DISJUNCTIVE IDENTITIES OF FINITE GROUPS

AND IDENTITIES OF REGULAR REPRESENTATIONS

Samuel M. Vovsi

Institute for Advanced Study

1. Introduction

The problem of describing identities of concrete algebraic structures is traditionally one of the most attractive in variety theory. Very often it is a difficult problem, and many deep works are entirely devoted to finding bases of identities of classical algebraic objects.

This paper was initiated by attempts to find bases of identities for certain representations of finite groups. Let F be the free group of countable rank on an alphabet x_{1}, x_{2}, …, and let KF be its group algebra over a field K. Recall that an element u = u(x_{1}, …, x_{n}) ∈ KF is called an identity of a representation ρ : G → Aut_{K}V (usually denoted by ρ = (V, G)) if for arbitrary g_{1}, …, g_{n} ∈ G the element u(g_{1}, …, g_{n}) ∈ KG annihilates V. It is well known that if ρ is a representation of a finite group, then its identities are finitely based, that is, are in a natural sense generated by some finite set of them [V, VN]. The problem is how to find such a finite basis explicitly. Of course, there is no hope to develop a universal method allowing to write down a basis of identities for an arbitrary representation, so we will be concerned with a much more concrete (but still very general)

Problem A. Let G be a finite group, charK ∤ |G|. Find a basis of identities for the regular representation Reg_{K}G = (KG, G) of G over K.

Why the regular representations are of particular interest? The reason is two-fold. First, it is easy to understand that the identities of Reg_{K}G are the identities of all representations of G over K. Second, they can be regarded as special ”weak” identities of the group algebra KG, for they are exactly the elements u(x_{1}, …, x_{n}) ∈ KF such that u(g_{1}, …, g_{n}) = 0 in KG for all g_{i} ∈ G.

An interesting approach to Problem A was suggested by Plotkin and Kushkuley [PK]. Consider a universally quantified formula

δ = ∀x_{1} … x_{n}[(f_{1} = 1) ∨ (f_{2} = 1) ∨ ⋯ ∨ (f_{n} = 1)]  (1)

where x_{1}, …, x_{n} are all variables involved in the f_{i}. Probably, the first systematic study of such formulas was done by Baker [B], who called them universal disjunctions of equations

Typeset by AMS-TEX

(UDE). We say that δ is a disjunctive identity (d-identity, for short) of a group G if it is identically true on G. It is obvious that if δ is a d-identity of G then

δ^{∗} = (f_{1}^{y_{1}} − 1) … (f_{n}^{y_{n}} − 1) ∈ KF

is an identity of every representation of G. A principal observation made in [PK] is that if one knows the d-identities of a finite group G, then one can easily obtain a basis of identities of the regular representation Reg_{K}G. To make this more precise, we recall a few simple definitions and facts.

Let δ be an arbitrary d-identity, that is, a universal sentence of the form (1). Usually we omit the quantor prefix and write this formula simply as δ = (f_{1} = 1) ∨ (f_{2} = 1) ∨ ⋯ ∨ (f_{n} = 1). A class of groups 𝔇 is called a d-variety if it is definable by a set of d-identities.^{1} This is equivalent to saying that 𝔇 is definable by positive universal sentences, because every such a sentence is obviously equivalent to a conjunction of d-identities. Applying a well known fact of model theory [G; p. 275], we see that a class of groups is a d-variety if and only if it is closed under taking subgroups, homomorphic images, and ultraproducts.

If 𝔛 is an arbitrary class of groups, then by dvar(𝔛) we denote the d-variety generated by 𝔛. In other words, dvar(𝔛) is the class of all groups satisfying all d-identities that are valid in all groups from 𝔛. Let G be a finite group and let HS(G) be the class of all sections of G. Being a finite class of finite algebras, HS(G) is axiomatizable by a single sentence. Therefore it is closed under ultraproducts and so is a d-variety. Thus we have established

Fact 1. The d-variety generated by a finite group is the class of all sections of this group.^{2}

In particular, this implies that if G and H are finite groups then

dvar(G) = dvar(H) ⟺ G ≃ H.

In other words, every finite group is uniquely determined by its d-identities.

A set of d-identities {δ_{i}} of a class of groups 𝔛 is called a basis of d-identities of 𝔛 if every group satisfying the δ_{i} belongs to dvar(𝔛) (that is, satisfies all d-identities of 𝔛). It is easy to see that every finite group has a finite basis of d-identities. Indeed, let G be group of order n. Consider the d-identity

ω_{n} = ⋁_{0≤i<j≤n}(x_{i} = x_{j})

and for every group H_{i} of order at most n and not belonging to dvar(G) choose a d-identity δ_{i} of G which is not valid in H_{i}. Then it is clear that a finite set of formulas {ω_{n}, δ_{1}, δ_{2}, … } is a basis of d-identities of G.^{3}.

1Initially, Plotkin and Kushkuley [PK] used the words “pseudoidentity” and “pseudovariety”. We would also prefer to use this (pseudo)terminology but, unfortunately, it has been already occupied by Eilenberg and Schützenberger [ES].

2A direct proof of this fact can be found, for example, in [PV; Theorem 13.2.1].

3All of the above is based on general results of model theory and is valid not only for groups, but for arbitrary algebraic systems

We now discuss the connection between d-identities of groups and identities of their regular representations. We will consider representations of groups over an arbitrary but fixed field K. As defined earlier, identities of representations are elements of the group algebra KF. A class of group representations V is called a variety if it consists of all representations satisfying a given set of identities. A variety is finitely based if it is definable by a finite set of identities. Such a set is called a (finite) basis of identities of V. If V is the variety generated by some class of representations X (notation: V = var(X)), then the same set of identities is also called a basis of identities of X.

Let 𝔙 be a variety of groups. Take the class of all representations ρ : G → Aut_{K}V such that G/Ker ρ ∈ 𝔙. It is a variety of group representations: if 𝔙 is definable by group identities f_{i}, then ω𝔙 is defined by the identities f_{i} − 1. A variety of group representations V is locally finite if it satisfies the following condition: if ρ = (V, G) is an arbitrary representation in V such that G is a finitely generated group and V is a finitely generated G-module, then V is finite-dimensional over K and the group G/Ker ρ is finite. This is equivalent to saying that there exists a locally finite variety of groups 𝔙 such that V ⊆ ω𝔙 (see, for example, [V2; §2.1]) . If in addition this 𝔙 can be chosen in such a way that exp 𝔙 is not divisible by the characteristic of the ground field K, the variety V is called ordinary, or nonmodular.

Example. Let ρ = (V, G) be an ordinary representation of a finite group G (that is, char K ∤ |G|). Then it is easy to see that the variety generated by ρ is locally finite and ordinary. It is known that this variety is finitely based [V1]. Moreover, the latter is true for arbitrary (not necessary ordinary) representations of finite groups [VN].

Let G be a group and let {δ_{i}} be a set of d-identities of G. This set is called a weak basis of d-identities of G if every irreducible linear (over K) group satisfying all the δ_{i} belongs to dvar(G) (i.e. is a section of G). Of course, every basis of d-identities is a weak basis, but the converse is not, in general, true. The following fact plays a key role.

Fact 2 [PK]. Let G be a finite group and let U be a set of identities of the regular representation Reg_{K}G = (KG, G), satisfying the following conditions:

(i) the variety of representations defined by U is locally finite and ordinary; (ii) there exists a weak basis {δ_{i}} of d-identities of G such that U ⊇ {δ_{i}^{∗}}.

Then U is a basis of identities of Reg_{K}G.

P r o o f of this statement is not difficult, and we will provide it here. Consider a d-identity δ = (f_{1} − 1) ∨ ⋯ ∨ (f_{n} − 1), and let δ^{∗} = (f_{1}^{y_{1}} − 1) … (f_{n}^{y_{n}} − 1). First let us show that if δ^{∗} is an identity of some faithful irreducible representation ρ = (V, H), then δ must be an identity of the acting group H. Assume the contrary; then there is a homomorphism ϕ : F → H such that f_{i}^{ϕ} ≠ 1 for every i. Denote f_{i}^{ϕ} = g_{i} and take 0 ≠ a_{1} ∈ V. Since V is an irreducible KH-module, we have a_{1}KH = V. Since H acts on V faithfully and g_{1} ≠ 1, there exists h_{1} ∈ H such that a_{1} · h_{1}^{−1}g_{1} ≠ a_{1} · h_{1}^{−1}, that is, a_{1} · (g_{1}^{h_{1}} − 1) ≠ 0. Next, denote a_{1} · (g_{1}^{h_{1}} − 1) = a_{2} and find h_{2} ∈ H_{2} such that a_{3} = a_{2} · (g_{2}^{h_{2}} − 1) ≠ 0. Repeating this argument, we find elements h_{3}, …, h_{n} ∈ H such that

a · (g_{1}^{h_{1}} − 1)(g_{2}^{h_{2}} − 1) … (g_{n}^{h_{n}} − 1) ≠ 0.

But this is impossible because δ^{∗} is an identity of ρ.

Now let G and U be as in the statement. Let V be the variety generated by Reg G and let W be the variety defined by U. Then V ⊆ W; we have to show that V = W. By (ii), W is locally finite and ordinary, and therefore it is generated by faithful irreducible representations of finite groups. Take in W such a representation ρ = (V, H); it remains to show that ρ ∈ V.

Since the δ_{i}^{∗} belong to U, they all are identities of ρ. By the above, the δ_{i} are d-identities of H. Since H is an irreducible linear group and {δ_{i}} is a weak basis of didentities of G, it follows that H ∈ dvar(G), that is, H is a section of G. But then Reg H satisfies all identities that are valid in Reg G. Hence Reg H belongs to V, and the same is true for any representation of H. □

Since it is very easy to find identities of the representation Reg_{K}G which guarantee that the corresponding variety is locally finite and ordinary, the established fact reduces solution of Problem A to finding a weak basis of d-identities of a given finite group G. These ideas were developed in [PK] in several directions; in particular, they were applied to the study of identities of the regular representations of the symmetric groups S_{n}, n ≤ 5.

The present paper is focused on finding actual (not weak) bases of d-identities.

Problem B. For a given finite group G, find a basis of its d-identities.

This problem is more difficult than its “weak sibling” and is of much greater independent interest. Indeed, as mentioned earlier, every finite algebra is uniquely determined by its disjunctive identities, and therefore a basis of d-identities of a finite group uniquely identifies it. What is especially important, Problem B is in a certain sense “finite”, and if the order of the group in question is small enough, it can be solved by a rather mechanical procedure.

The aim of this paper is to collect a sufficient experimental material. It consists of six sections. In §2 we compute bases of d-identities for several small groups (like D_{8}, Q_{8}, S_{4}), but in §3 — bases of d-identities for A_{5} and S_{5}. In particular, in §3 we sharpen some results from [PK] and eliminate a few gaps from that work. §4 is devoted to finding a basis of didentities for A_{6}; technically it is the most difficult part of the paper. In §5, contrary to the previous sections, we find bases of d-identities for an infinite series of groups, namely, for all dihedral groups D_{2n}. The results of this section are joint with Gregory Cherlin. Finally, in §6 we return to our initial Problem A and explicitly write down bases of identities for the (nonmodular) regular representations of certain groups discussed in Sections 2–5.

