Automorphism groups of finite groupoids

1David Hobby1Donald Silberger2Sylvia Silberger

Abstract

Given a finite set M of size n and a subgroup G of Sym(M), Gis pertinent iff it is the automorphism group of some groupoid 〈M ; ?〉.We examine when subgroups of Sym(M) are and are not pertinent.For instance, An, the alternating group on M , is not pertinent forn > 4. We close by indicating a natural extension of our ideas, whichrelates to a question of M. Gould.

1 Introduction

We begin the determination of the subgroups G of a finite symmetric groupSym(M) for which there is a groupoid on the universe M having G as itsautomorphism group. Call such G automorphically pertinent on M .

For each element n ∈ N := {1, 2, 3, . . .}, the symbol n denotes also theset n := {0, 1, . . . , n− 1}. We will often restrict our attention to pertinentgroups on the sets n for n ∈ N; nothing will be lost by doing so.

The expression G(n) denotes the family of all nn2

distinct groupoidson the universe n. Every groupoid V with |V | = n is isomorphic to atleast one element in G(n).

Our original interest in pertinent groups was combinatorially motivated:Into how many distinct isomorphism classes is G(n) partitioned? Thatis to say, our motivating interest lay in the number β(n) := | G(n)/ ' |.We eventually found that M. Harrison [11] had derived an exact formulafor β(n). As shown in Theorem 1.1 below, there is a relationship betweenβ(n) and the index of the automorphism groups of elements in G(n) assubgroups of Sym(n). Thus our original interest in β(n) led us to thestudy of automorphically pertinent groups.

There are many results of the sort, “Every finite group is the automor-phism group of a finite X”, where X is a structure in some given class.E.g., R. Frucht [7] showed this for the class of simple nondirected graphs,and G. Birkhoff [2] showed it for groups and for distributive lattices. Butsuch results often involve larger structures than necessary. Our aim here isto identify the particular subgroups of the symmetric group Sym(n) on

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the set n that are automorphism groups of elements in G(n). A focusupon groupoid automorphisms recommends itself since the automorphismsof groups, semigroups, and lattices are perforce groupoid automorphisms.

For example, consider the alternating group A5 on five objects. A5

is isomorphic to many permutation groups. Among these is the subgroupAlt(5) of Sym(5). Another is a permutation group obtained via Cayley’sTheorem as a subgroup H of Sym(60). We prove that Alt(5) fails tobe the automorphism group of any element in G(5), although M. Gould [8]has proved that H is the automorphism group of some groupoid on the set60.

We now introduce a terminological scheme for G(n) .

When j ∈ nn2then nj := 〈n; ?j〉 is the groupoid, on the universe n,

whose binary operation ?j : n × n → n is specified by its “multiplicationtable” – i.e., by its ?j table – which is constructed in the following fashion:

Let hn(j) denote the base n Hindu-Arabic numeral for j, where hn(j)may be prefixed by zeros to ensure that the word, hn(j) in the alphabet n,is exactly n2 digits long. Dissect hn(j) into n segments, each of lengthn. The i-th such segment is the i-th row of the binary operation tabledefining the groupoid nj.

Observe that every isomorphism from nj onto an ni is a permutationon the set n. Moreover, for every permutation f ∈ Sym(n) there is aunique jf ∈ nn

2for which f is an isomorphism from nj onto njf , where

the binary operation ?jf is given by

x ?jf y := f(f−1(x) ?j f−1(y)) for every 〈x, y〉 ∈ n× n.

So no isomorphism class [nj] := {ni : ni ' nj} ⊆ G(n) contains more than

n! elements. Since G(n)/ ' := {[nj] : j ∈ nn2} is a partition of G(n),

since n! does not divide nn2

if n ≥ 3, and finally since |[n0]| = n, weinfer for every integer n ≥ 3 that

nn2

n!< β(n) < nn

2

.

When j ∈ nn2we define α(nj) := |Aut(nj)|, where Aut(nj) is the

group of all automorphisms of nj. Since Aut(nj) is a subgroup of Sym(n),by Lagrange the integer α(nj) divides n!

Although Harrison gives the value of β(n), we do not know, for k afactor of n! , the number of distinct [nj] with |[nj]| = k. But Theorem 1.1,below, indicates a relationship between β(n) and the pertinent subgroupsof Sym(n).

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For j ∈ nn2let m(j) := min{i : ni ' nj}. Each class [nj] is uniquely

represented by an nm(j). If ni ' nm(j) then Aut(ni) ' Aut(nm(j)), andso α(ni) = α(nm(j)). Arrange the β(n) distinct integers m(j) into anascending sequence: t(0) < t(1) < · · · < t(β(n)− 1) .

Theorem 1.1.β(n)−1∑i=0

1

α(nt(i))=

nn2

n!.

Proof. Our claim is equivalent to the assertion

β(n)−1∑i=0

n!

α(nt(i))= nn

2

.

Every summand on the left side of this equality is an integer. The theoremfollows when we show that in fact

n!

α(nt(i))= |[nt(i)]|

for each i ∈ β(n). So choose an arbitrary such i. Let np ' nt(i) ' nq.There is an isomorphism f from np onto nq. For every pair of distinctautomorphisms {g, h} ⊆ Aut(np), clearly f ◦ g and f ◦ h are distinctisomorphisms from np onto nq. Conversely, if d is an isomorphism fromnp onto nq then f−1 ◦ d ∈ Aut(np), and d = f ◦ f−1 ◦ d. Thus, u is anisomorphism from np onto nq if and only if u ∈ {f ◦ g : g ∈ Aut(np)}.So the number of such isomorphisms is α(nt(i)) := |Aut(np)|. Thus theinteger n!/α(nt(i)) equals the number of distinct groupoids nj ' nt(i) .

The numbers β(n) := | G(n)/ ' | grow essentially as fast as nn2

does.

Our focus turns to the families {Aut(nj) : j ∈ nn2}.

2 Automorphically Pertinent Groups

Of special note are the pairs of groupoids S(n) := 〈n; �n〉 and S ′(n) :=〈n; �′n〉, whose binary operations are defined by x�n y := x and x�′n y := y.Observe that S(n) and S ′(n) are anti-isormorphic to each other, and thatthey are minimally commutative idempotent semigroups.

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If f ∈ Sym(n) then f(x �n y) = f(x) = f(x) �n f(y); likewise,f(x �′n y) = f(x) �′n f(y). Therefore Sym(n) = Aut(S(n)) = Aut(S ′(n)).Thus we obtain

Lemma 2.1. The group Sym(n) is pertinent for every n ∈ N.

We resort to a few relevantly simplified ideas from universal algebra andmodel theory. The texts [3] and [4] provide a more detailed exposition.

A structure is a set X together with sets of finitary operations andrelations on X. The type of such structures will be written by giving theoperations first, followed by the relations. When the operation and relationsymbols are unimportant, we will merely give their arities. In this case, thetype of groupoids is (2; ), and the type of groupoids with two unary relationsis (2; 1, 1).

The language of a type τ is the set of formulas Φ that can be madewith the symbols in τ . Henceforth we avoid distinguishing between formu-las involving relation symbols and their interpretations as statements aboutrelations.

Let C be a class of structures of type τ . When S and T are subsetsof τ , we say that S and T are interdefinable in C iff every operationand relation in T is expressible by a first order formula Φ in the languageof S, and likewise each operation and relation in S is expressible by aformula in the language of T .

For example, let C be the class of all groups, where the type τ consistsof a binary operation •, a unary operation ′ and a constant e. ThenS := {•, ′ , e} and T := {•} are interdefinable. Clearly the one element inT is expressible in terms of S. So what we need to see is how the elementsin S \ T are expressible in terms of • . But this is standard: In any groupwe have that y = e iff ∀z (y • z = z) and that x′ = y iff x • y = e.

When R is a k-ary relation and 〈x0, x1, . . . , xk−1〉 ∈ Xk, we may writeR(x0, . . . , xk−1) to make the statement that 〈x0, . . . , xk−1〉 ∈ R. We calla permutation f ∈ Sym(X) an automorphism of the relational structure〈X;S〉 iff for every k-ary relation R ∈ S we have that R(x0, . . . , xk−1)⇔R(f(x0), . . . , f(xk−1)). It is a fact that if f is an automorphism of 〈X;S〉and if Φ is a formula involving elements in S then Φ(x0, x1, . . .) ⇔Φ(f(x0), f(x1), . . .).

