m I ( E 3 ). It follows that (x, y ) ∈ X ⇔ [x][ y ] ∈ / I . In particular, X = G × G implies I = ∅. On the other hand, we know from [8, Lemma 1] that X = G × G exactly when ρ ∈ m0 ( E 3 ). Thus in any case I = ∅. Furthermore, clearly I = E 3 as σ (1, 1) = 1. Now if we take another factor set σ1 with domain X , it is not clear whether or not the factor set ρ1 of E 3 , which corresponds to σ1 according to Proposition 5.2, lies in m I ( E 3 ). The fact below answers this positively. Proposition 5.3. There is an isomorphism ι : C (G ) → Y ( E 3 ) \ { E 3 , ∅} of semilattices.

3 Since the union of two T -subsets is again a T -subset, it follows that C (G ) is a lattice. Nevertheless we shall use only the meet semilattice structure of C (G ).

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Proof. Take X ∈ C (G ) and

σ1 , σ2 ∈ pm X (G ). By Proposition 5.2 there exist ρ1 , ρ2 ∈ m( E 3 ) verifying

(x, y ) ∈ X = dom σi If

⇔



 [x], [ y ] ∈ dom ρi (i = 1, 2).

ρ1 ∈ m I ( E 3 ) and ρ2 ∈ m J ( E 3 ) we have [x][ y ] ∈ /I



⇔

 [x], [ y ] ∈ dom ρ1

⇔

[x][ y ] ∈ I

⇔



 [x], [ y ] ∈ dom ρ2

⇔

[x][ y ] ∈ / J

so that

[x][ y ] ∈ J .

(16)

We already observed that X = G × G exactly when ρi ∈ m0 ( E 3 ), i.e. I = J = 0. Assume X = G × G, then ∅ = I , J = 0. We wish to show that I = J , from (16) it is seen that

[x] ∈ I

⇔

[x] ∈ J .

We know that the non-zero elements of E 3 are of the form [1], [a], ea , ea [b], ea eb where 1 = a = b = 1. Therefore:

• • • •

If If If If

[a] ∈ I ⇒ [a] ∈ J . ea = [a][a−1 ] ∈ I ⇒ ea ∈ J . ea [b] = [a][a−1 ][b] = [a][a−1 b] ∈ I ⇒ ea [b] ∈ J . ea eb ∈ I ⇒ ea eb [b] = ea [b] ∈ I ⇒ ea [b] ∈ J ⇒ ea eb ∈ J .

Consequently, I ⊆ J , and similarly I ⊇ J , so that I = J . Thus we have a well-defined map

C (G )  X → ι( X ) = I ∈ Y ( E 3 ) \ { E 3 , ∅}. It is injective because if X → I and Y → I , then (x, y ) ∈ X ⇔ [x][ y ] ∈ / I ⇔ (x, y ) ∈ Y , which gives / I ∪ J which X = Y . Its surjectivity immediately follows from Proposition 5.2. Finally note that 1 ∈ means that I ∪ J ∈ Y ( E 3 ) \ { E 3 , ∅}, and the fact that our map is an isomorphism of semigroups is a direct consequence of (1). 2 Now let σ ∈ pm(G ) be a factor set with domain X and I = ι( X ). Then there is ρ ∈ m I satisfying (14) and (15). By Theorem 3.2 there exists a factor set ρ  ∈ mI equivalent to ρ . This means that there is a function ε : E 3 → K ∗ such that

ρ  (x, y ) = ρ (x, y )ε(x)ε( y )ε−1 (xy ) ∀x, y ∈ E 3 . Now if

σ  : G × G → K is defined by 

σ (a, b) = then by Proposition 5.2

ρ  ([a],[b])ρ  ([a−1 ],[a][b]) ρ  ([a−1 ],[ab])

if (a, b) ∈ dom σ ,

0

if (a, b) ∈ / dom σ ,

σ  ∈ pm(G ) and we readily check that

σ  (a, b) = σ (a, b)δ(a)δ(b)δ(ab)−1 ∀a, b ∈ G ,

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213

where δ : G → K ∗ is defined by δ(a) = ε ([a]), a ∈ G. Thus σ and σ  are equivalent, and we may assume without loss of generality that ρ ∈ mI ( E 3 ). By Theorem 4.6 there exists a map ν : E 3 × E 3 → K verifying (3)–(7). Then for (a, b) ∈ dom σ :

