Cofull Embeddings in Coset Monoids

James East^{∗}

School of Mathematics and Statistics, University of Sydney, Sydney, NSW 2006, Australia

July 13, 2005

Abstract

Easdown, East and FitzGerald (2004) gave a sufficient condition for a (factorizable inverse) monoid to embed as a cofull submonoid of the coset monoid of its group of units. We show that this condition is also necessary. This yields a simple description of the class of finite monoids which embed in the coset monoids of their group of units. We apply our results to give a short proof of the result of McAlister (1980) that the symmetric inverse semigroup on a finite set X does not embed in the coset monoid of the symmetric group on X. We also present examples which show that the word “cofull” may not be removed.

Keywords: Factorizable inverse monoid, coset monoid, symmetric inverse semigroup.

MSC: Primary 20M18; Secondary 20M30.

1 Factorizable Inverse Monoids and Coset Monoids

If M is a monoid then we denote by G_{M} the group of units of M. We say that a submonoid of M is cofull if it contains G_{M}. If N is another monoid and ψ : M → N an embedding, then we say that ψ is cofull if Mψ is a cofull submonoid of N.

If M is an inverse monoid then we denote by E_{M} the semilattice of idempotents of M. An inverse monoid M is factorizable if M = E_{M}G_{M}. The study of factorizable inverse monoids (henceforth FIMs) was initiated in [2]; for related studies see [4, 5, 6, 10] and references therein.

Let G be a group and denote by S(G) the join semilattice of all subgroups of G. The join H ∨ K of two subgroups H, K ∈ S(G) is defined to be ⟨H ∪ K⟩, the smallest subgroup of G containing HK. Now let

C(G) = {Hg | H ∈ S(G) , g ∈ G}

∗jamese@maths.usyd.edu.au

be the set of all cosets of all subgroups of G. An associative product ∗ is defined on C(G), for H, K ∈ S(G) and g, l ∈ G, by

(Hg) ∗ (Kl) = (H ∨ gKg^{−1})gl,

the smallest coset of G containing HgKl. The set C(G) is a FIM under ∗ with identity {1}, known as the coset monoid of G. The coset monoid was introduced in [8, 9]; see also [7]. Following is a collection of some elementary properties of coset monoids; these, along with other properties, are stated in [7].

Lemma 1 Let G be a group. Then

(i) E_{C(G)} = S(G);

(ii) G_{C(G)} = {{g} | g ∈ G} ≅ G; and

(iii) the subgroups of C(G) are precisely the sections of G (a section of G is a quotient of a subgroup of G). □

The following was proved in [7].

Theorem 2 (McAlister) Let X be a set. Then

(i) the symmetric inverse semigroup I_{X} embeds in the coset monoid C(G_{Y}) of the symmetric group G_{Y} on a set Y with |Y | = |X| + 1; and

(ii) I_{X} does not embed in C(G_{X}) if X is finite and nonempty. □

It follows by Theorem 2(i) and the Wagner-Preston Theorem that any inverse monoid (indeed any inverse semigroup) embeds in the coset monoid of some group. An interesting question that arises is “given an inverse monoid M, what is the minimum cardinality of a group G for which an embedding M → C(G) exists?” Now suppose that Ψ : M → C(G) is an embedding of an inverse monoid M in the coset monoid of a group G. By Lemma 1(iii), the image of G_{M} under Ψ is a section of G, showing that the cardinality of G is bounded below by the cardinality of G_{M}. Thus, another natural question arises: “Which inverse monoids M embed in C(G_{M})?” The goal of this article is to give necessary and sufficient conditions for an inverse monoid M to embed as a cofull submonoid of C(G_{M}).

Let M be an inverse monoid and write E = E_{M} and G = G_{M}. For e ∈ E let

G_{e} = {g ∈ G | eg = e}.

It is easy to check that each G_{e} is a subgroup of G, and that G_{e} ∨ G_{f} ⊆ G_{ef} for each e, f ∈ E. Define a map

ψ_{M} : E → S(G) : e ↦ G_{e} for each e ∈ E.

Let 𝒞 denote the class of factorizable inverse monoids M for which ψ_{M} is a semilattice embedding. The following was proved in [5].

Theorem 3 (Easdown, East, FitzGerald) A monoid M embeds as a cofull submonoid of C(G_{M}) if M ∈ 𝒞. □

This theorem was proved by showing that if M ∈ 𝒞, then the map

M → C(G) : eg ↦ G_{e}g for each e ∈ E and g ∈ G

is a cofull embedding. Our main goal is to show that the condition M ∈ 𝒞 is also necessary for a monoid to embed as a cofull submonoid of C(G_{M}). In addition we show that the word “cofull” may be removed within the class of finite (but not infinite) inverse monoids.