A key tool to which we will resort is given in the following

Lemma 2.2. Whenever the sets S and T of operations and relationsare interdefinable, the structures 〈n;S〉 and 〈n;T 〉 have exactly the sameautomorphisms.

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Proof. Note that k-ary operations on X can be treated as (k + 1)-aryrelations on X, so we may assume that the structures 〈n;S〉 and 〈n;T 〉have no operations and only relations.

We show that every automorphism of 〈n;S〉 is an automorphism of〈n;T 〉. The other direction is similar.

Let f be an automorphism of 〈n;S〉. Let R ∈ T be a k-ary relation.We must show that

∀〈x0, x1, . . . , xk−1〉 ∈ nk (R(x0, . . . , xk−1) ⇔ R(f(x0), . . . , f(xk−1))) .

Pick 〈x0, . . . , xk−1〉 ∈ nk. Since S and T are interdefinable, we haveR(x0, . . . , xk−1) ⇔ Φ(x0, . . . , xk−1) for some formula Φ in the language ofS. It can be verified that Φ(x0, . . . , xk−1) ⇔ Φ(f(x0), . . . , f(xk−1)), giventhat f is an automorphism of S. Furthermore,

Φ(f(x0), . . . , f(xk−1))⇔ R(f(x0), . . . , f(xk−1)).

The lemma follows.

Our first application of Lemma 2.2 is

Lemma 2.3. Let a subgroup G of Sym(n) be the group of automorphismsof an irreflexive binary relation on n. Then G is pertinent on n.

Proof. Let R be an irreflexive binary relation on n; that is to say, (∀x ∈n )(〈x, x〉 6∈ R). Suppose that G = Aut(〈n; R〉). We produce a binaryoperation ? on n that is interdefinable with R: Let x ? y := y if eitherx = y or 〈x, y〉 ∈ R, and let x ? y := x otherwise. In order to showinterdefinability we need to recover R from ? . Since R is irreflexive, thefollowing works: 〈x, y〉 ∈ R iff both x 6= y and x ? y = y. Lemma 2.2implies that Aut(〈n; R〉) = Aut(〈n; ?〉).

The conclusion of Lemma 2.3 holds also for reflexive R. The proof isthe same except for a minor change in how we retrieve R from ?.

Corollary 2.4 is a special case of Theorem 2.6. We offer the proof as asimpler introduction to the style of reasoning used in Theorem 2.6.

Corollary 2.4. Let G be the cyclic subgroup of Sym(n) generated by asingle cycle f := (c0 c1 · · · ck−1). Then G is pertinent.

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c0   c1   c2    .  .  .     ck  –  1    

       .  .  .     dn – k – 1 -­    

Figure1.1

Proof. Let d0 < d1 < · · · < dn−k−1 be the elements in n \ {c0, c1, . . . , ck−1}.We define the binary relation R on n by

R := {〈ci, ci+1〉 : i ∈ k − 1} ∪ {〈ck−1, c0〉} ∪ {〈di, di+1〉 : i ∈ n− k}.

Figure 1.1 depicts R as a digraph.

It is obvious that f is an automorphism of the structure 〈n; R〉. There-fore G ⊆ Aut(〈n; R〉), since G is the cyclic subgroup of Sym(n) generatedby f . It remains only to establish that Aut(〈n; R〉) ⊆ G. So pick anyg ∈ Aut(〈n; R〉). We must show that g is a power of f .

Since d0 is the only element in n for which there is no x ∈ n withR(x, d0), we get g(d0) = d0. So R(d0, d1) implies R(g(d0), g(d1)), givingR(d0, g(d1)) whence g(d1) = d1 since R(d0, y) if and only if y = d1.Continuing similarly we see that g(di) = di for every i ∈ n− k+ 1. Thus,since g ∈ Sym(n), we also have {g(ci) : i ∈ k} ∩ {di : i ∈ n − k + 1} = ∅.It follows that g(c0) = cm for some m ∈ k. By the definition of R wehave R(ci, y), iff y = cj with j = i + 1 mod k for every i ∈ k. Sosince R(c0, c1) gives R(g(c0), g(c1)), which yields R(cm, g(c1)), we haveg(c1) = cj where j = m + 1 (mod k). Continuing inductively we get thatg(ci) = cj where j = m + i (mod k). Hence g = fm as desired. Thecorollary follows by Lemma 2.3.

An immediate consequence of the corollary above is that for each k ≤ nthere exists a pertinent subgroup of Sym(n) of order k.

The next result utilizes an extended version of the Chinese RemainderTheorem. One finds this theorem listed as an exercise in [13] on Page 292among other places; to wit:

Proposition 2.5. Let m1,m2, . . . ,mr be positive integers. Let m be theleast common multiple of m1,m2, . . . ,mr, and let a, u1, u2, . . . , ur be any

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integers. Then there is exactly one integer u that satisfies the conditions

a ≤ u < a+m, u ≡ uj (mod mj), 1 ≤ j ≤ r

provided that

ui ≡ uj (mod gcd(mi,mj)) 1 ≤ i < j ≤ r;

and there is no such integer u when the latter condition fails to hold.

As usual, when f ∈ Sym(n) then supp(f) := {x : f(x) 6= x ∈ n}.

Theorem 2.6. Every cyclic subgroup of Sym(n) is pertinent.

Proof. Let f = f0f1 · · · fk−1 ∈ Sym(n) where the fi are the nontrivialcyclic components of f and where fi := (ci,0 ci,1 · · · ci,mi−1) is an mi-cycle

for each i ∈ k. Let n \ supp(f) = {di : i ∈ s} where s = n−∑k−1

j=0 mj.

We define the binary relation R on n as follows, where i and j rangeover all possibilities:

(1) R(di, y) if and only if y = di+1.

(2) R(ci,p, y) if and only if either

i. y = ci,p+1 mod mior

ii. y = cj,q where j > i and there exists z with both p ≡ z mod mi

and q ≡ z mod mj.

Clearly R is irreflexive. Figure 1.2 partially depicts R as a digraph.Missing from that picture are the majority of a welter of arrows from uppercircuits to lower circuits indicated in Specification (2)ii above. As we willsee, these downward arrows compel each automorphism of the structure topermute the vertices of the structure’s circuits in a coordinated fashion. Fur-thermore, each of the vertices in a given circuit has the same in-degree andout-degree as each of the other vertices in that circuit. This fact, and thefact that the existence of z for which z ≡ p mod mi and z ≡ q mod mj

is equivalent to the existence of z for which both f z(ci,0) = ci,p andf z(cj,0) = cj,q, ensure that f is itself an automorphism of R. Thus thecyclic group generated by f is a subset of the automorphism group of R.

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Figure1.2

The brunt of our proof lies ahead. That portion of our argument is aimedat showing that every automorphism of 〈n; R〉 is a power of f .

Fix an automorphism g of R.As in the proof of Corollary 2.4, we have that g(di) = di for all i ∈ s.Now we show g must map each of the k circuits onto itself. Observe

that these k circuits are the only circuits in the digraph R. So g mustpermute these k circuits; that is, if g maps ci,j to ci′,j′ then g mustmap each ci,u to a ci′,u′ . But, R gives us a linear order on these circuitssince there is an x ∈ {ci,0, ci,1, · · · , ci,mi−1} and y ∈ {cj,0, cj,1, · · · , cj,mj−1}for which xRy if and only if i ≤ j. Thus g must preserve this linearordering and take each of the k circuits onto itself.