σ (a, b) =

ν ([a][b], [a])ν ([a−1 ], [b])ν ([a−1 ][a][b], [a−1 ])ν ([a], [a][b]) . ν ([a−1 ][ab], [a−1 ])ν ([a], [ab])

Observe that one has ν ([a], [a][b]) = 1 since supp([a]) ⊆ supp([a][b]). Furthermore, [a−1 ]) = ν ([a−1 ][ab], [a−1 ]), which gives

σ (a, b) =

ν ([a−1 ][a][b],

ν ([a][b], [a])ν ([a−1 ], [b]) . ν ([a], [ab])

Using property (5) we obtain



 

1 = ν [a][b], [a]









 







ν a−1 [a][b], a−1 = ν [ab], [a] ν a−1 [ab], a−1 ,

which implies









ν [ab], [a] = ν [a][b], [a] ,

(17)

and, consequently,

σ (a, b) =

ν ([ab], [a])ν ([a−1 ], [b]) . ν ([a], [ab])

Now define λ : G × G  (a, b) → ν ([a], [b]) ∈ K . Then

σ (a, b) = and, keeping in mind the properties of

λ(a−1 , b)λ(ab, a) , λ(a, ab)

ν , we immediately see that

λ(a, b) = 0

⇔





a−1 [b] ∈ I ,

(18)

and

λ(a, 1) = λ(1, a) = 1,

if [a] ∈ / I.

Moreover, if [a−1 ][b] ∈ / I , then

              λ(a, b)λ b−1 a, b−1 = ν [a], [b] ν b−1 a , b−1 = ν [a], [b] ν b−1 [a], b−1 = 1, thanks to (17) and (5). Summarizing we have the next theorem.

(19)

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Theorem 5.4. Let σ ∈ pm X (G ). Then up to equivalence σ has the form

σ (a, b) =

λ(a−1 , b)λ(ab, a) , λ(a, ab)

∀(a, b) ∈ X ,

(20)

where λ : G × G → K is a map satisfying (18), (19) and

    λ(a, b)λ b−1a, b−1 = 1 ∀ a−1 , b ∈ X .

2

(21)

The converse of Theorem 5.4 is also true: Theorem 5.5. Let X ∈ C (G ) and λ : G × G → K be a function such that (18), (19) and (21) are satisfied. Then σ defined by (20) is a factor set of G with domain X . Proof. Let I = ι( X ). Using the properties of λ we want to find first a function ν : E 3 × E 3 → K verifying (3)–(6). We start by setting ν (x, y ) = 0 for all (x, y ) ∈ E 3 × E 3 such that x−1 y ∈ I , and if x−1 y ∈ / I we declare ν (x, y ) = 1 if either y is idempotent, or y has volume 3, or x = 1. Now suppose y = ( B , b) with B = {1, b}, b = 1 and x−1 y ∈ / I , and let A be the support of x. If x has volume 2, i.e. A = {1, a} for some a ∈ G, a = 1, we define ν (x, y ) = λ(a, b). Since [a−1 ] = [a−1 ]ea and x ∈ {[a], ea }, we readily see that x−1 y ∈ / I ⇒ [a−1 ][b] ∈ / I , so that λ(a, b) = 0. If | A | = 3, then in view of − 1 − 1 x y = 0, we have vol(x y ) = 3. Consequently, B ⊂ A, and we can write A = {1, a, b}. In this case we again define ν (x, y ) = λ(a, b). If [a−1 ][b] ∈ I then since y = [b] we have e x y = ({1, a, b}, 1)({1, b}, b) = [a][a−1 ][b] ∈ I and consequently x−1 y ∈ I which is an absurd. Hence [a−1 ][b] ∈ / I and again λ(a, b) = 0. It follows that ν satisfies the conditions (3) and (6). Observe next that taking b = a in (21) we obtain λ(a, a) = 1, provided that [a−1 ][a] ∈ / I . It follows that ν ([a], [a]) = 1 which proves (4). As to (5), it is enough to prove it for y = [b] with 1 = b ∈ G and x = [1], assuming of course x−1 y ∈ / I . From the above, [a−1 ][b] ∈ / I. If x has volume 2, i.e. its support is of the form {1, a}, we have x = [a] or x = ea , and in both cases ν (x, y ) = λ(a, b). Furthermore, y −1 x = ({1, b−1 , b−1 a}, c ), c ∈ {b−1 , b−1 a}. If a = b then ν ( y −1 x, y −1 ) = λ(a−1 , a−1 ) = 1, ν (x, y ) = λ(a, a) = 1, so that (5) is evidently satisfied. If a = b then ν ( y −1 x, y −1 ) = λ(b−1 a, b−1 ) and (5) follows by (21). Finally, if the support of x has three elements 1, a, b then ν (x, y ) = λ(a, b) and, since in this case the support of y −1 x is {1, b−1 , b−1 a} it follows again that ν ( y −1 x, y −1 ) = λ(b−1 a, b−1 ), obtaining (5) from (21). By Theorem 4.9 the map ρ : E 3 × E 3  (x, y ) → ν (xy , x)ν (x−1 , y ) ∈ K lies in mI . As it was computed in the proof of Theorem 5.4, taking (a, b) ∈ X ([a][b] ∈ / I ) we have