2 Cofull Embeddings

Our goal in this section is to show that a monoid M embeds as a cofull submonoid of C(G_{M}) if and only if M ∈ 𝒞.

Lemma 4 Any cofull submonoid of a factorizable inverse monoid is a factorizable inverse monoid.

Proof Suppose that N is a cofull submonoid of a FIM M, and choose m ∈ N. Then m = eg for some e ∈ E_{M} and g ∈ G_{M}. Since N is cofull, we have g^{−1} ∈ N and so m^{−1} = g^{−1}e = g^{−1}(eg)g^{−1} = g^{−1}mg^{−1} ∈ N, showing that N is inverse. We also have e = mg^{−1} ∈ N so that N is factorizable. □

Theorem 5 A monoid M embeds as a cofull submonoid of C(G_{M}) if and only if M ∈ 𝒞.

Proof The “if” part of the theorem is true by Theorem 3. To show the converse, it suffices to show that N ∈ 𝒞 for every cofull submonoid N of C(G_{M}). Write G = G_{M}, and \overline{G} = G_{C(G)} = \{\{g\} | g ∈ G\}. Now N is a FIM by Lemma 4, so it remains only to show that

ψ_{N} : E_{N} → S(G_{N})

is an embedding. Now E_{N} = S(G) ∩ N, and G_{N} = \overline{G} since N is cofull. Further, if H ∈ E_{N}, then Hψ_{N} = \overline{H} = \{\{h\} | h ∈ H\}. It follows that ψ_{N} is an embedding since \overline{H} ∨ \overline{K} = \overline{H ∨ K} for any subgroups H, K ∈ S(G). □

As a corollary, we have the following.

Theorem 6 A finite monoid M embeds in C(G_{M}) if and only if M ∈ 𝒞.

Proof Write G = G_{M}. Any embedding Ψ : M → C(G) gives rise to an embedding \overline{Ψ} : G → G_{C(G)} ≅ G since the image of G under Ψ, being finite, cannot be contained in a proper section of G. Since G is finite, \overline{Ψ} is an isomorphism whence Ψ is cofull, and we are done by Theorem 5. □

We now apply Theorem 6 to provide an alternative proof of Theorem 2(ii).

Corollary 7 (McAlister) Let X be a finite nonempty set. Then I_{X} does not embed in C(G_{X}).

Proof Put G = G_{X} = G_{I_{X}}. For A ⊆ X denote by id_{A} the identity map on A so that E_{I_{X}} = {id_{A} | A ⊆ X}. Then for each A ⊆ X,

G_{id_{A}} = {π ∈ G | aπ = a (∀a ∈ A)}

is the pointwise stabilizer of A, which we will denote by Stab(A). Now if x ∈ X, then Stab(X) = Stab(X \ {x}) = {id_{X}} so that ψ_{I_{X}} is not injective. We are now done by Theorem 6. □

We remark that if X is any set with |X| ≥ 2, then the map ψ_{I_{X}} is not a semilattice homomorphism since if x, y ∈ X with x ≠ y then, writing A = X \ {x} and B = X \ {y}, we have

G_{id_{A}} ∨ G_{id_{B}} = Stab(A) ∨ Stab(B) = Stab(A) = Stab(B) = {id_{X}}

while the transposition which interchanges x and y is in

Stab(X \ {x, y}) = Stab(A ∩ B) = G_{id_{A∩B}} = G_{id_{A}∘id_{B}}.

3 Other Embeddings

In this final section we consider examples of FIMs M which embed in C(G_{M}) but do not belong to 𝒞. These FIMs are necessarily infinite, and the embeddings are not cofull.

Example 8 Let X be an infinite set. Then the symmetric inverse semigroup I_{X} ∉ 𝒞 by the comments after the proof of Corollary 7. On the other hand, I_{X} does embed in the coset monoid of the symmetric group G_{X} = G_{I_{X}} by Theorem 2(i).

Now I_{X} (indeed E_{I_{X}}) is uncountable for any infinite set X. Our second example is a countable FIM M for which |E_{M}| = 3 and rank(G_{M}) = 1. Here for a group G we have written rank(G) for the minimal cardinality of a set which generates G (as a group).

