endomorphism monoids of free G-acts

and endomorphism monoids of free G-acts University of York Based on joint work with Igor Dolinka and Victoria Gould

Let E be a biordered set (equivalently, a set of idempotents E of a semigroup S). The free idempotent generated semigroup IG(E) is a free object in the category of semigroups that are generated by E, defined by IG(E) = E : ē f = ef, e, f E, {e, f } {ef, fe}. where E = {ē : e E}. Note It is more usual to identify elements of E with those of E, but it helps the clarity of our later arguments to make this distinction.

Facts 1 IG(E) = E. 2 The natural map φ : IG(E) S, given by ēφ = e, is a morphism onto S = E(S). 3 The restriction of φ to the set of idempotents of IG(E) is a bijection. 4 The morphism φ induces a bijection between the set of all R-classes (resp. L-classes) in the D-class of ē in IG(E) and the corresponding set in S = E(S). 5 The morphism φ is an onto morphism from H e to H e.

Maximal subgroups of IG(E) Work of Pastijn (1977, 1980), Nambooripad and Pastijn (1980), McElwee (2002) led to a conjecture that all these groups must be free groups. Brittenham, Margolis and Meakin (2009) Z Z can be a maximal subgroup of IG(E), for some E. Gray and Ruskuc (2012) Any group occurs as a maximal subgroup of some IG(E), a general presentation and a special choice of E are needed. Gould and Yang (2012) Any group occurs as a maximal subgroup of a natural IG(E), a simple approach suffices. Dolinka and Ruskuc (2013) Any group occurs as IG(E) for some band.

Maximal subgroups of IG(E) Given a special biordered set E, which kind of groups can be the maximal subgroups of IG(E)? Let S be a semigroup with E = E(S). Let e E. Our aim is to find the relationship between the maximal subgroup H e of IG(E) with identity e and the maximal subgroup H e of S with identity e. There is an onto morphism from H e to H e. Is H e = He, for some E and some e E?

Full transformation monoids and full linear monoids T n (PT n ) - full (partial) transformation monoid, E - its biordered set. Gray and Ruskuc (2012); Dolinka (2013) rank e = r < n 1, H e = He = Sr. Brittenham, Margolis and Meakin (2010) M n (D) - full linear monoid, E - its biordered set. rank e = 1 and n 3, H e = He = D. Dolinka and Gray (2012) rank e = r < n/3 and n 4, H e = He = GLr (D). Note rank e = n 1, H e is free; rank e = n, H e is trivial.

Independence algebras Sets and vector spaces over division rings are examples of independence algebras. Fountain and Lewin (1992) Let A be an independence algebra of rank n, where n N is finite. Let End A be the endomorpshim monoid of A. Then Gould (1995) S(End A) = {α End A : rank α < n} = E \ {I }. For any α, β End A, we have the following: (i) im α = im β if and only if α L β; (ii) ker α = ker β if and only if α R β; (iii) rank α = rank β if and only if α D β if and only if α J β.

Independence algebras The results on the biordered set of idempotents of T n and M n (D) suggest that it would be worth looking into the maximal subgroups of IG(E), where E = E(End A). The diverse method needed in the biordered sets of T n and M n (D) indicate that it would be very hard to find a unified approach to End A. It was pointed out by Gould that free G-acts provide us with another kind of independence algebras.

Free G-acts Let G be a group, n N, n 3. Let F n (G) be a rank n free left G-act. Recall that, as a set, F n (G) = {gx i : g G, i [1, n]}; identify x i with 1x i, where 1 is the identity of G; gx i = hx j if and only if g = h and i = j; the action of G is given by g(hx i ) = (gh)x i.

Endomorphism monoids of free left G-acts Let End F n (G) be the endomorphism monoid of F n (G) with E = E(End F n (G)). The rank of an element of End F n (G) is the minimal number of (free) generators in its image. An element α End F n (G) depends only on its action on the free generators {x i : i [1, n]}. For convenience we denote α by ( ) x1 x α = 2... x n w1 αx 1α w2 αx 2α... wn α, x nα where α T n, w α 1,, w α n G. Note End F n (G) = G S n and S(End F n (G)) = E \ {I }.

Endomorphism monoids of free left G-acts For any rank r idempotent ε E, where 1 r n, we have H ε = G Sr. How about the maximal subgroup H ε of IG(E)?