We have shown that g restricted to the i-th circuit of R is somepower of fi. So Aut(〈n; R〉) is a subgroup of the group generated bythe cyclic components of f . It remains only to show that there is someinteger m so that g restricted to the i-th circuit of R is equal to fmifor every i ∈ k. However, this stems directly from Proposition 2.5. For, letm be the least common multiple of m0,m1, . . . ,mk−1, and fix the integersa, u0, u1, . . . , uk−1 with a = 0 and with g(ci,0) = ci,ui

. Pick {i, j} ⊆ k− 1with i < j. By Specification (2)ii, we have that R(ci,0, cj,0). So, since gis an automorphism, R(g(ci,0), g(cj,0)); that is, R(ci,ui

, cj,uj). But then,

again by Specification (2)ii, there is some z so that ui ≡ z mod mi anduj ≡ z mod mj. Thus ui ≡ uj mod gcd(mi,mj). So, by 2.5, there existsu ∈ m− 1 with u ≡ uj mod mj for each j ∈ k. So g = fu.

When X0 and X1 are nonempty disjoint subsets of n, and G0 andG1 are subgroups of Sym(X0) and Sym(X1), respectively, then the inner

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direct product of G0 and G1 refers to the subgroup H := {f ∈ Sym(n) :f |X0 ∈ G0 ∧ f |X1 ∈ G1 ∧ supp(f) ⊆ X0 ∪X1} of Sym(n). The expressionG0 � G1 denotes the inner direct product of G0 and G1. This notationextends in the obvious fashion to that of G0 �G1 � · · ·�Gk.

Lemma 2.7. Let {X0, X1} be a partition of the set n. For each i ∈ 2let Gi be a pertinent subgroup of Sym(Xi). Then G0 �G1 is pertinent.

Proof. By hypothesis there exists for each i ∈ 2 a groupoid Di := 〈Xi; ?i〉such that Gi = Aut(Di). Let H = G0 � G1. We define the groupoid〈n; ?〉 thus: If 〈x, y〉 ∈ Xi ×Xi then x ? y := x ?i y; if 〈x, y〉 ∈ X0 ×X1

then x ? y := x; if 〈x, y〉 ∈ X1 × X0 then x ? y := y. We claim thatH = Aut(〈n; ?〉).

Let f ∈ H. Pick 〈x, y〉 ∈ n × n. If {x, y} ⊆ Xi for some i ∈ 2then f(x ? y) = (f |Xi)(x ?i y) = (f |Xi)(x) ?i (f |Xi)(y) = f(x) ? f(y). Butif 〈x, y〉 ∈ X0 ×X1 then 〈f(x), f(y)〉 = 〈(f |X0)(x), (f |X1)(y)〉 ∈ X0 ×X1.So f(x ? y) = f(x) = (f |X0)(x) = (f |X0)(x) ? (f |X1)(y) = f(x) ? f(y).Similarly, if 〈x, y〉 ∈ X1 × X0 then f(x ? y) = f(y) = f(x) ? f(y). Sof ∈ Aut(〈n; ?〉). We infer that H ⊆ Aut(〈n; ?〉).

Pretend there exists g ∈ Aut(〈n; ?〉) for which g|Xi /∈ SymXi forsome i ∈ 2. Without loss of generality assume there is a u ∈ X0 for which〈u, g(u)〉 ∈ X0 × X1. Then, since g ∈ Sym(n), there exists v such that〈v, g(v)〉 ∈ X1 ×X0. By definition of ? we have that u ? v = u, whenceg(u) ? g(v) = g(u) since g ∈ Aut(n). But g(u) ? g(v) = g(v) by definitionof ? since 〈g(u), g(v)〉 ∈ X1 ×X0. Contradiction!

Thus, for each g ∈ Aut(〈n; ?〉), we have g|Xi ∈ SymXi for each i ∈ 2.But then, for (x, y) ∈ Xi×Xi, we have g(x ?i y) = g(x ? y) = g(x) ? g(y) =g(x) ?i g(y) and thus g|Xi ∈ Gi for each i ∈ 2. This shows that g is inG0 �G1 = H, so Aut(〈n; ?〉) ⊆ H.

Therefore H = Aut(〈n; ?〉).

Theorem 2.8. Let {X0, X1, . . . , Xs−1} be a partition of the set n. Foreach p ∈ s let Gp be a pertinent subgroup of Sym(Xp). Then the innerproduct of all the Gp is pertinent.

Proof. Model the inductive step after the proof of Lemma 2.7.

Corollary 2.9. Let c0, c1, · · · , ck−1 be single cycles in Sym(n) with pair-wise disjoint supports. Let F = 〈c0, c1, · · · , ck−1〉 be the subgroup generatedby the k cycles. Then F is pertinent.

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Proof. Let S be the set of elements of n that are not elements of supp(ci)for any i ∈ k. For each i ∈ k let 〈ci〉 be the cyclic subgroup of Sym(n)generated by ci. These subgroups are all pertinent by Corollary 2.4. Nowobserve that F is precisely the inner product of ιS and the 〈ci〉. Theconclusion follows by Theorem 2.8.

Corollary 2.10. Let {k0, k1, . . . , ks−1} ⊆ N with k0 + k1 + · · · + ks−1 =n. Then there is a pertinent subgroup H of Sym(n) such that |H| =k0!k1! · · · ks−1! .

Proof. Partition n into s sets {X0, X1, . . . , Xs−1} with |Xi| = ki foreach i ∈ s. Let H be Sym(X0)� Sym(X1)� · · ·� Sym(Xs−1). Then byLemma 2.1 each Sym(Xi) is pertinent. So by Theorem 2.8, H is pertinent.But |H| = Πs−1

i=0 | Sym(Xi)| = k0!k1! · · · ks−1! The claim follows.

3 Transitivity and Pertinence

A permutation group G is k-transitive iff whenever 〈x0, x1, . . . , xk−1〉and 〈y0, y1, . . . , yk−1〉 are k-tuples of distinct elements in n, there is apermutation g ∈ G with g(xi) = yi for all i ∈ k. A transitive group ismerely a 1-transitive group. In general, the more transitive an automorphismgroup G = Aut(〈n; ?〉) is, the more restrictions G imposes on the structureof 〈n; ?〉.

A permutation group G is said to be semiregular iff its only elementwith fixed points is the identity. Equivalently, whenever g, h ∈ G are suchthat g(x) = h(x) for some x, then g and h are equal.

A permutation group is regular iff it is both semiregular and transitive.An equivalent formulation is that a permutation group is regular if and onlyif it is a permutation representation obtained by using Cayley’s Theorem.(Given a regular permutation group G on n, let e be some arbitraryelement of n. Define ψ : G −→ n by setting ψ(g) = g(e) for all g ∈ G,and define an operation • on n by x • y = (ψ−1(x))(y). It is theneasy to show that ψ is an isomorphism from G to 〈n; •〉; the conclusionfollows.) Gould proved in [8] that any countable permutation group producedby Cayley’s Theorem is the automorphism group of a groupoid, giving ournext lemma. We offer an elementary proof of our own.

Regular and semiregular groups tend to have small cardinalities. Lemma3.1 and Theorem 3.2 below seem to suggest that groups with a “small numberof elements” are pertinent. But we will also see in Theorem 3.7 that it is notenough to insist on a small number of group elements; those elements mustalso be “evenly distributed,” which semiregularity ensures.

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Lemma 3.1. Every regular finite permutation group is pertinent. Further-more, unless the group is trivial, it is the automorphism group of a groupoid〈n; ?〉 which satisfies x ? y 6= y, for all x, y ∈ n.

Proof. Let n be given, and let G be a regular permutation group that isa subgroup of Sym(n). Observe that G has n elements and that theycan be uniquely labelled by where they take 0. So g0 takes 0 to 0,and is obviously the identity. Continuing, we have gj(0) = j for all j ∈ n.With this correspondence established, we will work with the group elementsgj, writing their group operation of composition as simple concatenation.Notice that for i, j ∈ n, we have gigj(0) = gi(j); that is, gigj = ggi(j).