The results of this work demonstrate some general principles of finding bases for didentities of finite groups. It is not excluded that a certain computer-based system (like GAP or CAYLEY) can be involved in this process, and we hope to investigate such a possibility in the future.

It should be noted that we essentially use some ideas and results from [PK]. Unfortunately, this interesting work has never been published in full form and exists only as a preprint (part of it was expounded in [PV; §13]). We are grateful to B. I. Plotkin for making this preprint available.

2. Several simple examples

The aim of this section is to illustrate, on several simple examples, some basic principles of finding bases of d-identities of finite groups. We will consider the groups D_{8}, Q_{8}, A_{4}, and S_{4}.

THE GROUP D_{8}. The dihedral group of order eight has the following presentation:

D_{8} = ⟨a, b | a^{4} = b^{2} = 1, a^{b} = a^{−1}⟩.

Clearly it satisfies the d-identities

ω_{8} = ⋁_{0≤i<j≤8}(x_{i} = x_{j})  (2.1)

and

x^{4} = 1.  (2.2)

Since D_{8} contains only two elements of order > 2, it also satisfies the d-identity

(x_{1}^{2} = 1) ∨ (x_{2}^{2} = 1) ∨ (x_{3}^{2} = 1) ∨ (x_{1} = x_{2}) ∨ (x_{1} = x_{3}) ∨ (x_{2} = x_{3}).  (2.3)

Let us prove that the formulas (2.1)–(2.3) form a weak basis of d-identities of D_{8}. We have to show that every irreducible linear group G satisfying (2.1)–(2.3) belongs to dvar(D_{8}), i.e. is a section of D_{8}. Since G satisfies ω_{8}, it must be of order at most 8. The groups of orders 3,5,6,7 are excluded by (2.2), but the groups of orders 1,2,4 are all sections of D_{8}. There are five groups of order 8:

Z_{8}, Z_{4} × Z_{2}, Z_{2} × Z_{2} × Z_{2}, Q_{8}, and D_{8},

where, as usual, Z_{n} denotes the cyclic group of order n and Q_{8} the quaternion group. Z_{8} does not satisfy (2.2), each of the groups Z_{4} × Z_{2} and Q_{8} has three distinct elements of order 4 and so does not satisfy (2.3), and Z_{2}^{3} is not irreducible linear (an abelian irreducible linear group must be cyclic). This completes the proof.

However, the formulas (2.1)–(2.2) do not form a basis of d-identities of D_{8} because Z_{2}^{3} obviously satisfies all of them but is not a section of D_{8}. To eliminate Z_{2}^{3}, we have to introduce another d-identity. For brevity, for any variables x and y we will write x ∈ ⟨y⟩ to denote

(x = 1) ∨ (x = y) ∨ (x = y^{2}) ∨ (x = y^{3}).

Consider the following formula in the variables x_{1}, x_{2}, x_{3}:

[⋁_{i≠j} x_{i} ∈ ⟨x_{j}⟩] ∨ [ ⋁_{σ∈S_{3}} x_{σ(1)} ∈ ⟨x_{σ(2)}x_{σ(3)}⟩] ∨ [ ⋁_{σ∈S_{3}} x_{σ(1)}x_{σ(2)} ∈ ⟨x_{σ(2)}x_{σ(3)}⟩] (2.4)

and show that it is a d-identity of D_{8}. The group D_{8} is a semidirect product

D_{8} = A ⋋ B, where A ≃ Z_{4}, B ≃ Z_{2}.

Note that if x, y are elements of D_{8} not in A, then xy ∈ A. Now take arbitrary x_{1}, x_{2}, x_{3} ∈ D_{8}. If two of the x_{i} lie in A then the first clause of (2.4) holds, if one of them lies in A and the other two outside then the second clause holds, and if all three x_{i} lie outside then the third clause holds.

On the other hand, (2.4) is not valid in Z_{2}^{3}: it is enough to take a basis of this group for x_{1}, x_{2}, x_{3}. Thus we have proved

Proposition 1. Formulas (2.1)–(2.4) form a basis of d-identities of D_{8}. □

Although the above arguments are rather elementary, they give a clear idea about the procedure of finding a basis of d-identities of a finite group G. If |G| = n, then we first take ω_{n}, and then for every group H of order ≤ n find a d-identity of G that is not valid in H. It is clear that the complexity of this procedure increases drammatically with the order of the group in question. In general, it is noticeably easier to find a weak basis of d-identities because in this case we do not have to think about groups not admitting faithful irreducible representations. In particular, we do not have to think about groups of the form Z_{p}^{n} which usually create most of the troubles.

THE GROUP Q_{8}. Recall that

Q_{8} = {±1, ±i, ±j, ±k},

where

i^{2} = j^{2} = k^{2} = −1,
ij = −ji = k, jk = −kj = i, ki = −ik = j,

and the other multiplications are natural. Obviously Q_{8} satisfies (2.1) and (2.2). Every group satisfying these two d-identities has order 1, 2, 4, or 8. The groups of order 1, 2 and 4 are sections of Q_{8}, but Z_{8} does not satisfy (2.2). It remains to eliminate three groups of order 8: Z_{4} × Z_{2}, Z_{2}^{3} and D_{8}. Note that

(x_{1} ∈ ⟨x_{2}⟩) ∨ (x_{2} ∈ ⟨x_{1}⟩) ∨ (x_{1}^{2}x_{2}^{2} = 1),  (2.5)

where x ∈ ⟨y⟩ has the same meaning as before, is valid in Q_{8}. For if x_{1} ∉ ⟨x_{2}⟩ and x_{2} ∉ ⟨x_{1}⟩, then both x_{i} are different from ±1 and therefore are of order 4. Hence x_{1}^{2} = x_{2}^{2} = −1 and x_{1}^{2}x_{2}^{2} = 1.

Next we observe that (2.5) eliminates both Z_{4} × Z_{2} and D_{8}. Indeed in the first case we simply take a basis of Z_{4} × Z_{2} for x_{1} and x_{2}, but in the second case let x_{1} and x_{2} be the standard generators of D_{8} (with |x_{1}| = 4 and |x_{2}| = 2). Then x_{1} ∉ ⟨x_{2}⟩ and x_{2} ∉ ⟨x_{1}⟩ and x_{1}^{2}x_{2}^{2} ≠ 1.

To eliminate Z_{2}^{3}, consider the following formula in x_{1}, x_{2}, x_{3}:
(⋁_{i≠j} x_{i} ∈ ⟨x_{j}⟩) ∨ (x_{1}x_{2} = x_{3}) ∨ (x_{1}x_{2} = x_{3}^{−1}).  (2.6)

Choosing a basis of Z_{2}^{3} for x_{1}, x_{2}, x_{3}, we see that (2.6) is not valid in this group. Now let x_{1}, x_{2}, x_{3} ∈ Q_{8}. If one of the x_{i} is equal to ±1, then the first clause of (2.6) holds, otherwise all x_{i} are elements of order 4. There are three cyclic subgroups of order 4 in Q_{8}, generated by i, j and k, respectively. If two of the x_{i} belong to one of these subgroups then again the first clause holds, but if all three of them lie in different subgroups then either the second or the third clause holds. Thus (2.6) is a d-identity of Q_{8}, and we obtain

Proposition 2. Formulas (2.1), (2.2), (2.5) and (2.6) form a basis of d-identities of Q_{8}. □

THE GROUP A_{4}. The alternating group of degree 4 contains three involutions (12)(34), (13)(24), (14)(23); all other nonunit elements are cycles of length 3. Therefore A_{4} satisfies the d-identity

(x^{2} = 1) ∨ (x^{3} = 1)  (2.7)

and, of course, ω_{12}. The order of a group satisfying ω_{12} and (2.7) must be one of the numbers

1, 2, 3, 4, 6, 8, 9, 12.

All groups of orders 1,2,3 are sections of A_{4}. The cyclic group of order 4 does not satisfy (2.7), but the noncyclic one is a subgroup of A_{4}. There are two groups of order 6: Z_{6} and S_{3}. The first one does not satisfy (2.7); as to the second one, consider the formula

(x^{3} = 1) ∨ (x_{2}^{3} = 1) ∨ ((x_{1}x_{2})^{2} = 1).  (2.8)

Seting x_{1} = (12), x_{2} = (13), we see that it is not valid in S_{3}. But it is valid in A_{4}. Indeed, take arbitrary x_{1}, x_{2} ∈ A_{4}. Observe that

H = {1, (12)(34), (13)(24), (14)(23)}

is a normal subgroup of exponent 2 in A_{4}. If one of the x_{i} has order 3, we are done; otherwise both x_{1} and x_{2} belong to H, hence x_{1}x_{2} ∈ H and (x_{1}x_{2})^{2} = 1.

All groups of order 8 except Z_{2}^{3} have elements of order 4 and therefore do not satisfy (2.7). To eliminate Z_{2}^{3}, we note that the unique Sylow 2-subgroup H of A_{4} is a 2-dimensional vector space over GF (2), while Z_{2}^{3} is 3-dimensional over GF (2). Therefore the d-identity

⋁_{(k_{1},k_{2},k_{3})}(x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}} = 1),

where (k_{1}, k_{2}, k_{3}) runs over all nonzero {0, 1}-vectors of length 3, is valid in H but not in Z_{2}^{3}. Further, for every x ∈ A_{4} we have x^{3} ∈ H. It follows that

⋁_{(k_{1},k_{2},k_{3})}(x_{1}^{3k_{1}}x_{2}^{3k_{2}}x_{3}^{3k_{3}} = 1)  (2.9)

is a d-identity of A_{4} but not of Z_{2}^{3}.

To eliminate the groups of order 9, we first prove a simple lemma which will be repeatedly used in the sequel.

Lemma 1. Let α and β be two n-cycles in S_{n+1}. Then for some d ≤ n − 1 the permutation αβ^{d} is not an n-cycle.

P r o o f. a) Suppose that Supp(α) = Supp(β). Without loss of generality, let

α = (1, 2, 3, …, n), β = (1, b_{2}, b_{3}, …, b_{n})

where 2 ≤ b_{i} ≤ n. For any d, the permutation αβ^{d} fixes n + 1, hence it is enough to show that for some natural d ≤ n − 1 it also fixes the number 1. Let d = d_{β}(2, 1) be the distance from 2 to 1 in β, that is, d is the smallest number with 2(β^{d}) = 1. Then d ≤ n − 1 and 1(αβ^{d}) = (1α)β^{d} = 2β^{d} = 1, as required.

b) Suppose that Supp(α) ≠ Supp(β). We may assume that

α = (1, 2, 3, …, n), β = (b_{1}, b_{2}, …, b_{n−1}, n + 1)

where 1 ≤ b_{i} ≤ n − 1. Note that (n − 1)αβ^{d} = n for any d. Set d = d_{β}(1, n − 1), then n(αβ^{d}) = (nα)β^{d} = 1β^{d} = n − 1 and so the cycle form of αβ^{d} contains a transposition (n − 1, n). It follows that αβ^{d} is not an n-cycle. □

It is now easy to see that

(x_{1}^{4} = 1) ∨ (x_{2}^{4} = 1) ∨ ((x_{1}x_{2})^{4} = 1) ∨ ((x_{1}x_{2}^{2})^{4} = 1)  (2.10)

is a d-identity of S_{4} (and so of A_{4}). Indeed, if x_{1}, x_{2} ∈ S_{4} and x_{i}^{4} ≠ 1, then x_{1}, x_{2} are 3-cycles. By Lemma 1, either x_{1}x_{2} or x_{1}x_{2}^{2} is not a 3-cycle, and (2.10) follows. On the other hand, (2.10) can not be valid in a group of order 9.