ρ ([a], [b])ρ ([a−1 ], [a][b]) λ(a−1 , b)λ(ab, a) = = σ (a, b), λ(a, ab) ρ ([a−1 ], [ab]) and by Proposition 5.2

σ is a factor set of G with domain X . 2

Now we can obtain a characterization of the partial factor sets. Theorem 5.6. Let τ be a partial factor set of G with domain X . Then there is a partial factor set σ , equivalent to τ , such that for all (a, b) ∈ X

σ (a, b) =

1

σ (b−1 , a−1 ) 

(22)

,







σ (a, b) = σ b−1a−1 , a = σ b, b−1a−1 ,

(23)

σ (a, 1) = 1.

(24)

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215

Conversely, let σ : G × G → K be a partially defined map with domain X ∈ C (G ) such that (22)–(24) are satisfied for all (a, b) ∈ X . Then σ is a partial factor set of G. Proof. 1) Let (a, b) ∈ X and I = ι( X ). By Theorem 5.4 there is a factor set Since I is an ideal of E 3 (G ), we have





 





σ ∼ τ of the form (20).



/ I, b −1 a −1 , b −1 e a −1 , e b a −1 ∈

and therefore all the expressions in (22) and (23) are different from zero. We prove first (22). By (20)





σ (a, b)σ b−1 , a−1 =

λ(a−1 , b)λ(ab, a) λ(b, a−1 )λ(b−1 a−1 , b−1 ) . λ(a, ab) λ(b−1 , b−1 a−1 )

Now by (21)

    −1 −1   [b] ∈ λ a−1 , b λ b−1 a−1 , b−1 = 1 because a / I,  −1  −1   −1  ∈ / I, λ b, a λ(ab, a) = 1 because b a  −1   −1 −1 −1  = 1 because a [ab] = ea−1 [b] ∈ λ(a, ab)λ b , b a / I, and (22) follows. To prove (23) observe that by (20) and (21)





σ b −1 a −1 , a =

λ(ab, a)λ(b−1 , b−1 a−1 ) λ(a−1 , b)λ(ab, a) = = σ (a, b). λ(a, ab) λ(b−1 a−1 , b−1 )

Analogously σ (a, b) = σ (b, b−1 a−1 ), and (24) is immediate. 2) Note first that using (24) and taking b = 1 in (23), we obtain σ (a, 1) = σ (a−1 , a) = σ (1, a−1 ) = 1 K for all a ∈ G such that [a] ∈/ I = ι( X ). Consider the symmetric group S 3 with the following presentation:





S 3 = g , h  g 2 = h2 = ( gh)3 = 1 . Since X is T -invariant, we have



 



xy , y −1 , y −1 , x−1 ∈ X

for all (x, y ) ∈ X , and consequently, we may use the formulas (10) and (11) to obtain an action of S 3 in X . Explicitly, the action is given by the transformations





g : (x, y ) → xy , y −1 ,





h : (x, y ) → y −1 , x−1 .

Pick a pair (a, b) with a, b ∈ G, and let A (a,b) be the S 3 -orbit which contains (a, b). Then we have





 

 

 

 

A (a,b) = (a, b), b−1 a−1 , a , b, b−1 a−1 , b−1 , a−1 , ab, b−1 , a−1 , ab

 .