We define a successor function ′ on the group elements by setting g′j =gj+1 for all j < n− 1, and letting g′n−1 = g0. (The operation ′ : G→ Gmay have nothing to do with the group operation of G.) We define thebinary operation • on G by gi • gj = gi(g

−1i gj)

′ and the binary operation? on n by i ? j = k iff gi • gj = gk. Note that 0 ? j = j + 1 (mod n) forall j ∈ n, since g0 • gj = g0(g

−10 gj)

′ = g0g′j = g′j = gj+1, where addition is

modulo n.Let i, j, k ∈ n be arbitrary. Then gk(gi • gj) = gk(gi(g

−1i gj)

′) =(gkgi)(g

−1i (g−1

k gk)gj)′ = (gkgi)((gkgi)

−1(gkgj))′ = (gkgi)•(gkgj) = ggk(i)•ggk(j).

That is to say gk(i?j) = gk(i)?gk(j). Thus all of the gk are automorphismsof the groupoid 〈n; ?〉.

So {gj : j ∈ n} is a transitive subgroup of Aut(〈n; ?〉). We will showthat Aut(〈n; ?〉) is regular which will imply it has n elements and thusthat the two groups are equal. Since it has a transitive subgroup, it is clearlytransitive. From the above, we have that 0 ? j = j + 1 (mod n) for allj ∈ n. Thus if h is an automorphism that fixes 0 then h(1) = h(0 ? 0) =h(0) ? h(0) = 0 ? 0 = 1. Inductively we get that h(j) = j for each j ∈ n.So g0 is the unique automorphism of 〈n; ?〉 that fixes 0. Now supposeh1 and h2 are automorphisms of 〈n; ?〉 for which h1(i) = h2(i) for somefixed i ∈ n. Since G is transitive, there exists h3 ∈ G with h3(0) = i.But then h1(h3(0)) = h2(h3(0)). So, by the above, h1h3 = h2h3. That is,h1 = h2. Therefore Aut(〈n; ?〉) is regular.

To show our second claim, suppose i ? j = j for some i, j ∈ n. Thengi(g

−1i gj)

′ = gj. That is, (g−1i gj)

′ = g−1i gj. However, unless n = 1, no

element is equal to its own successor.

Theorem 3.2. Every semiregular finite permutation group is pertinent.

Proof. Let n be given, and let G be a subgroup of Sym(n) that actssemiregularly on n. By the previous lemma, we may assume that G is

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not transitive. Since the trivial group is obviously pertinent, we also assumethat G nontrivial.

Define Bi = {g(ai) : g ∈ G} to be the orbits of G, where we leta0, a1, . . . ak−1 be representatives of the Bi. These k orbits form a partitionof n. By semiregularity, we have that all of the Bi have the same numberof elements as G.

By construction, G acts transitively, and hence regularly, on each block.Thus the previous lemma gives us binary operations ?i on each Bi sothat Aut(〈Bi, ?i〉) = G|Bi for all i, and we have that x ?i y 6= y for allx, y ∈ Bi and i ∈ k.

We will define a global operation ? on n that agrees with each ?i.To aid in this, we define a binary relation R on n by setting xRy iffx = g(ai) and y = g(ai+1) for some g ∈ G and i ∈ k − 1. Note thatG “preserves” R by construction; that is to say, xRy if and only ifg(x)Rg(y) for each g ∈ G. Define the binary operation x ? y on n thus:x ? y = x ?i y if x and y are both in the same Bi, and for x and y indifferent blocks, x ? y = y if xRy else x ? y = x. Notice that x ? y = yif and only if xRy, and that xRy if and only if h(x)Rh(y) when h ∈ G.

We have that G is a subgroup of Aut(〈n, ?〉) by construction.We see that B0 = {y : ¬∃x(x ? y = y)}, and hence every element of

Aut(〈n, ?〉) maps B0 onto itself. But we can define B1 by B1 = {y :(∃x ∈ B0)(x ? y = y)}, and hence every element of Aut(〈n, ?〉) takes B1

onto itself. Continuing inductively, Bi = {y : (∃x ∈ Bi−1)(x ? y = y)}, andhence every element of Aut(〈n, ?〉) takes every Bi onto itself. Now leth ∈ Aut(〈n, ?〉) be given. We have that h(a0) = g(a0) for some g ∈ G.Since a0Ra1, we have h(a0)Rh(a1); that is, g(a0)Rh(a1). Further,h(a1) ∈ B1. But there is only one element x ∈ B1 such that g(a0)Rx,and that element is x = g(a1). Thus h(a1) = g(a1). Continuing, we getthat h(ai) = g(ai) for all i ∈ k. Recall that Aut(〈Bi, ?i〉) = G|Bi forall i, and that G|Bi acts regularly on Bi. Thus h is determined as anautomorphism by where it takes the ai, and so h = g ∈ G.

Proposition 3.3. Let G be a transitive subgroup of Sym(n), with G =Aut(nj) for some j. Let the function s : n → n, be s : x 7→ x ?j x.Then s ∈ Sym(n), and all of the cycles of s have the same length.

Proof. Pick any a, b ∈ n. Then b ?j b = c for some c ∈ n. SinceG is transitive, there exists f ∈ G with f(c) = a. Thus s(f(b)) =f(b) ?j f(b) = f(b ?j b) = f(c) = a. So Rng(s) = n. Hence s ∈ Sym(n).

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Note also for each g ∈ G and x ∈ n that g(s(x)) = g(x ?j x) =g(x) ?j g(x) = s(g(x)). Let c and c′ be distinct cyclic components ofthe permutation s. Let y ∈ supp(c), and let y′ ∈ supp(c′). That is,supp(c) = {sj(y) : j ∈ ω}. Since G is transitive, there exists h ∈ Gwith h(y) = y′. So h| supp(c) is a bijection of supp(c) onto the set{h(sj(y)) : j ∈ ω} = {sj(h(y)) : j ∈ ω} = {sj(y′) : j ∈ ω} = supp(c′).Therefore every cyclic component of s is of length | supp(c)|.

To simplify the following presentation, assume n ≥ 3. Then a permu-tation group G on the set n is primitive iff the unary algebra 〈n;G〉is simple. (This is not quite the standard definition, but suffices for ourpurposes.) Primitive groups are 1-transitive, and all 2-transitive groupsare primitive. Thus primitive groups form an intermediate class between1-transitive and 2-transitive groups.

It is interesting to note some results of Palfy, Szabo and Szendrei from[14]. This article culminated a series of papers on algebras with large au-tomorphism groups and considered algebras with primitive automorphismgroups. Lemma 1 from that paper strengthens our Proposition 3.3 and statesthat every algebra with a primitive automorphism group G is either idem-potent or of prime order and equal to the set of unary term functions of thealgebra.

An algebra is functionally complete iff every possible finitary operation onits universe is a polynomial function; that is, if it is obtainable by substitutingconstants in a term function of the algebra, vis [3]. A quick paraphrase of themain result of [14] is that “almost all algebras with primitive automorphismgroups are functionally complete.” The exact statement is a bit lengthy andlists some exceptions obtained from Rosenberg’s classification of preprimalclones. The interested reader should refer to the original paper. (Beware:“polynomial function” is there used to mean “term function”.) However,that paper’s Corollary 1 states that whenever a finite algebra A has |A|composite but not a power, then if Aut(A) is primitive, A is functionallycomplete. (Results of this type for transitive automorphism groups occuralso in [17].)

A groupoid H is called a quasigroup iff H satisfies both the leftcancellation and the right cancellation laws. An idempotent groupoid isone in which every element is idempotent. The following result is due toStein in [16]. That article also offers many additional results on idempotentquasigroups with transitive automorphism groups.

Lemma 3.4. Let G be a 2-transitive subgroup of Sym(n), and supposethat G = Aut(nj) for some j. Then one of the following statements holds

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for the groupoid nj.

(1) n = 2 and h2(j) ∈ {0011, 0101, 1010, 1100}.

(2) nj ∈ {S(n), S ′(n)}.

(3) nj is an idempotent quasigroup. Moreover, if x 6= y with {x, y} ⊆ n,then x ? y is neither x nor y.

Proof. We give an outline of a proof; this is Theorem 2.1 of [16]. If n is 1or 2, the proof is left to the reader. So assume n ≥ 3.

We claim that nj is idempotent. For if x ?j x = y 6= x, we take anautomorphism g of nj with g(x) = x and g(y) = z for some z 6= y,quickly obtaining a contradiction.