Finally, all groups of order 12 except A_{4} contain elements of order 6 and do not satisfy (2.7). Puting all this together, we have

Proposition 3. Formulas ω_{12} and (2.7)–(2.10) form a basis of d-identities of A_{4}. □

Note that (2.9) was only used to eliminate Z_{2}^{3}, but (2.10) — to eliminate Z_{3}^{2} (the only other group of order 9 is cyclic and does not satisfy (2.7)). Since none of these two groups admits a faithful irreducible representation, it follows that the formulas ω_{12}, (2.7) and (2.8) form a weak basis of d-identities of A_{4}.

THE GROUP S_{4}. First we introduce three new d-identities:

(x^{3} = 1) ∨ (x^{4} = 1),  (2.11)
(x_{1}^{6} = 1) ∨ (x_{2}^{6} = 1) ∨ (x_{1} = x_{2}) ∨ (x_{1}x_{2} = 1) ∨ ((x_{1}x_{2})^{3} = 1),  (2.12)

(x_{1}^{3} = 1) ∨ (x_{2}^{3} = 1) ∨ (x_{3}^{3} = 1)∨

∨ ((x_{1}^{−1}x_{2})^{3} = 1) ∨ ((x_{1}^{−1}x_{3})^{3} = 1) ∨ ((x_{2}^{−1}x_{3})^{3} = 1) ∨ θ, (2.13)

where θ denotes the formula (2.4).

Proposition 4. Formulas ω_{24} and (2.10)–(2.13) form a basis of d-identities of S_{4}.

P r o o f. a) First we show that these d-identities are valid in S_{4}. This was already verified for (2.10) and is obvious for ω_{24} and (2.11). Next, take x_{1}, x_{2} ∈ S_{4} such that x_{i}^{6} ≠ 1. Then the x_{i} are 4-cycles and a straightforward verification shows that, unless x_{1} = x_{2}^{±1}, x_{1}x_{2} is a 3-cycle. Hence (2.12) is valid in S_{4}.

Further, take arbitrary x_{1}, x_{2}, x_{3} ∈ S_{4} and show that they satisfy (2.13). If one of the x_{i} is a 3-cycle, we are done; otherwise all the x_{i} are 2-elements. Suppose they all belong to the same Sylow 2-subgroup of S_{4}. This subgroup is isomorphic to D_{8}; since θ is identically true in D_{8}, the x_{i} satisfy (2.13). Now suppose that two of the x_{i}, say x_{1} and x_{2}, belong to distinct Sylow 2-subgroups of S_{4}. The Sylow 2-subgroup of S_{4} are

H_{1} = ⟨H, (12)⟩, H_{2} = ⟨H, (13)⟩, H_{3} = ⟨H, (23)⟩,

where H is as above. Let for example x_{1}^{−1} = h_{1} (12), x_{2} = h_{2} (13) with h_{i} ∈ H. Then x_{1}^{−1}x_{2} = h_{3} (123) with h_{3} ∈ H. If (x_{1}^{−1}x_{2})^{3} ≠ 1, then x_{1}^{−1}x_{2} lie in some Sylow 2-subgroup of S_{4}, say x_{1}^{−1}x_{2} = h_{4} (23) with h_{4} ∈ H. But then (123)(23) = h_{3}^{−1}h_{4} ∈ H which is impossible because (123)(23) is an odd permutation. Thus (x_{1}^{−1}x_{2})^{3} = 1 and (2.13) holds.

b) Now we show that every group satisfying ω_{24} and (2.10)–(2.13) is a section of S_{4}. The order of a group satisfying ω_{24} and (2.11) must be among the numbers

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24.

All groups of order ≤ 4 are subgroups of S_{4}. The cyclic group of order 6 does not satisfy (2.11) but the noncyclic one is a subgroup of S_{4}. Among the five groups of order 8, Z_{8} does not satisfy (2.11); Z_{4} × Z_{2} does not satisfy (2.12) (set x_{1} = (a, 1), x_{2} = (a, b) where a, b is the natural basis of Z_{4} × Z_{2}); Z_{2}^{3} is eliminated by (2.13) (take a basis of this group for the x_{i}); D_{8} is a subgroup of S_{4}; and Q_{8} is eliminated by (2.12) (since there are two elements i and j of order 4 in Q_{8} whose product ij = k is also of order 4).

Every group of order 9 does not satisfy (2.10). The same is true for groups of order 18. Among the groups of order 12, A_{4} is a subgroup of S_{4} but the others contain elements of order 6 and do not satisfy (2.11). It is easy to see that every group of order 16 has an abelian section of order 8, and therefore is eliminated by either (2.11), or (2.12), or (2.13). It remains to consider groups of order 24.

Lemma 2. Let G be a group of order 24 without elements of order 6. Then G ≃ S_{4}.

P r o o f. Suppose there exists A ◁ G such that |A| = 3. Take in A an element a of order 3, and let S be a Sylow 2-subgroup of G. Then |S/C_{S}(A)| is 1 or 2. In both cases S has an element b of order 2 centralizing A. But then ab is an element of order 6, which is impossible. Thus O_{3}(G) = 1.

Let T be a Sylow 3-subgroup of G. Then T ⋪ G and so j = |G : N_{G}(T)| ≠ 1. Since j ≡ 1( mod 3) and j|8, we have j = 4, whence |N_{G}(T)| = 6. The group G acts on the right cosets of N = N_{G}(T) by right multiplication. This determines a homomorphism G → S_{4} with kernel K = ⋂_{g∈G} N^{g}. Since T ⋪ G, |T | = 3 and |N| = 6, it is clear that N ⋪ G. Therefore K is proper in N, and so |K| is 1 or 2 or 3. But |K| = 3 is impossible because O_{3}(G) = 1. If |K| = 2 then, since K, T ◁ N and K ∩ T = 1, we have KT ≃ Z_{6}, which is also impossible. Thus |K| = 1 and so G ≃ S_{4}. □

It follows that (2.11) is not valid in every group of order 24, except S_{4}. This completes the proof of Proposition 4. □

Note. It was claimed in [PK] that ω_{24}, (2.10), (2.11), plus the formula

[ ⋁_{1≤i≤9} x_{i}^{3} = 1] ∨ [ ⋁_{1≤i<j≤9}(x_{i}x_{j}^{−1})^{3} = 1]

form a weak basis of d-identities of S_{4} (this result was also announced in [P], p.80). This is not quite true, for it is easy to see that the quaternion group Q_{8} satisfies all these d-identities, has a faithful irreducible representation, but is not a section of S_{4}.

3. Disjunctive identities of A_{5} and S_{5}

In the previous section we considered groups of very small orders, and it is not surprising that our arguments were so short and elementary. In this section we deal with groups of slightly higher order, and immediately the difficulty of the proofs increases.

In the sequel, we often use the following immediate consequence from the Sylow Theorem: if a set of d-identities eliminates all groups of order p^{k} with p a prime, then the same set eliminates all groups of the order divisible by p^{k}. Also, we sometimes informally say that the permutation (123)(45) has a (cycle) form (∗ ∗ ∗)(∗∗), the permutation (15)(32) has form (∗∗)(∗∗), etc.

THE GROUP A_{5}. 1. Every nonunit element of A_{5} has one of the following three cycle forms: (∗ ∗ ∗ ∗ ∗), (∗ ∗ ∗), (∗∗)(∗∗). Therefore A_{5} satisfies the d-identity

(x^{2} = 1) ∨ (x^{3} = 1) ∨ (x^{5} = 1)  (3.1)

and, of course, ω_{60}. These two d-identities may both be valid in the groups of the following orders only:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60.

The groups of order 1,2,3,5 are all subgroups of A_{5}. The cyclic group of order 4 does not satisfy (3.1), but the noncyclic one is a subgroup of A_{5}. Similarly, the cyclic group of order 6 does not satisfy (3.1), but the noncyclic one (namely, S_{3}) is a subgroup of A_{5}: it can be generated, for example, by (123) and (12)(45). All groups of order 8, except Z_{2}^{3}, have elements of order 4 and do not satisfy (3.1).

2. The group Z_{2}^{3} requires more efforts. First we note that if

a = (a_{1}a_{2})(a_{3}a_{4}) and b = (b_{1}b_{2})(b_{3}b_{4})

are two elements of order 2 in A_{5}, then there are three possibilities:

(i) Supp(a) = Supp(b), say a = (12)(34) and b = (13)(24). Then ab = ba = (14)(23), so a and b are contained in the same Sylow 2-subgroup of A_{5}.

(ii) |Supp(a) ∩ Supp(b)| = 3 and a and b have a common cycle, say a = (12)(34) and b = (12)(35). Then ab = (345), so |ab| = 3.

(iii) |Supp(a) ∩ Supp(b)| = 3 and a and b have no common cycles, say a = (12)(34) and b = (15)(24). Then ab = (14325), so |ab| = 5.

Now consider the following formula in three variables x_{1}, x_{2}, x_{3}:

[⋁_{i} x_{i}^{15} = 1] ∨ [⋁_{i<j}(x_{i}x_{j})^{15} = 1] ∨ [ ⋁_{(i,j,k)≠0} x_{1}^{i}x_{2}^{j}x_{3}^{k} = 1]  (3.2)

where the last disjunction is taken over all nonzero {0, 1}-vectors (i, j, k). Take any x_{1}, x_{2}, x_{3} ∈ A_{5}. If the first two clauses of (3.2) are false then all x_{i} are elements of order 2 belonging to the same Sylow 2-subgroup H of A_{5}. Since H is a 2-dimensional vector space over GF (2), the last clause must hold. Thus (3.2) is valid in A_{5}. On the other hand, it is obviously false in Z_{2}^{3}.

Let G be any group whose order is divisible by 8 and which satisfies (3.1). Then G has a subgroup H of order 8; in view of (3.1), exp H = 2 and so H ≃ Z_{2}^{3}. This eliminates all groups of orders 16, 24, 32, 40 and 48.

3. There are two groups of order 9: Z_{9} and Z_{3} × Z_{3}. The first does not satisfy (3.1). To eliminate the second, take the formula

(x_{1}^{10} = 1) ∨ (x_{2}^{10} = 1) ∨ (x_{1} = x_{2}) ∨ ((x_{1}x_{2})^{10} = 1) ∨ ((x_{1}^{2}x_{2})^{2} = 1).  (3.3)

If x_{1}, x_{2} ∈ A_{5} and x_{i}^{10} ≠ 1, then x_{i} are cycles of length 3. Without loss of generality we assume that x_{1} = (123). If x_{2} moves the same symbols 1,2,3, then either x_{1} = x_{2} or x_{1}x_{2} = 1. If x_{2} moves only one of the symbols 1,2,3, then x_{1}x_{2} is a cycle of length 5 and (x_{1}x_{2})^{5} = 1. If x_{2} moves two of the symbols 1,2,3, then there are two essentially different cases: x_{2} = (124) and x_{2} = (214). In the first case

x_{1}x_{2} = (123)(124) = (14)(23) and so (x_{1}x_{2})^{2} = 1,

in the second case

x_{1}^{2}x_{2} = (132)(214) = (13)(24) and so (x_{1}^{2}x_{2})^{2} = 1.