Consider now an action of the cyclic group C 3 =  g  | ( g  )3 = 1 ⊂ S 3 , ( g  = gh) on G × G given by the transformation g  · (x, y ) = ( y −1 x−1 , x). We can write

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A (a,b) = C 3 · (a, b) ∪ C 3 · b−1 , a−1 , with





 

C 3 · (a, b) = (a, b), b−1 a−1 , a , b, b−1 a−1





C 3 · b −1 , a −1 =



 

 ,

 



b−1 , a−1 , ab, b−1 , a−1 , ab .

In view of (22) and (23) we see that σ is constant on C 3 · (a, b), as well as on C 3 · (b−1 , a−1 ), and in addition, σ (b−1 , a−1 ) = σ (a, b)−1 . Now for (a, b) ∈ X , the C 3 -orbit C 3 · (a, b) has only one element exactly when a = b and a3 = 1, in any other case |C 3 · (a, b)| = 3. Since (a, b) ∈ X is taken arbitrarily, the same holds for the orbit C 3 · (b−1 , a−1 ). So, if |C 3 · (a, b)| = 3, then A (a,b) has 3 elements if and only if C 3 · (a, b) = C 3 · (b−1 , a−1 ), and this occurs exactly when (a, b) ∈ {(b−1 , a−1 ), (ab, b−1 ), (a−1 , ab)}. This implies that one of the elements of C 3 · (b−1 , a−1 ) must have an entry equal to 1 and thus the value of σ is 1 K on all the elements of A (a,b) . Let us define a function ω : X  (a, b) → ωa,b ∈ K ∗ such that ωa3,b = σ (a, b) (remember that K is algebraically closed) and

ωa,b = ωb−1 a−1 ,a = ωb, b−1 a−1 = and when A (a,b) has 3 elements, set different orbits do not intersect. Define λ : G × G → K ∗ by

1

ωb−1 ,a−1

=

1

ωab, b−1

=

1

ωa−1 , ab

,

ωa,b = 1. We note that ω is a well-defined function because

λ(a, b) =

ωa−1 ,b if (a−1 , b) ∈ X , 0

if (a−1 , b) ∈ / X.

Then λ(a, b) = 0 ⇔ [a−1 ][b] ∈ I , so in order to prove that σ is a factor set of G, thanks to Theorem 5.5 it is enough to verify the equalities (19), (21) and (20). By definition of λ, λ(a, 1) = λ(1, a) = 1, for all / I so we have (19). a ∈ G, such that [a] ∈ Next, consider (u , v ) ∈ G × G such that [u −1 ][ v ] ∈ / I . Then [u −1 v ][ v −1 ] ∈ / I and

  ω u −1 , v λ(u , v )λ v −1 u , v −1 = ωu −1 , v ωu −1 v , v −1 = = 1,

ω u −1 , v

and this proves (21). Finally, to prove (20), we observe that

    λ(u −1 , v )λ(uv , u ) = λ u −1 , v λ(uv , u )λ v −1 , v −1 u −1 λ(u , uv ) = ωu , v ω v −1 u −1 ,u ω v , v −1 u −1 = ωu3, v = σ (u , v ), for each (u , v ) ∈ X . Thus

σ is a factor set with domain X . 2

Remark 5.7. By the way note that with the above notation, any S 3 -orbit of X of length 3 is of the form {(a, 1), (a−1 , a), (1, a−1 )} with a = 1.

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217

We denote by pmX (G ) the subgroup of pm X (G ) consisting of the functions satisfy (22)–(24).

σ : G × G → K which

Corollary 5.8. Let G be an arbitrary group and X a domain of a partial factor set of G. Then: (i) Any partial factor set from pm X (G ) is equivalent to an element from pmX (G ). (ii) The kernel N X of the natural epimorphism pmX (G ) → pM X (G ) consists of those which there are functions ρ : G → K ∗ such that

σ : G × G → K for

ρ (1) = 1 K , ρ (a) =

1

ρ (a−1 )

σ (a, b) =

(25)

∀a ∈ G with (a, 1) ∈ X ,

(26)

ρ (a)ρ (b)ρ (ab)−1 if (a, b) ∈ X ,

(27)

if (a, b) ∈ / X.