We next claim that if there are distinct {x, y} ∈ n with x ?j y equal tox or y, that nj is S(n) or S ′(n) respectively. This also follows easilyfrom the 2-transitivity of G. Suppose henceforth that nj is neither S(n)nor S ′(n).

Now suppose nj is not a quasigroup. Without loss of generality, assumethere exist elements x, y, and z in n with x ?j y = x ?j z = u andwith y 6= z. Using the previous claim, one shows that x, y and z areall distinct. Using the pigeonhole principle there exists v ∈ n \ {x} withx ?j w 6= v for all w ∈ n. Then using the 2-transitivity of G, one easilyobtains a contradiction.

Not every idempotent quasigroup has a 2-transitive automorphism group.However, we will now produce an abelian idempotent quasigroup 5j forwhich Aut(5j) is 2-transitive: Let Z5 be the usual field, 〈5; +, ·〉, anddefine the binary operation ?j : 5 × 5 → 5 in terms of the field operationsby setting x ?j y := (x + y) · 3. It is easily verified that 5j = 〈5; ?j〉 is anidempotent quasigroup. Let F denote the set of all functions f : 5 → 5such that f : x 7→ a ·x+b with a 6= 0. Plainly F ⊆ Sym(5). Furthermore,if f ∈ F is specified by f(x) = a · x+ b then f(x ?j y) = f((x+ y) · 3) =a · ((x + y) · 3) + b = 3 · a · x + 3 · a · y + b. Thus, since 3 + 3 = 1, we getf(x ?j y) = 3 · a · x+ 3 · b+ 3 · a · y + 3 · b = 3 · (f(x) + f(y)) = f(x) ?j f(y).Therefore F ⊆ Aut(5j). Finally, since Z5 is a field, we can see as followsthat Aut(5j) is 2-transitive: Let t 6= s and t′ 6= s′ be elements in 5.Then there exists a 6= 0 for which a · (s− t) = s′− t′. Now let b := t′−a · t.Then the function f : x 7→ a · x+ b satisfies t′ = f(t) and s′ = f(s).

Lemma 3.5. Let n ≥ 4. Let G be a 3-transitive subgroup of Sym(n),and let G = Aut(nj) for some j. Then nj ∈ {S(n), S ′(n)}.

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Proof. By Lemma 3.4, we are done unless nj is an idempotent quasigroupsuch that |{x, y, x ?j y}| = 3 whenever x and y are distinct elements ofn. So take it that nj satisfies that condition. Then, letting x and y bedistinct elements in n, we have that x ?j y ∈ n \ {x, y}. Since n ≥ 4,there exists z ∈ n \ {x, y, x ?j y}. By the 3-transitivity of G, there is anautomorphism f with f(x) = x, with f(y) = y, and with f(x ?j y) = z.Then x ?j y = f(x) ?j f(y) = f(x ?j y) = z, a contradiction.

Corollary 3.6. No 3-transitive proper subgroup of Sym(n) is pertinent.

Proof. Let G be a 3-transitive proper subgroup of Sym(n). Then n ≥ 4.Pretend that there exists nj ∈ G(n) such that G = Aut(nj). Then nj ∈{S(n), S ′(n)} by Lemma 3.5. But Aut(S(n)) = Aut(S ′(n)) = Sym(n).So, since the subgroup G of Sym(n) is proper, we have a contradiction.

Corollary 3.6 suggests a multitude of impertinent subgroups of Sym(n).Among these are the Alt(n) for n ≥ 5; for, each such Alt(n) is a 3-transitive proper subgroup of the corresponding Sym(n). (Indeed, Alt(n)is (n − 2)-transitive. This is proved in [10]. But one can verify this factalso by observing that, for each pair 〈s0, . . . , sn−3〉 and 〈t0, . . . , tn−3〉 ofinjective length-(n− 2) sequences in n, there are permutations f 6= g ofn such that f(si) = g(si) = ti for each i ∈ n − 2. These permutationsdiffer by a single transposition; so, one of them is an element in Alt(n).)

A more exhaustive list of 3-transitive groups can be found in [5], or onlineat [15]. Furthermore, notice that full 3-transitivity is not used in the proofof Lemma 3.5, but only the fact that G is 2-transitive and that given anydistinct x, y and w, there is an automorphism in G that fixes x andy while moving w.

We offer results producing additional impertinent groups:

Theorem 3.7. Let 2 < m < n. Let G be a 2-transitive subgroup ofSym(m). Let G′ := G� {ιn\m}, where ιn\m is the identity on n \m. IfG′ is pertinent then G is pertinent.

Proof. Let G′ be pertinent. Then G′ = Aut(〈n; ?〉) for some binaryoperation ? on the set n.

For p ∈ Sym(n) and 〈a, b〉 ∈ n2 we say that p preserves the producta ? b iff p(a ? b) = p(a) ? p(b). Note that p ∈ Aut(〈n; ?〉) if and only if ppreserves a ? b for every 〈a, b〉 ∈ n2.

Let S := Sym(m)� {ιn\m}. To facilitate our proof we show first that,if all elements in S preserve a ? b when a ? b is “possible” (i.e., welldefined), then {a, b, a ? b} 6⊆ m.

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Now choose any 〈a, b〉 ∈ n2.

Case 1: {a, b, a ? b} ⊆ n \m. Then g(a) ? g(b) = a ? b = g(a ? b) foreach g ∈ S.

Case 2: {a, b} ⊆ n \m, but the product a ? b is a well-defined elementin the set m. Since G is transitive, for each c ∈ m there exists fc ∈ Gwith fc(a ? b) = c. Let f ′c ∈ G′ be such that f ′c|m = fc. It follows thata ? b = fc(a) ? fc(b) = f ′c(a) ? f ′c(b) = f ′c(a ? b) = fc(a ? b) = c. Thus a ? b = cfor every c ∈ m. Since m > 1, we must infer that the product a ? b isnot well defined, contrary to hypothesis. So Case 2 cannot actually occur.

Case 3: a ∈ n \m, while {b, a ? b} ⊆ m with a ? b well defined.Subcase: b 6= a?b. Since G is 2-transitive we have for each c ∈ m\{b}

that there exists fc ∈ G such that fc(b) = b and fc(a ? b) = c. Letf ′c ∈ G′ be such that f ′c|m = fc. Then a ? b = f ′c(a) ? f ′c(b) = f ′c(a ? b) = c.Thus a ? b = c for every c ∈ m \ {b}, an impossibility since |m \ {b}| ≥ 2and since a ? b is well defined. The situation in this Subcase never occurs.

Subcase: b = a ? b. Pick c ∈ m. There exists fc ∈ G such thatfc(b) = c; let f ′c ∈ G′ be such that f ′c|m = fc. Then c = fc(a ? b) =f ′c(a ? b) = f ′c(a) ? f ′c(b) = a ? c. Thus a ? d = d for every d ∈ m. Hence,for every g ∈ S we have that g(a) ? g(b) = a ? g(b) = g(b) = g(a ? b). Soin this Subcase every element in S preserves the product a ? b.

Case 4: b ∈ n \m, while {a, a ? b} ⊆ m with a ? b well defined. Theargument parallels the argument in Case 3, and we omit it.

Case 5: {a, a?b} ⊆ n\m with a?b well defined, and b ∈ m. Then foreach c ∈ m there exists fc ∈ G with fc(b) = c, and there exists f ′c ∈ G′with f ′c|m = fc. So a ? b = f ′c(a ? b) = f ′c(a) ? f ′c(b) = a ? c. Thereforea ? b = a ? c for every c ∈ m. Consequently, for every g ∈ S we have thatg(a) ? g(b) = a ? g(b) = a ? b = g(a ? b) since g(b) ∈ m. Thus every elementin S preserves the product a ? b in this Case too.

Case 6: {b, a ? b} ⊆ n \m with a ? b well defined, and a ∈ m. Theargument here parallels the argument in Case 5 and is omitted.