It follows that (3.3) is valid in A_{5}. Clearly it is not valid in Z_{3} × Z_{3}, and so the groups of order 9 are eliminated. The same is true for groups of orders 18, 27, 36, 45 and 54.

4. There are two groups of order 10: Z_{10} and D_{10}. The first does not satisfy (3.1) but the second is a subgroup of A_{5}. There are five groups of order 12: Z_{12}, Z_{2} × Z_{2} × Z_{3}, A_{4}, D_{12} and also the group H = ⟨a, b | a^{3} = b^{4} = 1, a^{b} = a^{−1}⟩. All of them except A_{4} have elements of order 6 and therefore do not satisfy (3.1), but A_{4} is a subgroup of A_{5}. The only group of order 15 is cyclic. Every group G of order 20 has only one Sylow 5-subgroup, and so G = Z_{5} ⋋ T where |T | = 4. Since G may not have elements of order 4, we have T = Z_{2} × Z_{2}. But then C_{T} (Z_{5}) ≠ 1, and so G contains an element of order 10 which again contradicts (3.1). Thus the groups of order 10, 12, 15 and 20 are eliminated.

5. Let G be a group of order 30. It is solvable and has a composition series

1 ◁ A ◁ B ◁ G

with factors of orders 2, 3 and 5. Therefore the order of B is either 15, or 10, or 6. If |B| = 15, then B = Z_{15} and (3.1) fails. If |B| = 10, then either B ≃ Z_{10} or B ≃ D_{10}. The first case is again impossible, but in the second case A ≃ Z_{5} is normal subgroup of G and so G/A is an extension of a group B/A of order 2 by a group of order 3. But then G/A ≃ Z_{6} and (3.1) fails.

Finally, if |B| = 6, then we may assume that B ≃ S_{3}. In this case A ≃ A_{3} is characteristic in B and so A ◁ G. Therefore |G/A| = 15, whence G/A ≃ Z_{15}. This eliminates all groups of order 30. Similar arguments apply to solvable groups of order 60; the only nonsolvable group of order 60 is A_{5} itself.

6. To eliminate the groups of order 25, consider the formula

(x_{1}^{6} = 1) ∨ (x_{2}^{6} = 1) ∨ ((x_{1}x_{2})^{6} = 1)

∨ ((x_{1}x_{2}^{2})^{6} = 1) ∨ ((x_{1}x_{2}^{3})^{6} = 1) ∨ ((x_{1}x_{2}^{4})^{6} = 1). (3.4)

It is valid in A_{5}. Indeed, take x_{1}, x_{2} ∈ A_{5} with x_{i}^{6} ≠ 1. Then the x_{i} are 5-cycles. By Lemma 1, there exists d (1 ≤ d ≤ 4) such that x_{1}x_{2}^{d} is not a 5-cycle, that is (x_{1}x_{2}^{d})^{6} = 1. On the other hand, (3.4) fails in any group of order 25: take an element x_{2} ∈ G of order 5 and any element x_{1} not in ⟨x_{2}⟩. This eliminates the groups of order 25 and 50 and completes our argument.

Theorem 1. Formulas ω_{60} and (3.1)–(3.4) form a basis of d-identities of A_{5}. □

THE GROUP S_{5}. 1. We begin with two standard d-identities: ω_{120} and

(x^{4} = 1) ∨ (x^{5} = 1) ∨ (x^{6} = 1).  (3.5)

They both are valid in S_{5}, and if they are valid in some group G, then the order of G must be among the numbers

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48,
50, 54, 60, 64, 72, 75, 80, 81, 90, 96, 100, 108, 120.

Noting that all groups of order ≤ 6 are subgroups of S_{5}, consider the groups of order 8. Among them, Z_{8} does not satisfy (3.5), but D_{8} is a subgroup of S_{5}. To eliminate Z_{2} × Z_{4} and Q_{8}, we take the formula

(x_{1}^{30} = 1) ∨ (x_{2}^{30} = 1) ∨ ((x_{1}x_{2})^{15} = 1) ∨ ((x_{1}x_{2}^{2})^{15} = 1) ∨ ((x_{1}x_{2}^{3})^{15} = 1).  (3.6)

It is easy to see that it is not valid in either Z_{2} × Z_{4} or Q_{8}. We claim that it holds in S_{5}. Take x_{1}, x_{2} ∈ S_{5} such that x_{i}^{30} ≠ 1. Then both x_{1} and x_{2} are 4-cycles, and the claim will follow from

Lemma 3. If x_{1} and x_{2} are 4-cycles in S_{5}, then

((x_{1}x_{2})^{15} = 1) ∨ ((x_{1}x_{2}^{2})^{15} = 1) ∨ ((x_{1}x_{2}^{3})^{15} = 1).

P r o o f. Without loss of generality we may assume that x_{1} = (1234), and let x_{2} = (a_{1}a_{2}a_{3}a_{4}). Consider two cases.

(i) {a_{1}, a_{2}, a_{3}, a_{4}} = {1, 2, 3, 4}. We may assume that x_{2} = (1 ∗ ∗∗), and then there are 6 choices for x_{2}. A direct verification shows that, unless x_{2} = x_{1}^{±1}, x_{1}x_{2} is a 3-cycle and then (x_{1}x_{2})^{15} = 1. If x_{2} = x_{1}^{−1} then x_{1}x_{2} = 1, but if x_{2} = x_{1} then x_{1}x_{2}^{3} = x_{1}^{4} = 1.

(ii) {a_{1}, a_{2}, a_{3}, a_{4}} ≠ {1, 2, 3, 4}. We may assume that {a_{1}, a_{2}, a_{3}, a_{4}} = {2, 3, 4, 5} and x_{2} = (∗ ∗ ∗ 5). Again there are 6 choices for x_{2}. For each of them, a direct verification shows that x_{1}x_{2}^{k} is a 5-cycle for some k = 1, 2, 3. □

2. The next step is to eliminate Z_{2}^{3}. As usual, it is the most painful. Recall that every Sylow 2-subgroup of S_{5} is isomorphic to D_{8} and therefore contains precisely two 4-cycles which uniquely determine it. Namely, the Sylow 2-subgroup containing a cycle (abcd) consists of the elements

1, (abcd), (adcb), (ac), (bd), (ab)(cd), (ac)(bd), (ad)(bc).

We will denote it by Syl(abcd). From this description, it is easy to deduce that if y_{1}, y_{2}, y_{3} are 2-elements of S_{5} such that any two of them are contained in one Sylow subgroup, then all three of them are contained in one Sylow subgroup.

Take three variables x_{1}, x_{2}, x_{3} and denote x_{i}^{15} = y_{i}. Let θ(x_{1}, x_{2}, x_{3}) be the d-identity (2.4). Consider the formula

θ(y_{1}, y_{2}, y_{3}) ∨ [⋁_{i} y_{i} = 1] ∨ [⋁_{i≠j}(y_{i}y_{j})^{15} = 1]

∨ [ ⋁_{j≠i≠k}(y_{i}y_{j}y_{k})^{15} = 1] ∨ [ ⋁_{j≠i≠k}(y_{i}y_{j}y_{k}y_{j})^{15} = 1], (3.7)

where j ≠ i ≠ k means that j ≠ i and i ≠ k (but j and k may coincide). A basis of Z_{2}^{3} refutes (3.7) in this group. Our goal is to show that (3.7) is valid in S_{5}. This is trivially true if y_{i} = x_{i}^{15} = 1 for some i, so we assume that y_{i} ≠ 1 for all i. Hence the y_{i} are elements of order 2 or 4.

Next, suppose that y_{1}, y_{2}, y_{3} lie in the same Sylow 2-subgroup of S_{5}. Since this subgroup is isomorphic to D_{8} and θ is a d-identity of D_{8}, it follows that θ(y_{1}, y_{2}, y_{3}) is valid, and so is (3.7). Thus we may assume that none of the Sylow 2-subgroups of S_{5} contains all three y_{i} simultaneously. It follows from the above property of the Sylow 2subgroups that two of the elements y_{i} do not belong to the same Sylow 2-subgroup. From now on, we assume that these elements are y_{1} and y_{2}. There are three cases.

(i) Both y_{1} and y_{2} are 4-cycles. Then by Lemma 3 we have ((y_{1}y_{2})^{15} = 1)∨((y_{1}y_{2}^{2})^{15} = 1) ∨ ((y_{1}y_{2}^{3})^{15} = 1), and (3.7) holds.

(ii) Only one of the y_{1} and y_{2} is a 4-cycle. Let for example y_{1} = (1234). Then there are two possibilities.

a) y_{2} is a transposition. Note that y_{2} can not be equal to (13) or (24) because both these transpositions belong to Syl(1234). If Supp(y_{2}) ⊂ Supp(y_{1}) then y_{1}y_{2} is a 3-cycle (examples: (1234)(12) = (234), (1234)(14) = (123)). If Supp(y_{2}) ⊈ Supp(y_{1}) then, without loss of generality, we may assume that y_{2} = (15), whence y_{1}y_{2} = (12345). In all cases (y_{1}y_{2})^{15} = 1 and (3.7) holds.

b) y_{2} has form (∗∗)(∗∗). Since y_{2} ∉ Syl(1234), it must involve symbol 5. Essentially, there may be only three possibilities:

y_{2} = (15)(23) and then y_{2}y_{1}^{2} = (15342),
y_{2} = (15)(24) and then y_{2}y_{1}^{2} = (153),
y_{2} = (15)(34) and then y_{2}y_{1}^{2} = (15324).

In all cases (y_{2}y_{1}^{2})^{15} = 1 and (3.7) is valid.

(iii) None of the y_{1} and y_{2} is a 4-cycle. Then both y_{1} and y_{2} are involutions and therefore the subgroup they generate in S_{5} is a dihedral group:

⟨y_{1}, y_{2}⟩ = ⟨y_{1}y_{2}⟩ ⋋ ⟨y_{2}⟩ ≃ D_{2m},

where ⟨y_{1}y_{2}⟩ ≃ Z_{m}. Since y_{1} and y_{2} do not lie in the same Sylow 2-subgroup, y_{1}y_{2} may not be a 2-element. It follows that the order of y_{1}y_{2} is either 3, or 5, or 6. In the first two cases (y_{1}y_{2})^{15} = 1. In the last case y_{1}y_{2} = (∗ ∗ ∗)(∗∗), from which one can easily see that

y_{1} = (∗∗)(∗∗), y_{2} = (∗∗), Supp(y_{1}) ∪ Supp(y_{2}) = {1, 2, 3, 4, 5}.

Without loss of generality we may assume that y_{1} = (12)(34) and y_{2} = (15). Again we have two possibilities.

a) y_{3} is a transposition. If y_{2} and y_{3} overlap, then (y_{2}y_{3})^{3} = 1. Otherwise there are two essentially different cases:

y_{3} = (23) and then y_{1}y_{3}y_{2} = (13425),
y_{3} = (34) and then y_{1}y_{3}y_{2} = (125).