A presentation for H ε To specialise Gray and Ruškuc s presentation of maximal subgroups of IG(E) to our particular circumstance. Step 1 To obtain an explicit description of a Rees matrix semigroup isomorphic to the semigroup D 0 r = D r {0}, where D r = {α End F n (G) rank α = r}. Let I and Λ denote the set of R-classes and the set of L-classes of D r, respectively. Here we may take I as the set of kernels of elements in D r, and Λ = {(u 1, u 2,..., u r ) : 1 u 1 < u 2 <... < u r n} [1, n] r. Let H iλ = R i L λ.

A presentation for H ε Assume 1 I Λ with 1 = (x 1, x i ) : r + 1 i n I, 1 = (1,, r) Λ. So H = H 11 is a group with identity ε = ε 11. A typical element of H looks like ( ) x1 x α = 2... x r x r+1... x n w1 αx 1α w2 αx 2α... wr α x rα w1 αx 1α... w1 αx 1α where α T n, w α 1,, w α r G. Abbreviate α as ( ) x1 x α = 2... x r w1 αx 1α w2 αx 2α... wr α. x rα In particular, ( ) x1 x ε = ε 11 = 2... x r. x 1 x 2... x r

A presentation for H ε For any α D r, ker α induces a partition {B1 α,, Br α } on [1, n] with a set of minimum elements l1 α,, lr α such that l1 α < < lr α. Put Θ = {α D r : x l α j α = x j, j [1, r]}. Then it is a transversal of the H-classes of L 1. For each i I, define r i as the unique element in Θ H i1. We say that r i lies in district (l r i 1, l r i 2,, l r i r ) (of course, 1 = l r i 1 ). For each λ = (u 1, u 2,..., u r ) Λ, define ( x1 x q λ = q (u1,,u r ) = 2 x r x r+1 x n x u1 x u2 x ur x u1 x u1 ).

A presentation for H ε We have that D 0 r = D r {0} is completely 0-simple, and hence D 0 r = M 0 (H; I, Λ; P), where P = (p λi ) and p λi = (q λ r i ) if rank q λ r i = r and is 0 else. Note [ ] εiλ ε iµ ε kλ ε kµ is a singular square p 1 λi p λk = p 1 µi p µk.

A presentation for H ε Step 2 Define a schreier system of words {h λ : λ Λ} inductively, using the restriction of the lexicographic order on [1, n] r to Λ. Put h (1,2,,r) = 1; For any (u 1, u 2,..., u r ) > (1, 2,, r), take u 0 = 0 and i the largest such that u i u i 1 > 1. Then Define (u 1,..., u i 1, u i 1, u i+1,..., u r ) < (u 1, u 2,..., u r ). h (u1,,u r ) = h (u1,,u i 1,u i 1,u i+1,,u r )α (u1,,u r ), where ( ) x α (u1,,u r ) = 1 x u1 x u1 +1 x u2 x ur 1 +1 x ur x ur +1 x n x u1 x u1 x u2 x u2 x ur x ur x ur x ur

A presentation for H ε Facts 1 εh (u1,...,u r ) = q (u1,...,u r ). 2 h (u1,,u r ) induces a bijection from L (1,,r) onto L (u1,,u r ) in both End F n (G) and IG(E). Hence {h λ : λ Λ} forms the required schreier system for the presentation for H = H ε. Step 3 Define a function Note p ω(i),i = ε. ω : I Λ, i ω(i) = (l r i 1, l r i 2,..., l r i r ). Put K = {(i, λ) I Λ : H iλ is a group}.

A presentation for H ε Proposition Let E = E(End F n (G)). Then the maximal subgroup H of ε in IG(E) is defined by the presentation with generators: and defining relations Σ: (R1) f i,λ = f i,µ (h λ ε iµ = h µ ); (R2) f i,ω(i) = 1 (i I ); (R3) f 1 i,λ f i,µ = f 1 k,λ f k,µ p 1 µi p µk ). P = F : Σ F = {f i,λ : (i, λ) K} ( [ ] εiλ ε iµ ε kλ ε kµ is singular i.e. p 1 λi p λk =

A presentation for H ε Note If rank ε = n 1, then H ε is free, as no non-trivial singular squares exist; if rank ε = n, then H ε is trivial. How about H ε, where 1 rank ε n 2?

A presentation for H ε Given a pair (i, λ) K, we have a generator f i,λ and an element 0 p λi P. To find the relationship between these generators f i,λ and non-zero elements p λi P.