Case 7: a ? b ∈ n \ m and is well defined, and {a, b} ⊆ m. Pickany g ∈ S Then {g(a), g(b)} ⊆ m. Thus, since G is 2-transitive thereexists f ∈ G such that f(a) = g(a) and such that f(b) = g(b) whethera = b or not, and there exists f ′ ∈ G′ with f ′|m = f . Thereforeg(a ? b) = a ? b = f ′(a ? b) = f ′(a) ? f ′(b) = g(a) ? g(b). Again every elementin S preserves the product a ? b.

The cases above show that if some h′ ∈ S fails to preserve a well-defined

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product a? b ∈ n then {a, b, a ? b} ⊆ m. To prove that G is pertinent, wemodify the binary operation ? : n×n→ n, thus defining a binary operation� : m × m → m for which G = Aut(〈m; �〉): For 〈x, y〉 ∈ m × m, letx � y := x ? y when x ? y ∈ m, and let x � y := x when x ? y ∈ n \m.

The mapping, Sym(m)→ S given by h 7→ h′ := h∪ ιn\m, is a bijection.Pick any h ∈ Sym(m) \ G. Then h′ ∈ S. By Cases 1–7, there exists

〈xh, yh〉 ∈ m×m with h′(xh ? yh) 6= h′(xh) ? h′(yh) and with xh ? yh ∈ m.

So h(xh � yh) = h(xh ? yh) = h′(xh ? yh) 6= h′(xh) ? h′(yh) = h(xh) � h(yh).

Hence h ∈ Sym(m) \Aut(〈m; �〉). Therefore Aut(〈m; �〉) ⊆ G.Now pick any g ∈ G and 〈u, v〉 ∈ m×m. Then g′ ∈ G′ := Aut(〈n; ?〉).Case: u ? v ∈ m. Then g(u) ? g(v) = g′(u) ? g′(v) = g′(u ? v) ∈ m. So

{g(u), g(v), g(u)?g(v)} ⊆ m. Hence g(u)�g(v) = g(u)?g(v) = g′(u)?g′(v) =g′(u ? v) = g(u ? v) = g(u � v). Thus g preserves u � v.

Case: u ? v ∈ n \m. Then u � v := u. Furthermore, g(u) ? g(v) =g′(u) ? g′(v) = g′(u ? v) = u ? v ∈ n \m, whence g(u) � g(v) = g(u) since inthis case we got that 〈g(u), g(v), g(u) ? g(v)〉 ∈ m2 × (n \m). So here toog(u � v) = g(u) � g(v).

Thus in each case g preserves u � v. So g ∈ Aut(〈m; �〉). ThereforeG ⊆ Aut(〈m; �〉). Hence the group G = Aut(〈m; �〉) is pertinent.

Suppose m < n and H is a subgroup of Sym(n) such that everyelement in H takes m onto m. Then by the restriction to H of m,written H |m, we mean the subgroup {h|m : h ∈ H} of Sym(m). Theprevious theorem showed that sometimes a group with a restriction that isnot pertinent must also not be pertinent. The next theorem gives an examplewhere that is not the case.

Theorem 3.8. Let m ∈ {1, 2, 3, . . .}, let n := m+m2 + · · ·+mm−1 +mm,and let G be any subgroup of Sym(m). Then there exists a pertinentsubgroup H of Sym(n) with H |m = G.

Proof. Let m and G be given. We may take it that m ≥ 2.Partition the set n into the sets A1, A2, . . . , Am with A1 := m and

with |Aj| = mj for every j. For each j ∈ {1, 2, . . . ,m− 1} let fj be abijection from A1 × Aj onto Aj+1.

The binary operation ? : n× n→ n is defined thus:

For 〈x, y〉 ∈ A1 ×Aj let x ? y := fj(x, y). For 〈x, y〉 ∈ Am ×Am with

x = g(0) ? (g(1) ? (g(2) ? (g(3) ? · · · ? (g(m− 2) ? g(m− 1)) · · · )))

for some g ∈ G, let x?y := y. For all other 〈x, y〉 ∈ n×n, let x?y := x.

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Since m ? m = A1 ? A1 = A2, no element in A1 is idempotent. ButAj ?Aj = Aj, and each x ∈ Aj is idempotent, when 2 ≤ j ≤ m. So everyf ∈ Aut(〈n; ?〉) maps m onto itself, and an easy inductive argument showsthen that f maps Aj onto itself for each j ∈ {2, 3, . . . ,m}.

We use a result of Armbrust and Schmidt in [1]. They show that everypermutation group on m is the automorphism group of a structure 〈m; ρ〉,where ρ is an m-ary relation. To see this, let ~k := 〈0, 1, . . . ,m− 1〉, andfor all q ∈ Sym(m) let q(~x) := 〈q(x0), q(x1), . . . , q(xi)〉 whenever ~x :=〈x0, x1, . . . , xi〉 where 0 ≤ i < m. Our claim now is that G = Aut(〈m; ρ〉),where ρ := {g(~k) : g ∈ G}, an m-ary relation on m.

Pick any ~r ∈ ρ. Then ~r = g(~k) for some g ∈ G. Let q ∈ Sym(m).

Then q(~r) = q(g(~k)) = qg(~k). Note that q(~r) ∈ ρ if and only if qg ∈ G.But qg ∈ G if and only if q ∈ G. Thus q preserves ρ if and only ifq ∈ G. Thus G = Aut(〈m; ρ〉) as alleged.

For 1 ≤ i ≤ m we define si : Ai1 → Ai thus: When ~x := 〈x1, x2, . . . , xi〉,let si(~x) := x1 ? (x2 ? (x3 ? (. . . ? (xi−1 ? xi) . . .))). Straightforward inductiongives Ai = {si(~x) : ~x ∈ Ai} and si(r(~x)) = r(si(~x)), for all ~x ∈ Ai1and for all r ∈ Aut(〈n; ?〉). These two facts imply that every element inAut(〈n; ?〉) maps {sm(~r) : r ∈ ρ} onto itself.

Finally, we claim that Aut(〈n; ?〉)|m = G.

Pick q ∈ Aut(〈n; ?〉). Then q(sm(~k)) = q(sm(ιm(~k))) ∈ {sm(~r) : r ∈ ρ},since ιm(~k) ∈ ρ. So for some g ∈ G we have sm(g(~k)) = q(sm(~k)) =

sm(q(~k)) for some g ∈ G. But because sm(q(~k)) = sm(g(~k)) we seeinductively from the definition of ? and the bijectivity of the functions fithat q(j) = g(j) for every j ∈ m; that is to say, g = q|m. Thus we inferthat Aut(〈n; ?〉)|m ⊆ G.

Again using the fact that fi : A1 × Ai → Ai+1 is a bijection for everyi < m, by induction we get that every x ∈ n can be uniquely expressed byx = si(~x) for some i ∈ {1, 2, . . . ,m} and some ~x ∈ Ai1. Fix h ∈ G, anddefine qh : n → n by qh : x 7→ si(h(~x)) where x = si(~x) with ~x ∈ Ai1.Inductively one can show that qh ∈ Aut(〈n; ?〉). Notice that qh|m = h. Itfollows that G ⊆ Aut(〈n; ?〉)|m. So G = Aut(〈n; ?〉)|m.

Here is an example showing how our next theorem could be applied. In{0, 1, 2, 3, 4, 5}, let A = {0, 1, 2} and B = {3, 4, 5}. Then G = 〈(0 1 2)〉 isa transitive subgroup of Sym(A) and H = 〈(3 4 5)〉 is a transitive subgroupof Sym(B). Let K be the subgroup of Sym(A) � Sym(B) consistingof all the even permutations. Then K has 18 elements, half as many asSym(A) � Sym(B), and contains G � H. It is easy to check that K isnot a direct product; the theorem implies that K is not pertinent.

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Theorem 3.9. Let {A,B} be a partition of the set n with |A| ≥ 2 ≤ |B|.Let G and H be transitive subgroups of Sym(A) and of Sym(B),respectively. Let K be a subgroup of Sym(A)�Sym(B) for which G�H ⊆K. If K is not an inner product, then K is not pertinent.

Proof. Suppose K is pertinent. Then there is a binary operation ? : n×n→n with K = Aut(〈n; ?〉).