In both cases (y_{1}y_{3}y_{2})^{15} = 1.

b) y_{3} has form (∗∗)(∗∗). Suppose first that Supp(y_{3}) = {1, 2, 3, 4}. If y_{3} = y_{1}, then y_{1}y_{3} = 1 and there is nothing to prove. All other possibilities are essentially equivalent to the case when y_{3} = (13)(24). Then we have y_{1}y_{2}y_{3}y_{2} = (14532) and so (y_{1}y_{2}y_{3}y_{2})^{15} = 1. Now suppose that Supp(y_{3}) ≠ {1, 2, 3, 4}. Then we have two essentially different cases (y_{2} can now be disregarded):

y_{3} = (15)(34) and then y_{1}y_{3} = (125),
y_{3} = (15)(24) and then y_{1}y_{3} = (14325).

In both cases (y_{1}y_{3})^{15} = 1. Thus we have proved that (3.7) is a d-identity of S_{5}. This concludes the consideration of groups of order 8.

3. Consider the d-identity

(x_{1}^{20} = 1) ∨ (x_{2}^{20} = 1) ∨ (x_{1}^{2} = x_{2}^{2}) ∨ ((x_{1}^{2}x_{2}^{2})^{5} = 1) ∨ ((x_{1}^{4}x_{2}^{2})^{2} = 1).  (3.8)

Note that if δ(x_{1}, x_{2}) is the formula (3.3), then δ(x_{1}^{2}, x_{2}^{2}) is precisely (3.8). Since (3.3) is a d-identity of A_{5} and x_{i}^{2} ∈ A_{5} for any x_{i} ∈ S_{5}, we see that (3.8) is valid in S_{5}. On the other hand, it obviously fails in the groups of order 9 and so in the groups of orders 18, 27, 36, 45, 54, 72, 81, 90, 108.

4. Among the two groups of order 10, the cyclic one does not satisfy (3.5), but the noncyclic one is a subgroup of S_{5}. Among the groups of order 12, Z_{12} does not satisfy (3.5), but A_{4} and D_{12} are subgroups of S_{5}. To eliminate Z_{2} × Z_{6}, consider the d-identity

(x_{1}^{15} = 1) ∨ (x_{2}^{15} = 1) ∨ (x_{1}^{4} = 1) ∨ (x_{2}^{4} = 1) ∨ ((x_{1}x_{2})^{15} = 1) ∨ ((x_{1}x_{2})^{4} = 1). (3.9)

Let x_{1}, x_{2} ∈ S_{5} and suppose that the first four clauses of (3.9) are false. Then both x_{1} and x_{2} are of order 6, i.e. have form (∗ ∗ ∗)(∗∗). But then x_{1}x_{2} belongs to A_{5}, and so either (x_{1}x_{2})^{4} = 1 or (x_{1}x_{2})^{15} = 1. Thus (3.9) holds in S_{5}. On the other hand, it fails in Z_{2} × Z_{6}: if x_{1} = (1, b) and x_{2} = (a, b), where a, b is a natural basis of Z_{2} × Z_{6}, then x_{1}, x_{2}, x_{1}x_{2} are all elements of order 6.

The remaining group of order 12 is

G = ⟨a, b | a^{3} = b^{4} = 1, a^{b} = a^{−1}⟩ = Z_{3} ⋋ Z_{4}.

It is eliminated by the following d-identity from [PK] (see also [P], p.81):

(x_{1}^{10} = 1) ∨ (x_{2}^{10} = 1) ∨ (x_{1}^{4} = x_{2}^{4}) ∨ (x_{1}^{6} = x_{2}^{6})

∨ ((x_{1}^{2}x_{2}^{2})^{3} = 1) ∨ ((x_{1}^{2}x_{2}^{2})^{4} = 1) ∨ ((x_{1}^{2}x_{2}^{2})^{5} = 1). (3.10)

Leting x_{1} = a, x_{2} = b, we see that it fails in G. It is proved in [PK] that (3.10) holds in S_{5} (if the first four clauses of (3.10) fail, then one of the x_{i} has order 3 or 6 and the other has order 4; all such possibilities can be checked one by one).

5. Now we consider groups of orders 15, 30, 60, 120. The only group of order 15 is Z_{15} which does not satisfy (3.5). Every solvable group of order 15m with (15, m) = 1 has a subgroup of order 15, and therefore all solvable groups of orders 30, 60, 120 are also eliminated. The only nonsolvable group of order ≤ 60 is A_{5} which is a subgroup of S_{5}. Finally, there are only three nonsolvable groups of order 120, namely S_{5}, SL(2, 5) and A_{5} × Z_{2}, but the latter two have elements of order 10 and do not satisfy (3.5).

6. Every group of order 16 has an abelian section A of order 8. If exp A ≤ 4, then A is either Z_{2} × Z_{4} or Z_{2}^{3} and hence does not satisfy (3.6) or (3.7), respectively. Otherwise A does not satisfy (3.5). This eliminates all groups of orders 16, 32, 48, 64, 80, 96.

7. Every group G of order 20 has a unique Sylow 5-subgroup. It follows that G = H ⋋ T, where H ≃ Z_{5}, |T | = 4. Let C = C_{T} (H). If C ≠ 1, then G has an element of order 10 and does not satisfy (3.5). If C = 1, then it is easy to see that T ≃ Z_{4} and G is isomorphic to the subgroup of S_{5} generated by (12345) and (2354).

Similarly, every group of G order 40 can be presented as G = H ⋋ T, where H ≃ Z_{5}, |T | = 8. Since Aut Z_{5} ≃ Z_{4}, we see that C_{T} (H) ≠ 1 and so G must have an element of order 10.

8. The next step is to eliminate groups of orders divisible by 25. This can be done by the d-identity

(x_{1}^{12} = 1) ∨ (x_{2}^{12} = 1) ∨ ((x_{1}x_{2})^{12} = 1)

∨ ((x_{1}x_{2}^{2})^{12} = 1) ∨ ((x_{1}x_{2}^{3})^{12} = 1) ∨ ((x_{1}x_{2}^{4})^{12} = 1), (3.11)

using the same arguments as for the group A_{5} (cf. (3.4)). Moreover, it is easy to see that (3.11) is valid even in S_{6}. Indeed, if x_{1}, x_{2} ∈ S_{6} and x_{i}^{12} ≠ 1, then x_{i} are 5-cycles, and the rest follows from Lemma 1.

9. It remains to consider the groups of order 24. We will prove that every group G of order 24 satisfying the above d-identities is a section of S_{5}. We may assume that G has no elements of order 12 (by (3.5)) and that all of its sections of order 8 are isomorphic to D_{8} (because the other groups of order 8 have been eliminated).

The proof is similar to that of Lemma 2. Note that the number of Sylow 3-subgroups of G may be 1 or 4. In the first case

G = A ⋋ B where A ≃ Z_{3}, B ≃ D_{8}.

Let C = C_{B}(A), then |B/C| is 1 or 2. If |B/C| = 1, then G = A × B and G must have an element of order 12, which is impossible. If |B/C| = 2, then |C| = 4 and ⟨A, C⟩ = A × C. If C ≃ Z_{4}, we have an element of order 12. If C ≃ Z_{2} × Z_{2}, then ⟨A, B⟩ ≃ Z_{3} × Z_{2} × Z_{2} = Z_{6} × Z_{2}, which was eliminated earlier.

Suppose now that the number of Sylow 3-subgroups of G is 4. Let T be one of these subgroups and let N = N_{G}(T), K = ∩_{g∈G}N^{g}. Repeating the proof of Lemma 2, we eliminate all cases except |K| = 2. Then |G/K| = 12 and, since G/K ⊆ S_{4} and the only subgroup of order 12 of S_{4} is A_{4}, we obtain G/K ≃ A_{4}. Let B/K be the unique Sylow 2-subgroup of G/K. Then |B| = 8 and G = B ⋋ H with |H| = 3. By initial assumption, B ≃ D_{8}. Therefore the group H/C_{H}(B) embeds into Aut D_{8}, which is a 2-group. It follows that H/C_{H}(B) = 1, whence G = B × H ≃ D_{8} × Z_{3}, and we again have an element of order 12, which is impossible. This completes the proof.

Theorem 2. Formulas ω_{120} and (3.5)–(3.11) form a basis of d-identities of S_{5}. □

Note. The first study of d-identities of S_{5} was done by Plotkin and Kushkuley [PK] (see also [P; Section 2.5.1]). They exibited a list of seven formulas and claimed that it was a weak basis of d-identities of S_{5}. This result contains useful information and has partially influenced our arguments. One should note, however, that it has two gaps. First, the list provided in [PK] and [P] does not eliminate the quaternion group Q_{8}, which is an irreducible linear group, but is not a section of S_{5}. Second, one of the formulas in that list, namely

(x_{1}^{3} = 1) ∨ (x_{2}^{3} = 1) ∨ (x_{1}^{5} = 1) ∨ (x_{2}^{5} = 1) ∨ (x_{1}^{4} = x_{2}^{4})

∨ ((x_{1}x_{2})^{3} = 1) ∨ ((x_{1}x_{2})^{4} = 1) ∨ ((x_{1}x_{2})^{5} = 1) (3.12)

is not valid in S_{5} at all: take, for example, x_{1} = (123)(45), x_{2} = (14)(25).

4. Disjunctive identities of A_{6}

The procedure of computing a basis of d-identities of A_{6} is more comlicated. If we did it ”from scratch”, it would be substantially longer. Fortunately, a number of intermediate steps have been already done in the previous sections. We begin with stating the final result.

Theorem 3. The following formulas form a basis of d-identities of A_{6}:

ω_{360} = ⋁_{0≤i<j≤360}(x_{i} = x_{j}),  (4.1)

(x^{3} = 1) ∨ (x^{4} = 1) ∨ (x^{5} = 1),  (4.2)
[ ⋁_{1≤i≤9} x_{i}^{20} = 1] ∨ [⋁_{i<j} x_{i} = x_{j}] ∨ [⋁_{i<j}(x_{i}x_{j})^{20} = 1] ∨ [⋁_{i<j}(x_{i}x_{j}^{2})^{20} = 1],  (4.3)

(x_{1}^{30} = 1) ∨ (x_{2}^{30} = 1) ∨ ((x_{1}x_{2})^{15} = 1) ∨ ((x_{1}x_{2}^{2})^{15} = 1) ∨ ((x_{1}x_{2}^{3})^{15} = 1) , (4.4)

θ(x_{1}, x_{2}, x_{3}) ∨ [⋁_{i} x_{i}^{15} = 1] ∨ [⋁_{i≠j}(x_{i}x_{j})^{15} = 1]

∨ [⋁_{i≠j}(x_{i}x_{j}^{2})^{15} = 1] ∨ [⋁_{i≠j}(x_{i}x_{j}^{3})^{15} = 1], (4.5)

(x_{1}^{12} = 1) ∨ (x_{2}^{30} = 1) ∨ ((x_{1}x_{1}^{x_{2}})^{12} = 1)

∨ ((x_{1}x_{1}^{3x_{2}})^{12} = 1) ∨ ((x_{1}^{2}, x_{1}^{x_{2}})^{12} = 1) , (4.6)

(x_{1}^{12} = 1) ∨ (x_{2}^{12} = 1) ∨ ((x_{1}x_{2})^{12} = 1)

∨ ((x_{1}x_{2}^{2})^{12} = 1) ∨ ((x_{1}x_{2}^{3})^{12} = 1) ∨ ((x_{1}x_{2}^{4})^{12} = 1) . (4.7)

Notes. 1. The d-identities (4.4) and (4.7) were introduced in the previous section (as (3.6) and (3.11)) and are repeated here for convenience.