Identity generators Lemma If (i, λ) K and p λi = ε, then f i,λ = 1 H. Idea. The proof follows by induction on λ Λ, ordered lexicographically. Here we make use of our particular choice of schreier system and function ω.

Generators corresponding to the same columns or rows Lemma If p λi = p µi, then f i,λ = f i,µ. The proof is straightforward. Lemma If p λi = p λj, then f i,λ = f j,λ. Idea. For any i, j I, suppose that r i and r j lie in districts (1, k 2,, k r ) and (1, l 2,, l r ), respectively. We call u [1, n] a mutually bad element of r i with respect to r j, if there exist m, s [1, r] such that u = k m = l s, but m s; all other elements are said to be mutually good with respect to r i and r j. We proceed by induction on the number of bad elements.

Connectivity of elements in the sandwich matrix Definition Let i, j I and λ, µ Λ such that p λi = p µj. We say that (i, λ), (j, µ) are connected if there exist i = i 0, i 1,..., i m = j I and λ = λ 0, λ 1,..., λ m = µ Λ such that for 0 k < m we have p λk i k = p λk,i k+1 = p λk+1 i k+1.

Connectivity of elements in the sandwich matrix The following picture illustrates that (i, λ) = (i 0, λ 0 ) is connected to (j, µ) = (i m, λ m ): p λ0 i 0 p λi1 p λ1 i 1 p λ1 i 2 p λm 1 i m 1 p λm 1 i m p λmi m Lemma Let i, j I and λ, µ Λ be such that p λi = p µj where (i, λ), (j, µ) are connected. Then f i,λ = f j,µ.

The result for n 2r + 1 Lemma Let n 2r + 1. Let λ = (u 1,, u r ) Λ, and i I with p λi H. Then (i, λ) is connected to (j, µ) for some j I and µ = (n r + 1,, n). Consequently, if p λi = p νk for any i, k I and λ, ν Λ, then f i,λ = f k,ν. We may define f φ = f i,λ, if p λi = φ H. Lemma Let r n/3. Then for any φ, θ H, f φθ = f θ f φ and f φ 1 = f 1 φ Note Every element of H appears in P. Theorem Let r n/3. Then H = H, f φ φ 1.

The result for r n 2 For larger r this strategy will fail... :-( Two main problems: for r n/2, not every element of H lies in P; we lose connectivity of elements in P, even if r = n/2. However, for r n 2 all elements with simple form ( ) x φ = 1 x 2 x k 1 x k x k+1 x k+m 1 x k+m x k+m+1... x r, x 1 x 2 x k 1 x k+1 x k+2 x k+m ax k x k+m+1... x r where k 1, m 0, a G, lie in P.

The result for r n 2 Lemma Let ε φ = p λi where λ = (u 1,, u r ) and i I. Then (i, λ) is connected to (j, µ) where µ = (1,, k 1, k + 1,, r + 1) and j I. Lemma Let p λi = p νk have simple form. Then f i,λ = f k,ν. Our aim here is to prove that for any α H, if i, j I and λ, µ Λ with p λi = p µj = α H, then f i,λ = f j,µ. This property of α is called consistency. Note All elements with simple form are consistent.

The result for r n 2 How to split an arbitrary element α in H into a product of elements with simple form? Moreover, how this splitting match the products of generators f i,λ in H.

The result for r n 2 Definition Let α H. We say that α has rising point r + 1 if x m α = ax r for some m [1, r] and a 1 G ; otherwise, the rising point is k r if there exists a sequence with 1 i < j 1 < j 2 < < j r k r x i α = x k, x j1 α = x k+1, x j2 α = x k+2,, x jr k α = x r and such that if l [1, r] with x l α = ax k 1, then if l < i we must have a 1 G.

The result for r n 2 Fact The only element with rising point 1 is the identity of H, and elements with rising point 2 have either of the following two forms: ( ) x1 x (i) α = 2 x r, where a 1 G ; ax 1 x 2 x r ( x1 x (ii) α = 2 x k 1 x k x k+1 x r x 2 x 3 x k ax 1 x k+1 x r k 2. ), where Note Both of the above two forms are the so called simple forms; however, elements with simple form can certainly have rising point greater than 2, indeed, it can be r + 1. Lemma Let α H have rising point 1 or 2. Then α is consistent.