0

(iii) pmX (G ) is isomorphic to ( K ∗ )s , where s = s(G , X ) is the cardinality of the set of S 3 -orbits of X with 2 or 6 elements. (iv) For any domain Y ∈ C (G ) such that Y ⊇ X , the structural homomorphism (see Theorem 5.1 and the subsequent paragraph) ψ XY : pM Y (G ) → pM X (G ) is an epimorphism. In particular, pM X (G ) is an epimorphic image of pM G ×G (G ). Proof. (i) This is a direct consequence of Theorems 5.4 and 5.6. (ii) If σ ∈ N X ⊆ pmX (G ) then there exists ρ : G → K ∗ satisfying (27). Since (1, 1) ∈ X and σ (1, 1) = 1 K , it follows that ρ (1) = 1 K . Take any (a, b) ∈ X . Since X is T -invariant, (a−1 , a) ∈ X . Then 1 = σ (a−1 , a) = ρ (a−1 )ρ (a) and (26) is fulfilled. Conversely, take an arbitrary X in C (G ), consider a function ρ : G → K ∗ obeying to (25), (26), and define σ : G × G → K by (27). Then obviously σ (a, 1) = 1 for any (a, 1) ∈ X , and for arbitrary (a, b) ∈ X we have







 

 

σ (a, b)σ b−1 , a−1 = ρ (a)ρ (b)ρ (ab)−1 ρ b−1 ρ a−1 ρ b−1a−1 

σ (a, b) = ρ (a)ρ (b)ρ (ab)−1 = ρ b−1a

 −1



ρ (a)ρ b

 −1 −1

− 1

= 1,  = σ b −1 a −1 , a , 

and analogously σ (a, b) = σ (b, b−1 a−1 ). Hence σ ∈ pmX (G ), σ ∼ 1, so σ ∈ N X . (iii) We shall restrict ourselves to the S 3 -orbits of X of length 2 and 6. It will be convenient to view these orbits as follows. Consider the set

A = (G \ 1) × (G \ 1) \ , where  = {(x, x−1 ) | x ∈ G }. Take the action of S 3 in A given in the proof of Theorem 5.6. We have that X  = X ∩ A is an S 3 -set. Consider a set I with cardinality equal to s and pick {(ai , b i )}i ∈I a full set of representatives of the orbits of X  . Then we can write X  as a disjoint union of the form

X =



A (ai ,bi ) ,

i ∈I

where A (ai ,bi ) is the S 3 -orbit of (ai , b i ) ∈ X  . Observe that each of these orbits has 2 or 6 elements, and it follows from the proof of Theorem 5.6 that σ ∈ pmX (G ) is completely determined by its values

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on the representatives. Now take x = (xi )i ∈I and denote by σx (ai , bi ) = xi . Then the map



K∗

s

σx the factor set in pmX (G ) such that

 x → σx ∈ pmX (G )

is an isomorphism of multiplicative groups. (iv) We can extend σ ∈ pmX (G ) to Y setting σ (x, y ) = 1 for all (x, y ) ∈ Y \ X . Then this extended σ obviously satisfies (22)–(24) and by Theorem 5.6 it belongs to pmY (G ). 2 Let G be a group, X an element of C (G ) and ( K ∗ )s  x → σx ∈ pmX (G ) be the isomorphism as given in the proof of (iv) of Corollary 5.8. Set





LX = x ∈ K∗

s         σx ∼ 1 = x ∈ K ∗ s  σx ∈ N X

and













N ( X ) = cls(σx ) ∈ pM G ×G (G )  σx | X ∼ 1 = cls(σx ) ∈ pM G ×G (G )  σx | X ∈ N X . We summarize in the next theorem the most important information about the structure of the partial Schur multiplier obtained above. Theorem 5.9. Let G be a group, X ∈ C (G ) and s = s(G , X ) the cardinality of the set consisting of the S 3 -orbits of X with length 2 or 6. Then with the above notation there are isomorphisms





s pM G ×G (G )/ N ( X ) ∼ = K ∗ /L X . = pM X (G ) ∼ = pmX (G )/ N X ∼

2

We shall show in the next section by means of an example that the kernel N ( X ) may be trivial. Consequently, there may exist components of the partial Schur multiplier with X = G × G which are isomorphic to pM G ×G (G ). 6. An example: cyclic groups It is assumed in this section that G = C n . We shall describe the component pM G ×G (G ) of the partial Schur multiplier pM (G ) over K . Let σ be an element of pmG ×G (G ) and write σi j = σ (i , j ) (0  i , j  n − 1). Then the equalities (22) and (23) have the following form:

σi j =

1

σn− j,n−i

,

σi j = σn−i− j,i = σ j,n−i− j ,

(28) (29)

where the indexes are taken modulo n. Proposition 6.1. Let n  4. Then with the above notation σ is uniquely determined by its values σi j , where

 1i

n−1 3

and these values can be chosen arbitrarily in K ∗ .