2. As usual, θ(x_{1}, x_{2}, x_{3}) is the formula (2.4).

3. In (4.3), the indices i, j run over the numbers 1,2,...,9; in (4.5) they run over the numbers 1,2,3.

P r o o f. A. First we prove that the d-identities (4.1) – (4.6) are valid in A_{6}. This is obvious for (4.1) and (4.2).

A1. To show that (4.3) holds in A_{6}, we first prove an auxiliary fact similar to Lemma 1.

Lemma 4. Let α and β be elements of order 3 in S_{6} not belonging to the same Sylow 3-subgroup. Then either αβ or αβ^{2} is not an element of order 3.

P r o o f is straightforward. Each Sylow 3-subgroup of A_{6} is generated by two disjoint 3-cycles. For example, the Sylow 3-subgroup generated by (123) and (456) contains the following nonunit elements:

(123), (132), (456), (465),
(123) (456), (123) (465), (132) (456), (132) (465).

It follows that every Sylow 3-subgroup of A_{6} is uniquely determined by any of its nonunit elements.

Let α and β be 3-cycles. Then s = |Supp(α) ∩ Supp(β)| = 1 or 2, because otherwise α and β lie in the same Sylow 3-subgroup. If s = 1, say α = (123) and β = (145), then αβ = (12345) is of order 5. If s = 2, then we may assume that α = (123) and Supp(β) = {1, 2, 4}. Then there are only two choices for β and for each of them either αβ or αβ^{2} is (14)(23), i.e. is of order 2.

Now let both α and β be of the form (∗ ∗ ∗) (∗ ∗ ∗). Let α = (123) (456) and β = (b_{1}b_{2}b_{3}) (b_{4}b_{5}b_{6}). Since α and β do not lie in one Sylow 3-subgroup, the set {b_{1}, b_{2}, b_{3}} is neither {1, 2, 3} nor {4, 5, 6}. Without loss of generality, we may assume that {b_{1}, b_{2}, b_{3}} = {1, 2, 4}. Then there are four choices for β:

β_{1} = (124) (356),  β_{2} = (124) (365),
β_{3} = (142) (356),  β_{4} = (142) (365).

Since β_{1}^{2} = β_{4} and β_{2}^{2} = β_{3}, it remains to notice that αβ_{2} = (14326) and αβ_{4} = (26) (34).

In the case when only one of the α, β is a 3-cycle, the proof is similar and we omit it. □

Now we can prove that (4.3) is a d-identity of A_{6}. Let x_{1}, x_{2}, …, x_{9} ∈ A_{6}. If the first clause of (4.3) fails, then all x_{i} are of order 3. If, in addition, the second clause fails, then they are pairwise different. Among nine different elements of order 3 in A_{6}, there must be two, x_{i} and x_{j}, that do not belong to the same Sylow 3-subgroup. But then, by Lemma 4, either the third or the fourth clause of (4.3) is valid.

A2. It was proved in the previous section that (4.4) (that is, (3.6)) is valid in S_{5}. Now we show that it is valid in A_{6}. If x_{1}, x_{2} ∈ A_{6} and x_{i}^{30} ≠ 1, then both x_{1} and x_{2} have form (∗ ∗ ∗∗) (∗∗). Let x_{1} = (1234) (56) and x_{2} = (a_{1}a_{2}a_{3}a_{4}) (a_{5}a_{6}). There are three possibilities.

(i) {a_{1}, a_{2}, a_{3}, a_{4}} = {1, 2, 3, 4}. In this case the proof coincides with that for the group S_{5}.

(ii) {a_{1}, a_{2}, a_{3}, a_{4}} contains exactly three of the numbers 1,2,3,4. Suppose these numbers are 2,3,4 and x_{2} = (∗ ∗ ∗5) (16). Then there are six choices for x_{2}, and for each of them one can directly verify that

(x_{1}x_{2}^{k})^{15} = 1 for some k ≤ 3.

(iii) {a_{1}, a_{2}, a_{3}, a_{4}} contains exactly two of the numbers 1,2,3,4. Then there are six essentially different choices for x_{2}:

(5612) (34),  (5613) (24),  (5621) (34),
(5162) (34),  (5163) (34),  (5261) (34).

Again a direct verification shows that for each of these choices (x_{1}x_{2}^{k})^{15} = 1 for some k ≤ 3.

A3. It will be more difficult to prove the validity of (4.5). First we describe the structure of Sylow 2-subgroups of A_{6}. Just as in S_{5}, every such subgroup is isomorphic to D_{8} and is uniquely determined by each of the elements of order 4 it contains. Namely, the Sylow 2-subgroup containing the permutation α = (abcd) (ef) consists of the elements

1, (abcd) (ef), (adcb) (ef),
(ac) (ef), (bd) (ef), (ab) (cd), (ac) (bd), (ad) (bc).

This subgroup will be denoted by Syl(α). It is easy to see that the Sylow 2-subgroups of A_{6} possess the same property as those of S_{5}: if x_{1}, x_{2}, x_{3} are 2-elements of A_{c} and any two of them lie in one Sylow 2-subgroup, then all three of them lie in one Sylow 2-subgroup.

Choose x_{1}, x_{2}, x_{3} ∈ A_{6} such that x_{i}^{15} ≠ 1. Then the x_{i} are 2-elements. If they lie in one Sylow 2-subgroup, then θ(x_{1}, x_{2}, x_{3}), being a d-identity of D_{8}, must be valid. Otherwise two of the x_{i}, say x_{1} and x_{2}, do not belong to the same Sylow 2-subgroup. Then there are three possibilities.

(i) Both x_{1} and x_{2} are of order 4. We may assume that x_{1} = (1234) (56) and x_{2} = (abcd) (ef). Consider three cases.

a) {a, b, c, d} = {1, 2, 3, 4}. In this case one can repeat the proof of Lemma 3, part (i), and show that

((x_{1}, x_{2})^{15} = 1) ∨ ((x_{1}x_{2}^{2})^{15} = 1) ∨ ((x_{1}x_{2}^{3})^{15} = 1) .  (4.8)

b) {a, b, c, d} contains exactly one of the numbers 5,6. Assuming that x_{2} = (5∗∗∗) (64), we have six choices for x_{2}, and for each of them one can directly verify that (4.8) is valid.

c) {a, b, c, d} contains both 5 and 6. Without loss of generality, we may assume that either x_{2} = (516∗) (∗∗) or x_{2} = (561∗) (∗∗). This leads to six different choices for x_{2}

(5612) (34),  (5613) (24),  (5614) (23),
(5162) (34),  (5163) (24),  (5164) (23),

and again a direct verification shows that (4.8) is valid.

(ii) Only one of the x_{1} and x_{2} is of order 4. Assume that x_{1} = (1234) (56). Then x_{2} has form (∗∗) (∗∗) and, since x_{2} ∉ Syl(x_{1}), it follows that Supp(x_{2}) ≠ {1, 2, 3, 4}. Hence Supp(x_{2}) contains at least one of the numbers 5 and 6.

a) Let Supp(x_{2}) contain precisely one of these numbers, say 5. Repeating the proof of Theorem 2, Step 2, (ii), b), we obtain that (x_{2}, x_{1}^{2})^{15} = 1, whence (4.5) holds.

b) Now suppose that x_{2} = (56) (∗∗). Since (13)(56) and (24)(56) lie in Syl(x_{1}), there are four choices for x_{2}.

(56) (12),  (56) (14),  (56) (13),  (56) (34).

In each case x_{1}x_{2} is a 3-cycle, so (x_{1}x_{2})^{3} = 1 and (4.5) holds.

c) Finally, let x_{2} = (5∗) (6∗). There are only two essentially different choices for x_{2}: (51)(62) and (51)(63). In the first case, x_{2}x_{1}^{2} = (153) (264), in the second case x_{2}x_{1} = (164) (235), so (4.5) is valid again.

(iii) None of the x_{1} and x_{2} is of order 4. Then they both are involutions and therefore ⟨x_{1}, x_{2}⟩ is isomorphic to a dihedral group:

⟨x_{1}, x_{2}⟩ = ⟨x_{1}x_{2}⟩ ⋋ ⟨x_{2}⟩ ≅ D_{2m}

where ⟨x_{1}x_{2}⟩ ≅ Z_{m}. Since x_{1} and x_{2} are not contained in the same Sylow 2-subgroup, x_{1}x_{2} is not a 2-element. Therefore (x_{1}x_{2})^{15} = 1 and (4.5) holds.

A4. To show that (4.6) holds in A_{6}, take x_{1}, x_{2} ∈ A_{6} such that x_{1}^{12} ≠ 1 and x_{2}^{30} ≠ 1. Then x_{1} is a 5-cycle and x_{2} has form (∗ ∗ ∗∗) (∗∗). Without loss of generality, we may assume that x_{1} = (12345), then x_{1}^{x_{2}} is a 5-cycle involving 6. The rest follows from

Lemma 5. Let α and β be 5-cycles in S_{6} such that Supp(α) ≠ Supp(β). Then one of the permutations αβ, αβ^{3}, α^{2}β is not a 5-cycle.

P r o o f. Unfortunately, we could not find a nice proof of this fact (which certainly exists), and simply considered all possible cases. First, we may assume that α = (12345) and β = (b_{1}b_{2}b_{3}b_{4}6), where b_{i} ≠ 5. Thus there are 24 choices for (b_{1}b_{2}b_{3}b_{4}). Next, we notice that

if β = (1 ∗ ∗ 26) then αβ = (16) (∗ ∗ ∗∗),
if β = (2 ∗ ∗ 36) then αβ = (26) (∗ ∗ ∗∗),
if β = (3 ∗ ∗ 46) then αβ = (36) (∗ ∗ ∗∗).

It remains to consider 18 possibilities for (b_{1}b_{2}b_{3}b_{4}), which can be done by direct verification. For example:

if β = (12346) then αβ = (136) (245),
if β = (13246) then αβ^{3} = (26) (45),
if β = (21346) then α^{2}β = (1435) (26),

and so on. □

Lemma 5 implies that (4.6) is a d-identity of A_{6}. From the previous section, we know that (4.7) is also a d-identity of A_{6}. This completes the first part of the proof of Theorem 3.

B. Now we have to show that if a group G satisfies the d-identities (4.1)–(4.6), then it is a section of A_{6}. Let |G| = n; it follows from (4.1) and (4.2) that

n ≤ 360 and n = 2^{k}3^{ℓ}5^{m}.