Example 9 Let G = ⟨x⟩ be the infinite cyclic group generated by x, and let G^{y} be the semigroup obtained by adjoining a zero y to G. Let M = (G^{y})^{z} be the semigroup obtained by adjoing a new zero z to G^{y}. It is easy to check that M is a FIM with G_{M} = G and E_{M} = {1, y, z}. We also have G_{y} = G_{z} = G so that M ∉ 𝒞. Now define

Ψ : M → C(G) : \{ \begin{matrix}x & ↦ & \{x^{2}\} \\ y & ↦ & ⟨x^{2}⟩ \\ z & ↦ & G. \end{matrix}

Then one may easily check that Ψ is an embedding.

Our final example is also a countable FIM M although in this case we have |E_{M}| = 2 and rank(G_{M}) = 2.

Example 10 Let G = ⟨x, y⟩ be the free group freely generated by {x, y}. Define a homomorphism

φ : G → G : x ↦ x^{2}, y ↦ y^{2},

and put K = ⟨x^{2}, y^{2}⟩, the image of φ. Let B = G/N where N is the normal closure in G of {xyxy^{−1}x^{−1}y^{−1}}. So B has presentation ⟨x, y | xyx = yxy⟩ and is isomorphic to the braid group on 3 strings; see [1]. It is well known that Nx^{2} and Ny^{2} generate a free subgroup of B of rank 2; see for example [3]. It follows that N ∩ K = {1}.

Now let E = {0, 1} which we consider as a semilattice under multiplication, and put M = E × G. So M is a FIM with E_{M} = (E, 1) ≅ E and G_{M} = (1, G) ≅ G, and M ∉ 𝒞 since G_{(1,1)} = G_{(0,1)} = {(1, 1)}. Define

Ψ : M → C(G) : \{ \begin{matrix}(1, g) & ↦ & \{gφ\} & for each g ∈ G \\ (0, g) & ↦ & N(gφ) & for each g ∈ G. \end{matrix}

Then Ψ is a homomorphism since N is normal in G and φ is a homomorphism. To show that Ψ is injective, suppose that e_{1}, e_{2} ∈ E and g_{1}, g_{2} ∈ G such that

(e_{1}, g_{1})Ψ = (e_{2}, g_{2})Ψ.

Then we clearly must have e_{1} = e_{2}. Suppose first that e_{1} = e_{2} = 1. Then

{g_{1}φ} = (e_{1}, g_{1})Ψ = (e_{2}, g_{2})Ψ = {g_{2}φ}.

It then follows that g_{1} = g_{2} since φ is injective and so (e_{1}, g_{1}) = (e_{2}, g_{2}). Finally, suppose that e_{1} = e_{2} = 0. Then

N(g_{1}φ) = (e_{1}, g_{1})Ψ = (e_{2}, g_{2})Ψ = N(g_{2}φ)

from which it follows that (g_{1}g_{2}^{−1})φ = (g_{1}φ)(g_{2}φ)^{−1} ∈ N. But then (g_{1}g_{2}^{−1})φ = 1 since N ∩ K = {1}, and so g_{1}g_{2}^{−1} = 1 since φ is injective, whence g_{1} = g_{2} and (e_{1}, g_{1}) = (e_{2}, g_{2}). This completes the proof that Ψ is injective.

While the monoids M considered in Examples 9 and 10 had different values of |E_{M}| and rank(G_{M}), they shared the property that |E_{M}| + rank(G_{M}) = 4. It turns out that 4 is the minimum value of |E_{M}| + rank(G_{M}) for any FIM M ∉ 𝒞 which embeds in C(G_{M}).

Proposition 11 Suppose that M ∉ 𝒞 is a FIM for which there exists an embedding Ψ : M → C(G_{M}). Then |E_{M}| + rank(G_{M}) ≥ 4.

Proof Write E = E_{M} and G = G_{M} and suppose that |E| + rank(G) ≤ 3. Since M ∉ 𝒞, we have |E| ≥ 2, and since Ψ is injective, we have rank(G) ≥ 1. It then follows that |E| = 2 and rank(G) = 1. Write E = {1, e} where 1 is the identity of M. Since M ∉ 𝒞 we must have G_{e} = {1}. Since M is infinite, G must be an infinite cyclic group generated by x say, and since Ψ is an embedding, we have xΨ = {x^{i}} and eΨ = ⟨x^{j}⟩ for some i, j ∈ ℤ \ {0}. But then ex^{j} ≠ e since G_{e} = {1}, yet

(ex^{j})Ψ = (eΨ) ∗ (x^{j}Ψ) = ⟨x^{j}⟩ ∗ {x^{ij}} = ⟨x^{j}⟩x^{ij} = ⟨x^{j}⟩ = eΨ

contradicting the injectivity of Ψ. This completes the proof. □

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