 ,

i  j  n − 2i − 1,

(30)

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219

Proof. We shall show that every σkm can be expressed using (28) and (29) in terms of the values σi j , where i and j satisfy (30). Since σ0m = σk0 = 1, one can suppose that k, m  1. If k + m = n, we obtain from (29) that σkm = 1. If k + m > n then (n − m) + (n − k) < n, and since in view of (28) σkm is determined by σn−m,n−k , we may consider only those pairs (k, m) for which

0 < k < n,

0 < m < n,

0 < k + m < n.

On the set of these pairs an action (i , j ) → (n − i − j , i ) of the cyclic group of order 3 is defined. Its orbits are of the form

  (i , j ), (n − i − j , i ), ( j , n − i − j ) ,

(31)

and contain three elements each, except the case n ≡ 0 (mod 3), in which there exists an orbit consisting of a single element: {(n/3, n/3)}. In view of (29) it is clear that choosing a representative (i , j ) in each orbit we obtain a set of values {σi j } which uniquely determines the map σ . Choose as a representative of an orbit the pair (i , j ) with the least value of the first coordinate. If there are two such pairs then the orbit is of the form



 (i , i ), (n − 2i , i ), (i , n − 2i ) ,

in this case we choose (i , i ). If in (31) all first coordinates are equal then we have the particular case n ≡ 0 (mod 3) in which the orbit consists of a single element. The chosen representatives satisfy the inequalities

i  j,

i  n − i − j,

i.e.

i  j  n − 2i . But since

σii = σi,n−2i , we can suppose i  j  n − 2i − 1, and this implies i   n−3 1 . 2

Then for n  4 the number of the independent values of 1  n−  3

Pn =





1  n−  3

(n − 2i − 1 − i + 1) =

i =1

 i =1

(n − 3i ) =

σi j is computed as follows: n−1



3

n−



3 n−1 2

3

 −

3 2

 .

Set P 1 = P 2 = 0 and P 3 = 1. Notice that each of the orbits in (31) coincides with C 3 · (a, b) or C 3 · (b−1 , a−1 ) from the proof of Theorem 5.6. Consequently, P n = s(G , G × G ). Thus by Theorem 5.9 we obtain the next corollary. Corollary 6.2. There is an isomorphism



K∗

Pn

/ L G ×G  μ L G ×G → cls(σμ ) ∈ pM G ×G (G ),

where the values of σμ ∈ pmG ×G (G ) on the orbits in (31) are determined by μ. Thus in order to calculate pM G ×G (G ) it is sufficient to find L G ×G .

2

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1 Proposition 6.3. Let σ ∈ pmG ×G (G ). If σ ∼ 1 then σ is uniquely determined by its values σ1 j , 1  j   n− , 2 which can be chosen in K ∗ arbitrarily.

Proof. Since

σ ∼ 1 and σ ∈ pmG ×G (G ) there is a map ρ : G → K ∗ such that

ρi ρ j ρi + j

σi j =

(32)

for all i , j, and by (25) and (26) we have:

ρ0 = ρi ρn−i = 1 K , for all i. Set

π j = σ11 . . . σ1, j−1 =

(33)

ρ1j ρ j , j  2. Then

σi j =

πi + j , πi π j

with i , j  2. Note that (32) implies σi j = σ ji Thus all (keeping in mind that σ10 = σ1,n−1 = 1). By (29)

(34)

σi j are expressed in terms of σ1k , 1  k  n − 2

σ1k = σn−k−1,1 = σ1,n−k−1 . Therefore

σ is determined by the values

σ11 , σ12 , . . . , σ1, n−1  . 2

(35)

Notice that the elements in (35) are indexed by pairs belonging to distinct S 3 -orbits, as they satisfy (30). 1 Now take these values arbitrarily in K ∗ and write π j = σ11 . . . σ1, j −1 , j = 2, . . . ,  n+ . Suppose 2

1 first that n is even (the case of an odd n is fairly similar) and write n = 2m. Then  n+  = m. We 2 determine first ρ1 , . . . , ρm such that