B1. We show that in fact n must be a divisor of 360, that is, k ≤ 3, ℓ ≤ 2, m ≤ 1. Since the d-identity (4.7) is not valid in any group of order 25, it follows that m ≤ 1. To show that ℓ ≤ 2 assume that a group A of order 27 satisfies the d-identities (4.1)–(4.7). Then (4.2) implies that exp A = 3. We can choose in A nonunit elements x_{1}, x_{2}, …, x_{9} such that x_{i} ≠ x_{j}^{±1} for i ≠ j. Then x_{i}x_{j} ≠ 1 and x_{i}x_{j}^{2} ≠ 1 (since exp A = 3, we have x_{j}^{2} = x_{j}^{−1}), and so (4.3) fails. It follows that ℓ ≤ 2.

To show that k ≤ 3, consider first the abelian groups of order 8. Among them, D_{8} is a subgroup of A_{6}, Z_{8} does not satisfy (4.2), Z_{4} × Z_{2} and Q_{8} does not satisfy (4.4) (see the proof of Theorem 2, Step 1), and Z_{2}^{3} does not satisfy (4.5) (take a basis of Z_{2}^{3} for the x_{1}, x_{2}, x_{3}). Repeating the proof of Theorem 2, Step 6, we eliminate all groups of order 16. It follows that k ≤ 3.

Thus, if a group G of order n satisfies the d-identities (4.1)–(4.7), then

n = 2^{k}3^{ℓ}5^{m}, where k ≤ 3, ℓ ≤ 2, m ≤ 1,

that is, n is one of the numbers

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24,
30, 36, 40, 45, 60, 72, 90, 120, 180, 360.

B2. All groups of order ≤ 6, except Z_{6}, are subgroups of A_{6}, but Z_{6} does not satisfy (4.2). The groups of order 8 have been already taken care of. The cyclic groups of order 9 and 10 do not satisfy (4.2), but the noncyclic ones are subgroups of A_{6}. All groups of order 12, except A_{4}, have elements of order 6 and do not satisfy (4.2), but A_{4} is a subgroup of A_{6}. Repeating the proof of Theorem 2, Step 5, and noting that S_{5} does not satisfy (4.2), we eliminate all groups of orders 15, 30, 60, and 120. Lemma 2 and (4.2) eliminate all groups of order 24 except S_{4}, but S_{4} is a subgroup of A_{6}.

B3. Groups of order 18. Every group G of order 18 splits into a semidirect product

G = A ⋋ B, where |A| = 9 and |B| = 2.

If A ≃ Z_{9}, then (4.2) fails, so we assume that A ≃ Z_{3} × Z_{3}. Let 1 ≠ b ∈ B. If a^{b} ≠ a^{−1} for some a ≠ a ∈ A, then c = aa^{b} ≠ 1 and c^{b} = a^{b}a = aa^{b} = c. Hence cb is an element of order 6 and so G is eliminated by (4.2). If a^{b} = a^{−1} for all a ∈ A, then G is isomorphic to the subgroup of A_{6} generated by (123), (456) and (12) (45).

B4. Groups of order 36. Every group G of order 36 is solvable and therefore if H is a maximal normal subgroup of G, then |G : H| = 2 or |G : H| = 3.

(i) Let |G : H| = 3. Since all groups of order 12 except A_{4} have elements of order 6, we have H ≃ A_{4}. Hence

G = A ⋋ T, where A ≅ Z_{2} × Z_{2} and |T | = 9.

Since |Aut(A)| = 6, we see that C_{T} (A) is nontrivial. It follows that G must have elements of order 6, which is impossible.

(i) Let |G : H| = 2. Then |H| = 18 and, as established in B3, H = A ⋋ B with A ≃ Z_{3} × Z_{3}, B ≃ Z_{2}, and a^{b} = a^{−1} for all a ∈ A. Therefore A must be normal in G, whence G = A ⋋ T, |T | = 4. If C_{T} (A) ≠ 1, then G again has elements of order 6. Hence C_{T} (A) = 1.

a) If T ≃ Z_{2} × Z_{2}, then

G = A ⋋ (⟨b_{1}⟩ × ⟨b_{2}⟩), where |b_{i}| = 2.

As shown in B3, both b_{i} must act on A as follows: a^{b_{i}} = a^{−1} for all a ∈ A. Then b_{1}b_{2}^{−1} centralizes A, which is impossible.

b) If T ≃ Z_{4}, then G = A ⋋ ⟨b⟩, where |b| = 4. Since b^{2} is an involution, the same argument as in B3 shows that

∀a ∈ A : a^{b_{2}} = a^{−1}.  (4.9)

Take any 1 ≠ a_{1} ∈ A. If a_{1}^{b} = a_{1}, we get an element of order 12. If a_{1}^{b} = a_{1}^{−1}, then a_{1}^{b^{2}} = a_{1}, contradicting (4.9). It follows that a_{1}^{b} ≠ a_{1}^{±1}, and then a_{1} and a_{1}^{b} form a basis of A = Z_{3} × Z_{3}. Denoting a_{1}^{b} = a_{2}, we have

G = (⟨a_{1}⟩ × ⟨a_{2}⟩) ⋋ ⟨b⟩,

where |a_{i}| = 3, |b| = 4, a_{1}^{b} = a_{2}, a_{2}^{b} = a_{1}^{b_{2}} = a_{1}^{−1}. But this group is isomorphic to the subgroup of A_{6} generated by a_{1} = (123), a_{2} = (456) and b = (1425)(36), for

a_{1}^{b} = (5241) (36) (123) (1425) (36) = (456) = a_{2},
a_{2}^{b} = (5241) (36) (456) (1425) (36) = (132) = a_{1}^{−1}.

This completes the consideration of groups of order 36.

B5. Groups of orders 45, 90 and 180. It follows directly from the Sylow Theorem that any group G of order 45 is a direct product G = A × B, where |A| = 9. and |B| = 5. Hence G contains an element of order 15 and does not satisfy (4.2). Every group of order 90 is solvable. Since 90 = 45 · 2 and (45, 2)=1, this group has a subgroup of order 45 and therefore does not satisfy (4.2) either. The same argument works for solvable groups of order 180. There is only one unsolvable group of order 180, namely A_{5} × Z_{3}, which also contains elements of “wrong” orders.

B6. Groups of orders 20 and 40. Every group G of order 20 can be presented as

G = A ⋋ B, where A = Z_{5}, |B| = 4.

If C_{B}(A) ≠ 1, we get an element of order 10, which is impossible. Otherwise B ≃ Aut(Z_{5}) and G can be defined by the presentation

G = ⟨a, b | a^{5} = b^{4} = 1, a^{b} = a^{2}⟩.

But this group does not satisfy the d-identity (4.6): set x_{1} = a and x_{2} = b. As to the groups of order 40, they are eliminated literally as in Theorem 2, Step 7.

B7. Groups of order 72. Let G be a group of order 72 satisfying (4.1)–(4.7). Then G is solvable and if N is a maximal normal subgroup of G, then |G : N| = 2 or |G : N| = 3.

(i) Let |G : N| = 2. Then |N| = 36 and, as established in B4,

N = A ⋋ T, where A = Z_{3} × Z_{3}, T = Z_{4}, C_{T} (A) = 1

(recall that all other cases led to elements of order 6). Hence we have A◁G and |G/A| = 8. All groups of order 8, except D_{8}, were eliminated by (4.2), (4.4) and (4.5). So G/A ≃ D_{8} and, since (|A|, |G/A|) = 1, we have

G = A ⋋ D_{8} = (Z_{3} × Z_{3}) ⋋ D_{8}.

But then G has as a subgroup (Z_{3} × Z_{3}) ⋋ (Z_{2} × Z_{2}). Repeating B4 (ii), we get an element of order 6, which is impossible.

(ii) Let |G : N| = 3. Then |N| = 24 and, by Lemma 2, N ≃ S_{4}. Let b ∈ G/N. Clearly |b| is divisible by 3, and it follows from (4.2) that |b| = 3. Therefore

G = N ⋋ ⟨b⟩ ≃ S_{4} ⋋ ⟨b⟩.

If b centralizes N, we again have an element of order 6, otherwise ⟨b⟩ ⊂ Aut(S_{4}). Since all automorphisms of S_{4} are inner, it follows that b acts on S_{4} as conjugation by some 3-cycle, say (123):

∀α ∈ S_{4} : α^{b} = (321) α (123).

But then one can verify that the element g = (12) · b of the group G = S_{4} ⋋ ⟨b⟩ has order 6.

B8. Groups of order 360. Every solvable group of order 360 = 72·5 has a subgroup H of order 72. It has just been proven that if H satisfies (4.1)–(4.7), it must have elements of order 6.

Let G be a nonsolvable group of order 360. If G is simple, it is isomorphic to A_{6}. Otherwise it has composition series

1 ◁ A ◁ B ◁ G  (4.10)

with factors Z_{2}, Z_{3} and A_{5}. If |G/B| = 2, then B is a nonsolvable group of order 180. We already know that there is only one such group A_{5} × Z_{3}, which has elements of order 15 and 12. If |G/B| = 3, then B is a nonsolvable group of order 120, i.e. one of the groups S_{5}, SL(2, 5), A_{5} × Z_{2}. Each of them has elements of order ≥ 6 and is excluded by (4.2).

Finally, let G/B ≃ A_{5}. Then |B| = 6 and, since elements of order 6 are prohibited, we have B ≃ S_{3}. If C = C_{G}(B), then G/C acts faithfully on B. Since Aut(S_{3}) ≃ S_{3}, we get |G/C| ≤ G. Thus C must have elements of order 5, whence G must have elements of order 10 and 15. This concludes the proof. □

5. Disjunctive identities of the dihedral groups

Let D_{2m} = ⟨a, b | a^{m} = b^{2} = 1, a^{b} = a^{−1}⟩ be the dihedral group of order n = 2m. In this section we find bases of d-identities of the groups D_{2m} for any m. Since we have to deal with an infinite family of groups, rather than a concrete single group, the structure of the proof is different from that in the previous sections.

1. We begin with the following observation.

Lemma 5. A group G satisfies the d-identity

(x^{2} = 1) ∨ (y^{2} = 1) ∨ (xy = yx)  (5.1)

if and only if G = A ⋋ B, where A is abelian and B is either trivial or cyclic of order 2 acting by inversion on A.

P r o o f. Clearly if G has the described structure, it satisfies (5.1). On the other hand, let G satisfy (5.1). If exp G = 2, there is nothing to prove. Otherwise, let A be the subgroup generated by the non-involutions of G. By (5.1), A is abelian. If b ∈ G\A, then, as A ∩ bA = ∅, bA consists of involutions. Hence for each a ∈ A we have ba · ba = 1, whence bab = a^{−1}.

We have A ⊆ C_{G}(A) ⊆ G. Then \overline{G} = G/C_{G}(A) acts on A faithfully. By the above, each element of \overline{G} must act on A by inversion. It follows that |G/C_{G}(A)| = 2. Since each b ∈ C_{G}(A)\A acts on A trivially and simultaneously be inversion, we have A = C_{G}(A). Thus |G/A| = 2 and so G = A ⋋ Z_{2}. □

Lemma 6. Let G_{1} and G_{2} be finite groups satisfying (5.1). Suppose that every abelian section of G_{2} is isomorphic to a section of G_{1} and that G_{1} is nonabelian. Then G_{2} itself is isomorphic to a section of G_{1}.