Pick any a ∈ A and b ∈ B. First, take it that a ? b ∈ A.Fix any y ∈ B. The transitive subgroup H has an element h for which

h(b) = y. Since {ιA} �H ⊆ K, we have that h′ := ιA ∪ h ∈ K. Hencea?y = a?h(b) = h′(a)?h′(b) = h′(a?b) = a?b. That is, a?y = a?b for everyy ∈ B. Now, instead, fix x ∈ A. By the transitivity of G there existsgx ∈ G such that gx(a) = x. Thus, since g′x := gx ∪ ιB ∈ G � {ιB} ⊆ K,we have that x ? b = g′x(a) ? g′x(b) = g′x(a ? b) = gx(a ? b). That is, foreach x ∈ A there exists gx ∈ G such that x ? b = gx(a ? b) ∈ A. Thus,when 〈a, b, a ? b〉 ∈ A × B × A, there is a function fA : A → A suchthat x ? y = fA(x) for every 〈x, y〉 ∈ A × B. A similar argument for〈a, b, a ? b〉 ∈ A×B ×B gives us that there is a function fB : B → B suchthat x ? y = fB(y) for every 〈x, y〉 ∈ A×B.

Arguing as in previous paragraph we see that either there is a functiongB : B → B such that x ? y = gB(x) for every 〈x, y〉 ∈ B × A or there isa function gA : A→ A such that x ? y = gA(y) for every 〈x, y〉 ∈ B × A.

We have four cases, since we have either fA or fB, and also have eithergA or gB. Since all cases are similar, we exhibit detail in only one of them;namely, the case involving the functions fA and gA: In this case we letG′ := Aut(〈A; ?A, fA, gA〉) and H ′ := Aut(〈B; ?B〉). Claim: K = G′�H ′.

Pick k ∈ K.Let 〈x, y〉 ∈ A × A. Then k(x ?A y) = k(x ? y) = k(x) ? k(y). Since

k ∈ Sym(A)� Sym(B), we have {k(x), k(y)} ⊆ A. It follows that k(x ?Ay) = k(x) ?A k(y). Similarly k(x ?B y) = k(x) ?B k(y). The functions fAand gA are irrelevant when 〈x, y〉 ∈ (A×A)∪ (B×B). We have partiallyverified that k ∈ G′ �H ′.

Now let 〈x, y〉 ∈ A × B. Then k(fA(x)) = k(x ? y) = k(x) ? k(y) =fA(k(x)). A similar argument shows that k(gA(y)) = gA(k(y)) when〈x, y〉 ∈ B×A. Only one of the functions fA and gA is relevant in each ofthe circumstances where 〈x, y〉 ∈ (A× B) ∪ (B × A), and neither ?A nor?B is relevant there. So k ∈ G′ �H ′. Therefore K ⊆ G′ �H ′.

Let γ ∈ G′ � H ′. We can write γ as a composition γ = αβ ofelements α ∈ G � {ιB} and β ∈ {ιA} � H. If 〈x, y〉 ∈ A × A thenγ(x) ? γ(y) = α(x) ? α(y) = α(x ? y) = γ(x ? y) since α|A ∈ Aut(〈A; ?A〉).

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Similarly γ(x ? y) = β(x ? y) = β(x) ? β(y) = γ(x) ? γ(y) if 〈x, y〉 ∈ B ×B,since β |B ∈ Aut(〈B; ?B〉). Now take 〈x, y〉 ∈ A×B. Then γ(x) ? γ(y) =α(x) ? β(y) = fA(α(x)). But, since α ∈ G′, we have that fA(α(x)) =α(fA(x)), whence γ(x)?γ(y) = α(fA(x)) = α(x?y) = γ(x?y). Similarly, if〈x, y〉 ∈ B×A then γ(x)?γ(y) = gA(α(y)) = α(gA(y)) = α(x?y) = γ(x?y).So γ ∈ Aut(〈n; ?〉) = K. Thus K = G′ �H ′.

Recall that when G is a group and when S ⊆ G then 〈S〉 denotesthe subgroup of G generated by S.

Theorem 3.10. Let n ∈ {1, 2, 3, 4}. Then every subgroup of Sym(n) ispertinent.

Proof. The reader may wish to verify that we examine, below, all of therelevant subgroups of the Sym(n). By Lemma 2.1, every Sym(n) ispertinent.

For n ≤ 3, every proper subgroup of Sym(n) is cyclic. By Theorem2.6, all of these are pertinent. So henceforth take it that n = 4.

Up to a permutation of the elements of 4, the non-cyclic subgroups ofSym(4) are as follows:

(1) 〈{(0 1), (2 3)}〉 ' Z2 × Z2

(2) 〈{(0 1)(2 3), (0 2)(1 3)}〉 ' Z2 × Z2

(3) 〈{(0 1), (0 1 2)}〉 ' S3

(4) 〈{(0 2), (0 1 2 3)}〉 = D4

(5) 〈{(0 1 2), (1 2 3)}〉 = A4

(6) Sym(4)

For (1), observe that 〈{(0 1), (2 3)}〉 is a direct product of the cyclicgroups generated by (0 1) and (2 3). Each of these cyclic groups is pertinentby Theorem 2.6, so their product is pertinent by Lemma 2.7.

For (2), note that the group acts regularly on {0, 1, 2, 3}, so it is pertinentby Lemma 3.1.

For (3), we have that 〈{(0 1), (0 1 2)}〉 is pertinent on {0, 1, 2}. Also,the 1-element group on {3} is cyclic, and hence pertinent. Their innerproduct is 〈{(0 1), (0 1 2)}〉 on {0, 1, 2, 3}, which is pertinent by Lemma2.7.

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For (4), although the group D4 is usually defined as the symmetries ofa square, all that matters for us is the edge relation of the square; i.e., thebinary relation R := {〈0, 1〉, 〈1, 0〉, 〈1, 2〉, 〈2, 1〉, 〈2, 3〉, 〈3, 2〉, 〈3, 0〉, 〈0, 3〉}.Clearly, the symmetries of the square are exactly the elements in Sym(4)which preserve R. So D4 is the automorphism group of the structure withthis irreflexive binary relation. So, by Lemma 2.3 we have that D4 is alsothe automorphism group of a groupoid 4j.

For (5), we use Lemma 2.2 on a geometric representation of Alt(4). Tothis purpose, consider a regular vertex-labeled tetrahedron T := T (0, 1, 2, 3).Exactly half of the 4! symmetries of T fail to invert T . Therefore, thegroup of all non-inverting symmetries of T is Alt(4) itself, since Alt(4) isthe only concrete manifestation of a subgroup of Sym(4) that has index 2.

Since each 3-membered set of vertices determines a face, all permutationsin Sym(4) take faces to faces. The permutations in Alt(4) are preciselythose which do not change the orientation of faces, but take clockwise orientedfaces to other clockwise oriented faces. So consider the ternary relation Con 4 defined as follows: C(x, y, z) is true if and only if x, y and z arethe three vertices on some face of the tetrahedron, and a path from x to yto z to x goes clockwise around that face. We have that Alt(4) is theautomorphism group of the structure 〈4;C〉.

We define the groupoid operation as follows: x ?j x := x for everyx ∈ 4. For vertices x 6= y of T , the edge xy is shared by exactly twofaces, xyz and xyz′ . Viewed from outside T , the directed edge x → ydetermines a counterclockwise directed circuit around one of those faces, anda clockwise circuit around the other face. We define x ?j y := z if and onlyif the clockwise circuit goes around the triangle xyz . In other words, whenx 6= y then x ?j y is the unique z for which C(x, y, z) holds.

This shows that ?j is definable from C. But we can recover C from?j by defining C(x, y, z) to be true iff x, y and z are all distinct andx ?j y = z. Thus 4j and 〈4;C〉 are interdefinable. So by Lemma 2.2 weget that Alt(4) = Aut(4j). Finally, Lemma 2.1 establishes (6).

The groupoid 4j, above, for which Alt(4) = Aut(4j), is specified byits binary operation table, h4(j) = 0231310213202013.

We note in passing that this groupoid h4(j) := 0231310213202013 has aninteresting structure. It is an idempotent quasigroup in which x?j y = y ?j xif and only if x = y, and in which (x ?j y) ?j z = x ?j (y ?j z) if and only ifx = z.