πj =

ρ1j , ρj

(36)

j = 2, . . . , m, and the other ρi ’s will be defined using (33). Note that the equality ρi ρ−i = 1 gives us ρm ρm2 = 1, and we set ρm = 1. Then we need to have πm = ρ1m = ρ1m , and thus we define ρ1 = ω, where ω ∈ K ∗ is such that ωm = πm (ω exists because K is algebraically closed). Then the values of ρ j , j = 2, . . . , m − 1, one determines by (36). So we obtained a function ρ : G → K ∗ which satisfies (33). j The arbitrarily chosen values σ11 , σ12 , . . . , σ1,m−1 fit (32), in fact for 1  j  m − 1, ρ j = ω π j . Then

σ1 j =

π j+1 ωρ j ρ1 ρ j π j = ρ j+1 = ρ j+1 , j  2, and evidently

(25) and (26), and by Corollary 5.8

σ11 = π2 =

ρ1 2 ρ2 . Note that (33) means that

ρ verifies

ρ determines via (32) (or (27)) a factor set σ ∼ 1. 2 n −1

Proposition 6.3 means that L G ×G = ( K ∗ ) 2  , and thus, taking into account that the pairs (1, j ), 1 1  j   n− , belong to distinct S 3 -orbits, we come to the next: 2 Corollary 6.4. pM G ×G (G ) ∼ = ( K ∗ )T n , where T n = P n −  n−2 1 .

2

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221

Note that for 1  n  5, we have that T n = 0, and thus

pM C n ×C n (C n ) ∼ = C1. It follows by (iii) of Corollary 5.8 that all components in the partial Schur multiplier of C n are trivial. Hence in order to determine the partial Schur multipliers of these groups, we only need to know the cardinality of C (C n ).

• For G = C 1 , we have C (C 1 ) = {C 1 × C 1 }, then pM (C 1 ) ∼ = C1. In this case the only domain is G × G. • For G = C 2 , we have C (C 2 ) = { X 1 , X 2 }, where X 1 = {(1, 1)} and X 2 = G × G. The domains form the chain

G×G

X1 and,

pM (C 2 ) = Y 1 ∪ Y 2 ,

where Y 1 ∼ = C1 ∼ = Y 2.

• For G = C 3 = a | a3 = 1 we find that C (C 3 ) = { X 1 , X 2 , X 3 }, with





X 1 = (1, 1) ,









 

 



X 2 = (1, 1), (1, a), 1, a2 , (a, 1), a2 , 1 , a, a2 , a2 , a , X3 = G × G . In this case we also have a chain

G×G

X2

X1 and,

pM (C 3 ) = Y 1 ∪ Y 2 ∪ Y 3 ,

where Y 1 ∼ = Y2 ∼ = Y3 ∼ = C1.

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• For G = C 4 = a | a4 = 1 we have that C (C 4 ) = { X 1 , X 2 , X 3 , X 4 , X 5 }, and





X 1 = (1, 1) ,









 

 



X 2 = (1, 1), (1, a), 1, a3 , (a, 1), a3 , 1 , a, a3 , a3 , a ,



 



 

X 3 = (1, 1), 1, a2 , a2 , 1 , a2 , a2



,

X4 = X2 ∪ X3, X5 = G × G . The semilattice structure is

G×G

X4

X2

X3

X1 and,

pM (C 4 ) = Y 1 ∪ Y 2 ∪ Y 3 ∪ Y 4 ∪ Y 5 ,

where Y 1 ∼ = Y2 ∼ = Y3 ∼ = Y4 ∼ = Y5 ∼ = C1.

• For G = C 5 = a | a5 = 1, we have





X 1 = (1, 1) ,









 

 



X 2 = (1, 1), (1, a), 1, a4 , (a, 1), a4 , 1 , a, a4 , a4 , a ,





 

 

 

 

 

 

 

 

X 3 = (1, 1), 1, a2 , 1, a3 , a2 , 1 , a3 , 1 , a3 , a2 , a2 , a3



,

X4 = X2 ∪ X3,





 

X 5 = (a, a), a, a3 , a3 , a , a4 , a4 , a2 , a4 , a4 , a2 X6 =



∪ X4,            a, a2 , a2 , a , a3 , a4 , a4 , a3 , a3 , a3 , a2 , a2 ∪ X 4 ,



X7 = G × G . The semilattice structure is

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223

G×G

X6

X5

X4

X2

X3

X1 and the partial Schur multiplier is

pM (C 5 ) = Y 1 ∪ Y 2 ∪ Y 3 ∪ Y 4 ∪ Y 5 ∪ Y 6 ∪ Y 7 , with Y 1 ∼ = Y2 ∼ = Y3 ∼ = Y4 ∼ = Y5 ∼ = Y6 ∪ Y7 ∼ = C1. • For G = C 6 = a | a6 = 1. Note that by Corollary 6.4, pM C 6 ×C 6 (C 6 ) ∼ = K ∗ . We show that there exists another component isomorphic to this one. Consider the T -set:



 



 

 



 



 

 



X = (1, 1), (1, a), 1, a2 , 1, a3 , 1, a4 , 1, a5 , (a, 1), a2 , 1 , a3 , 1 , a4 , 1 , a5 , 1 ,



 



 

 

 

 

 





 

 

 

 

 



 

 

 



a, a5 , a2 , a4 , a3 , a3 , a4 , a2 , a5 , a , a, a2 , a3 , a , a2 , a3 , a4 , a5 , a3 , a4 , a5 , a3 , a, a3 , a2 , a , a3 , a2 , a3 , a5 , a4 , a3 , a5 , a4 ,



 

a2 , a2 , a4 , a4

 .

Hence

X  = X ∩ A = A (a,a2 ) ∪ A (a,a3 ) ∪ A (a2 ,a2 ) , where A and A (ai ,a j ) are as in the proof of Corollary 5.8. Then values in (a, a2 ), (a, a3 ) and (a2 , a2 ). Write





σ a, a2 = μ1 , and denote





σ ∈ pmX (C 6 ) is determined by its 



σ a, a3 = μ2 , and σ a2 , a2 = μ3

σ = σ(μ1 ,μ2 ,μ3 ) . There is an epimorphism 

ϕ : K∗

3

 (μ1 , μ2 , μ3 ) → cls(σ(μ1 ,μ2 ,μ3 ) ) ∈ pM X (C 6 ).

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We need to determine the kernel of ϕ . Obviously, it coincides with L X from Theorem 5.9. If (μ1 , μ2 , μ3 ) ∈ L X , there exists ρ : G → K ∗ , such that

σ(μ1 ,μ2 ,μ3 ) (x, y ) = ρ (x)ρ ( y )ρ (xy )−1 , for all (x, y ) ∈ X . Recall that 1 = ρ (x)ρ (x−1 ) when (x, 1) ∈ X by (26). Hence





μ1 = σ(μ1 ,μ2 ,μ3 ) a, a2 =

  ρ (a)ρ (a2 ) ρ (a)ρ (a3 ) = = σ(μ1 ,μ2 ,μ3 ) a, a3 = μ2 , 3 4 ρ (a ) ρ (a )

and consequently





ker ϕ = L X ⊆ (μ1 , μ2 , μ3 ) ∈ K ∗ Now take (μ1 , μ2 , μ3 ) ∈ ( K ∗ )3 with

 

ρ (1) = ρ a3 = 1 K , where

3    μ1 = μ2 .

μ1 = μ2 = μ. Define ρ : G → K ∗ by  

ρ a2 =

1K

ρ (a4 )

= ω,

ρ (a) =

1K

ρ (a5 )

=

μ , ω

ω3 = μ3 . Then it is readily verified that

σ(μ1 ,μ2 ,μ3 ) (x, y ) = ρ (x)ρ ( y )ρ (xy )−1 for all (x, y ) ∈ X . It follows that L X = {(μ1 , μ2 , μ3 ) ∈ ( K ∗ )3 | μ1 = μ2 }, and we conclude



pM X (C 6 ) ∼ = K∗

3

/ ker ϕ ∼ = K ∗.

Notice that pM X (C 6 ) ∼ = pM C 6 ×C 6 (C 6 ), which gives an example of a non-total component = K∗ ∼ isomorphic to the total one. Remark 1. Theorem 4 in [8] says that if G is a finite group then any element of the semilattice C (G ) of the domains is a union of so-called elementary domains (see the terminology of [8]) and, conversely, any union of elementary domains is again an element of C (G ). In particular, C (G ) is a lattice (which was already observed above). Remark 2. Notice that if G is a finite group then C (G ) is a chain exactly when |G |  3. Indeed, suppose that C (G ) is a chain. Then it follows by Remark 1 that C (G ) consists of elementary domains only, and, consequently, by [20, Theorem 5.1], the order of G is at most 3. Acknowledgments The authors thank the referee for many useful comments, in particular, for suggesting the present version of item (iv) of Corollary 5.8, which initially was stated only for Y = G × G.

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