The result for r n 2 Lemma Every α P is consistent. Further, if α = p λj then where p λt,i t f j,λ = f i1,λ 1 f ik,λ k, is an element with simple form, t [1, k]. Idea. We proceed by induction on rising points. For any α H with rising point k 3, we have α = βγ for some β H with rising point no more than k 1 and some γ H with simple form. Further, this splitting matches the products of corresponding generators in H. We may denote all generators f i,λ with p λi = α by f α, where (i, λ) K.

The result for r n 2 Our eventual aim is to show H = H = G S r. Definition We say that for φ, ϕ, ψ, σ P the quadruple (φ, ϕ, ψ, σ) is singular if φ 1 ψ = ϕ 1 σ and we can find i, j I, λ, µ Λ with φ = p λi, ϕ = p µi, ψ = p λj and σ = p µj.

The result for r n 2 Proposition Let H be the group given by the presentation Q = S : Γ with generators: S = {f φ : φ P} and with the defining relations Γ : (P1) f 1 φ f ϕ = f 1 ψ f σ where (φ, ϕ, ψ, σ) is singular; (P2) f ɛ = 1. Then H is isomorphic to H.

The result for r n 2 Put ( x1 x ι a,i = i 1 x i x i+1 x r x 1 x i 1 ax i x i+1 x r for 1 k r 1. Put ( ) x (k k +1 k +m) = 1 x k 1 x k x k+m 1 x k+m x k+m+1 x r x 1 x k 1 x k+1 x k+m x k x k+m+1 x r ) ; and we denote (k k + 1) by τ k.

The result for r n 2 The group H = G S r has a presentation U = Y : Υ, with generators Y = {τ i, ι a,j : 1 i r 1, 1 j r, a G} and defining relations Υ: (W 1) τ i τ i = 1, 1 i r 1; (W 2) τ i τ j = τ j τ i, j ± 1 i j; (W 3) τ i τ i+1 τ i = τ i+1 τ i τ i+1, 1 i r 2; (W 4) ι a,i ι b,j = ι b,j ι a,i, a, b G and 1 i j r; (W 5) ι a,i ι b,i = ι ab,i, 1 i r and a, b G; (W 6) ι a,i τ j = τ j ι a,i, 1 i j, j + 1 r; (W 7) ι a,i τ i = τ i ι a,i+1, 1 i r 1 and a G.

The result for r n 2 Recall that and further decomposition gives H = f φ : φ P, H = f τi, f ιa,j : 1 i r 1, 1 j r, a G.

The result for r n 2 Find a series of relations (T 1) (T 6) satisfied by these generators: (T 1) f τi f τi = 1, 1 i r 1. (T 2) f τi f τj = f τj f τi, j ± 1 i j. (T 3) f τi f τi+1 f τi = f τi+1 f τi f τi+1, 1 i r 2. (T 4) f ιa,i f ιb,j = f ιb,j f ιa,i, a, b G and 1 i j r. (T 5) f ιb,i f ιa,i = f ιab,i, 1 i r and a, b G. (T 6) f ιa,i f τj = f τj f ιa,i, 1 i j, j + 1 r. (T 7) f ιa,i f τi = f τi f ιa,i+1, 1 i r 1 and a G. Note A twist between (W5) and (T5). Lemma The group H with a presentation Q = S : Γ is isomorphic to the presentation U = Y : Υ of H, so that H = H.

The result for r n 2 Theorem Let End F n (G) be the endomorphism monoid of a free G-act F n (G) on n generators, where n N and n 3, let E be the biordered set of idempotents of End F n (G), and let IG(E) be the free idempotent generated semigroup over E. For any idempotent ε E with rank r, where 1 r n 2, the maximal subgroup H of IG(E) containing ε is isomorphic to the maximal subgroup H of End F n (G) containing ε and hence to G S r. Note If r = n, then H is trivial; if r = n 1, then H is free. If r = 1, then H = G and so that: Corollary Every group can be a maximal subgroup of a naturally occurring IG(E).

The result for r n 2 If G is trivial, then End F n (G) is essentially T n, so we deduce the following result: Corollary Let n N with n 3 and let IG(E) be the free idempotent generated semigroup over the biordered set E of idempotents of T n. For any idempotent ε E with rank r, where 1 r n 2, the maximal subgroup H of IG(E) containing ε is isomorphic to the maximal subgroup H of T n containing ε, and hence to S r.