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4 Further Possibilities

It would be natural to generalize our results to the following framework.We consider arbitrary types of mixed operational and relational structures,writing them with the arities of operations first, followed by the arities ofrelations. We define a preorder 4 on this set of types by setting σ 4 τ ifffor all natural numbers n every automorphism group of a structure on theset n of type σ is also the automorphism group of a structure of type τ .

We can easily define “extra” operations or relations in a trivial way thatwill not reduce the number of automorphisms, thus σ 4 τ whenever σ isa reduct of τ . We can replace operations by the relations that are theirgraphs, and the result does not affect automorphisms.

So for example, we have that (1; ) 4 (; 2), since a single unary operationcan be replaced by the binary relation that is its graph. In this case, theformer is strictly below the latter. The group D5 on the set 5 is theautomorphism group of a binary relation structure, but not of an algebrawith one unary function.

It is also possible to encode a number of relations in a function of greaterarity. For example, we can code the binary relations ρ0, ρ1 and ρ2 inthe 4-ary function f by setting f(w, x, y, z) to be w if wρjx and yotherwise, where j = 0 if z = w, j = 1 if z = y, and j = 2 if z isneither w nor y.

This example requires n ≥ 3 so that one can always have w, y and zdistinct, but that is a minor point since the cases n = 1 and n = 2 arefairly trivial. Thus (; 2, 2, 2) 4 (4; ). The above idea generalizes, and canbe used to show σ 4 τ in more instances.

For another example, consider structures on n with only unary relations.With one unary relation that is true on the set A, the automorphism groupof the structure is obviously Sym(A)� Sym(n \ A).

Structures with more unary relations have automorphism groups of theform Sym(A0)� Sym(A1)� · · ·� Sym(Ak−1), where {A0, A1, . . . Ak−1} isa partition of n. Note that once 2m ≥ n, having more than m unaryrelations in a type does not allow one to obtain any additional automorphismgroups on the set n.

These products of symmetric groups are not always automorphism groupsof structures with only unary operations. It is impossible to get even G =Sym({0, 1, 2}) � Sym({3, 4, 5}), the subgroup of Sym(6), as an automor-phism group of such a structure.

On the other hand, our Lemmas 2.1 and 2.7 give us that these innerproducts of symmetric groups are all automorphism groups of groupoids. So

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we have that the type consisting of a single binary operation is above all thetypes (; 1), (; 1, 1), (; 1, 1, 1), . . . in 4.

Yet some groups remain difficult to obtain. Let n ≥ 3, and considerAlt(n). As before, it is (n − 2)-transitive, and the methods of Lemma 3.5show that it is not the automorphism group of any structure that has onlyoperations of arity less than or equal to n − 3 and relations of arity lessthan or equal to n− 2.

However, the result of Armbrust and Schmidt used in the proof of Theo-rem 3.8 shows that Alt(n) is the automorphism group of a structure with asingle n-ary relation. Thus the type (;n) is strictly above all types of theform (n − 3, n − 3, . . . n − 3;n − 2, n − 2, . . . n − 2). (This was essentiallynoted by Jonsson in [12]).

This preorder on types also provides a natural statement of a questionposed by Gould in [9]; he asks for all m, k ≥ 1 if the type with m+1 k-aryoperations is strictly above the type with m k-ary operations.

For k = 1, this is true for all m ≥ 1. We will show this for m = 1,and our method clearly generalizes. Let f be a single unary operation on6, and consider the subgroup G of Sym(6) generated by (01), (23) and(45), also called 〈(01)〉 � 〈(23)〉 � 〈(45)〉. The reader is asked to verifythat there are only two ways that 〈(ij)〉 can be a factor in an inner directproduct equal to Aut(〈6; f〉). Either we have f(i) = j and f(j) = i,or we have f(i) = i and f(j) = j . Thus our best attempt to haveG = Aut(〈6; f〉) is to define f something like this: f(0) = 1, f(1) = 0,f(2) = 3, f(3) = 2, f(4) = 4, and f(5) = 5. Unfortunately, this does notquite work: (02)(13) is an additional automorphism of 〈6; f〉. So G is notthe automorphism group of a structure with a single unary operation. Butit is the automorphism group of a structure with two unary operations, forwe may use the above f and add g, where g(2) = 3, g(3) = 2, g(4) = 5,g(5) = 4, g(0) = 0, and g(1) = 1.

But for k = 2, the question remains open. Our examples of non-pertinentgroups are given by Corollary 3.6, Theorem 3.7 and Theorem 3.9. All of theirproofs may be easily modified to show that the groups in question are notautomorphism groups for structures with any number of binary operations.For all we know, it could be true that the type with one binary operation isabove the types with m binary operations, for all m. A nice statement ofa generalization of the question runs as follows.

Question: Is the preorder 4 actually a partial order?

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5 Acknowledgments

Our work benefited from our conversations with David Clark, Jacqueline R.Grace, William V. Grounds, and Allan J. Silberger.

References

[1] M. Armbrust and J. Schmidt, Zum Cayleyschen Darstellungssatz. Math-ematische Annalen 154 (1964), 70–73.

[2] G. Birkhoff, On the groups of automorphisms. Rev. Un. Mat. Argentina11(1946), 155 – 157.

[3] Stanley N. Burris and H. P. Sankappanavar, “A Course in UniversalAlgebra”. Springer-Verlag, 1981. ISBN 3-540-90578-2. (Also freely avail-able online at http://www.thoralf.uwaterloo.ca/htdocs/ualg.html)

[4] Chang and Kiesler, “Model Theory”. Elsevier Science Publishers V.P.,1992. ISBN 0 444 88054 2

[5] J. Dixon and B. Mortimer, “Permutation Groups”. New York: Springer-Verlag, 1996.

[6] F. G. Frobenius, Uber die Congruenz nach einem aus zwei endlichenGruppen gebildeten Doppelmodul. J. reine angew. Math. 101(1887),273 – 299.

[7] R. Frucht, Herstellung von Graphen mit vorgegebener abstrakterGruppe. Compositio Mathematica 6(1939), 239 – 250.

[8] M. Gould, A note on automorphisms of groupoids. Algebra Universalis,2 (1972), pp. 54-56.

[9] M. Gould, Automorphism groups of algebras of finite type. CanadianJournal of Mathematics 6(1972), 1065 – 1069.

[10] Marshall Hall, “The Theory of Groups”. Macmillan, 1959. (Theorem5.4.2.)

[11] M. Harrison, The number of isomorphism types of finite algebras. Proc.Amer. Math. Soc. 17(1966), No. 3, 731 – 737.

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[12] B. Jonsson, Algebraic structures with prescribed automorphism groups.Coll. Mathematicum XIX(1968), No. 1, 1 – 4.

[13] D. Knuth, “The Art of Computer Programming, Volume 2 Seminumer-ical Algorithms”. Addison-Wesley, 3rd Edition 1998. ISBN 0-201-89684-2. (Exercise 4.3.2.3)

[14] P.P. Palfy, L. Szabo, and A. Szendrei, Automorphism groups and func-tional completeness. Algebra Universalis 15 (1982), 385 – 400.

[15] Todd Rowland and Eric W. Weisstein, “Transitive Groups”. From Math-World – A Wolfram Web Resource.

http://mathworld.wolfram.com/TransitiveGroup.html

[16] S. Stein, Homogeneous quasigroups. Pacific J. Math 14 (1964), No. 3,1091 – 1102.

[17] L. Szabo, Algebras with transitive automorphism groups. Algebra Uni-versalis 31 (1994), 589 – 598.

Addresses1David Hobby and Donald Silberger. Department of Mathematics, State

University of New York, New Paltz NY 12561 – U.S.A.Emails: hobbyd@newpaltz.edu or silbergd@newpaltz.edu

2Sylvia Silberger. Department of Mathematics, Hofstra University, Hemp-stead NY 11550 – U.S.A.

Email: matsbs@hofstra.edu

2000 Mathematics Subject Classification: Primary: 20N02,08A35 Secondary:20B25,20B20Keywords: groupoids, automorphism groups

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