P r o o f. We may assume that G_{2} is also nonabelian. By Lemma 5, G_{i} = A_{i} ⋋ B_{i} where A_{i} is abelian and B_{i} ≃ Z_{2} acts on A_{i} by inversion. So it suffices to show that A_{2} is isomorphic to a section of A_{1}, i.e. that for each prime p the p-component of A_{2} is isomorphic to a section of the p-component of A_{1}. For p odd this follows directly from the hypothesis.

Let T_{i} be the 2-component of A_{i}. Since T_{2} is abelian, we have

T_{2} ≃ H/K, where K ◁ H ⊆ G_{1} = A_{1} ⋋ B_{1}.

Note that every subgroup X of G_{1} not containing in A_{1} has the form X = (X ∩ A_{1}) ⋋ ⟨x⟩, where x is any element from X\A_{1}. Therefore if K ⊈ A_{1}, then T_{2} ≃ H/K ≃ (H ∩ A_{1})/ (K ∩ A_{1}) is a section of A_{1} and so of T_{1}. If K ⊆ A_{1} and H ⊆ A_{1}, then T_{2} is again a section of A_{1}. Finally, if K ⊆ A_{1} but H ⊈ A_{1}, then T_{2} ≃ ((H ∩ A_{1})/K) ⋋ ⟨h⟩ = \overline{A} ⋋ ⟨h⟩, where ⟨h⟩ ≃ Z_{2} acts on \overline{A} by inversion. Since T_{2} is abelian, we have exp T_{2} = 2. But then ⟨T_{2}, B_{2}⟩ = T_{2} × B_{2} is abelian and hence is isomorphic to a section of G_{1} = A_{1} ⋋ B_{1}. It follows that T_{2} must be isomorphic to a section of T_{1}. □

2. Our aim is to write down a set of disjunctive identities of D_{2m} = Z_{m} ⋋Z_{2} such that the only groups they hold in are sections of D_{2m}. The first two d-identities are ω_{2m} and (5.1). Any group in which they are valid is a finite group of the form A ⋋ Z_{2}, or abelian (Lemma 5). Next, we will add to ω_{2m} and (5.1) several new d-identities of D_{2m} such that the only abelian groups all these d-identities hold in are sections of D_{2m}. Then by Lemma 6, any group in which all these d-identities hold will be a section of D_{2m}.

We assume that m is even. The case of odd m is simpler and will be discussed at the end of the section.

Lemma 7. Let A be a finite abelian group of an even exponent m. If A has no sections of the form

Z_{2}^{3}, Z_{4} × Z_{2}, Z_{p}^{2}, Z_{p} × Z_{2}^{2},

where p is an odd prime divisor of m, then A is isomorphic to a section of D_{2m}.

P r o o f. Let A = A_{2} × A_{p_{1}} × … × A_{p_{s}} be the decomposition of A into primary components. Since A has no sections of the form Z_{p}^{2}, where p is an odd divisor of m, each A_{p_{i}} must be cyclic. Hence A = A_{2} × Z_{k}, where k is an odd divisor of m. If A_{2} is cyclic, then A is a section of Z_{m}, otherwise it follows from the hypothesis that A ≃ Z_{2} × Z_{2}, which is a section of D_{2m}. □

3. Since m is even, the identity

x^{m} = 1  (5.2)

is valid in D_{2m}. Every group satisfying (5.2) has exponent m, and so, for each of the four groups of Lemma 7, it remains to find a d-identity of D_{2m} that is not valid in this group.

(i) The group Z_{p}^{2}. Let m = p^{k}m_{p}, where p is an odd prime and (p, m_{p}) = 1. Consider the formula

(x_{1}^{2} = 1) ∨ (x_{2}^{2} = 1) ∨ (x_{1}^{m_{p}} ∈ ⟨x_{2}^{m_{p}}⟩) ∨ (x_{2}^{m_{p}} ∈ ⟨x_{1}^{m_{p}}⟩),  (5.3)

where α ∈ ⟨β⟩ denotes

(α = β) ∨ (α = β^{2}) ∨ … ∨ (α = β^{p^{k}−1}).

It is valid in D_{2m} = Z_{m} ⋋ Z_{2} because if x_{1}^{2} ≠ 1 and x_{2}^{2} ≠ 1, then x_{i} ∈ Z_{m} = Z_{p^{k}m_{p}}. Hence x_{i}^{m_{p}} ∈ Z_{p^{k}} and so one of the subgroups ⟨x_{i}^{m_{p}}⟩ is contained in the other. On the other hand, (5.2) fails in Z_{p}^{2}: take a basis of this group for the x_{i}.

(ii) The group Z_{4} × Z_{2}. Take the d-identity (5.3) with m_{p} = m_{2} (that is, m = 2^{k}m_{2} with m_{2} odd). Again it is valid in D_{2m}. But it is not valid in Z_{4} × Z_{2}: if a and b are generators for Z_{4} and Z_{2}, respectively, and x_{1} = (a, b), x_{2} = (a, 1), then

(x_{1}^{2} ≠ 1) & (x_{2}^{2} ≠ 1) & (x_{1}^{m_{2}} ∉ ⟨x_{2}^{m_{2}}⟩) & (x_{2}^{m_{2}} ∉ ⟨x_{2}^{m_{2}}⟩).

(iii) The group Z_{2}^{3}. This group is eliminated by the following generalization of the d-identity (2.4)

[⋁_{i≠j} x_{i}^{m_{2}} ∈ ⟨x_{j}^{m_{2}}⟩]

∨ [ ⋁_{σ∈S_{3}} x_{σ(1)}^{m_{2}} ∈ ⟨(x_{σ(2)}x_{σ(3)})^{m_{2}}⟩] ∨ [ ⋁_{σ∈S_{3}} x_{σ(1)}^{m_{2}} ∋ ⟨(x_{σ(2)}x_{σ(3)})^{m_{2}}⟩]  (5.4)

∨ [ ⋁_{σ∈S_{3}} (x_{σ(1)}x_{σ(2)})^{m_{2}} ∈ ⟨(x_{σ(2)}x_{σ(3)})^{m_{2}}⟩]

using the same arguments as in §2.

(iv) The group Z_{p} × Z_{2}^{2} (where p is an odd divisor of m). Let m = 2^{k_{0}}p^{k_{1}}p_{2}^{k_{2}} … p_{s}^{k_{s}}, then m_{2} = p^{k_{1}}p_{2}^{k_{2}} … p_{s}^{k_{s}}, m_{p} = 2^{k_{0}}p_{2}^{k_{2}} … p_{s}^{k_{s}}. Consider the following formula in three variables x_{1}, x_{2}, x_{3}:

((x_{1}^{m_{p}})^{x_{2}} = x_{1}^{−m_{p}}) ∨ ((x_{1}^{m_{p}})^{x_{3}} = x_{1}^{−m_{p}})
∨ (x_{2}^{m_{2}} ∈ ⟨x_{3}^{m_{2}}⟩) ∨ (x_{3}^{m_{2}} ∈ ⟨x_{2}^{m_{2}}⟩) .  (5.5)

It is valid in D_{2m} = Z_{m} ⋋ Z_{2}. Indeed, since m is even, we have exp D_{2m} = m and so x^{m_{p}} is a p-element and x^{m_{2}} is a 2-element for any x ∈ D_{2m}. If the first two clauses of (5.4) are false, then both x_{2} and x_{3} belong to Z_{m}. But then both x_{2}^{m_{2}} and x_{3}^{m_{2}} are 2-elements, and the rest is obvious. On the other hand, a natural basis of the group Z_{p} × Z_{2} × Z_{2} refutes the d-identity (5.5) in this group.

Combining all of the above, we obtain

Theorem 4. Let m be even. Then the following formulas form a basis of d-identities of D_{2m}:

ω_{2m}, (5.1), (5.2), (5.4);

(5.3) for all prime divisors p of m;

(5.5) for all odd prime divisors p of m. □

4. The case of an odd m is simpler. First, instead of (5.2), one has to take the d-identity

(x^{2} = 1) ∨ (x^{m} = 1).  (5.2’)

Next, instead of Lemma 7, we notice that if A is a finite abelian group satisfying (5.2’), which is not isomorphic to a section of D_{2m}, then A has a section isomorphic to either Z_{2}^{2} or Z_{p}^{2}, where p is an odd prime dividing m. The latter is eliminated by (5.3), but the former – by the d-identity

(x_{1}^{m} = 1) ∨ (x_{2}^{m} = 1) ∨ ((x_{1}x_{2})^{m} = 1).  (5.6)

Thus, a basis of d-identities of the group D_{2m} with an odd m consists of the formulas ω_{2m}, (5.1), (5.2’), (5.3) for all primes dividing m, and (5.6).

6. Identities of regular representations

Using the method of [PK], outlined in the introduction, we can now easily write down bases for the identities of regular representations of the groups considered in the previous sections. Since this theory is applicable to ordinary representations only, the characteristic of the ground field K is always chosen to be coprime with the order of the group in question.

Recall that if

δ = (f_{1} = 1) ∨ … ∨ (f_{n} = 1)

is a disjunctive identity of a group G, then δ^{∗} denotes the corresponding identity of the regular representation Reg_{K}G, that is,

δ^{∗} = (f_{1}^{y_{1}} − 1) … (f_{n}^{y_{n}} − 1).

It will be convenient to extend this notation to the numbers of d-identities as well. For example, (2.10)* will denote the element of KF corresponding to the d-identity (2.10), namely the element

(x_{1}^{4y_{1}} − 1) (x_{2}^{4y_{2}} − 1) ((x_{1}x_{2})^{4y_{3}} − 1) ((x_{1}x_{2}^{2})^{4y_{4}} − 1).

Proposition 5. Let charK ≠ 2. Then the following elements of KF form a basis of identities of Reg_{K}D_{8}:

∏_{0≤i<j≤8} ((x_{i}x_{j}^{−1})^{y_{ij}} = 1) ,  (6.1)

x^{4} − 1,  (6.2)
(x_{1}^{2y_{1}} − 1)(x_{2}^{2y_{2}} − 1)(x_{3}^{2y_{3}} − 1)((x_{1}x_{2}^{−1})^{y_{4}} − 1)((x_{1}x_{3}^{−1})^{y_{5}} − 1)((x_{2}x_{3}^{−1})^{y_{6}} − 1). (6.3)

P r o o f. Clearly these elements are (2.1)*, (2.2)* and (2.3)*, respectively. Since (2.1)– (2.3) form a weak basis of d-identities of D_{8}, it suffices to show that the variety of group representations defined by (6.1)–(6.3) is locally finite and ordinary. But this variety is obviously contained in ωB_{4}, where B_{4} is the Burnside variety of groups of exponent 4. Since B_{4} is locally finite and charK ≠ 2, the result follows from Fact 2 in Introduction. □

It is not harder to find bases for the identities of the regular representations of Q_{8} and A_{4}. For example, we know that the formula ω_{12}, (2.7) and (2.8) form a weak basis of d-identities of A_{4}. To find a basis of identities of Reg_{K} A_{4} (provided charK ≠ 2, 3), we take ω_{12}^{∗}, (2.7)*, (2.8)* and also the identity x^{6} − 1. The latter defines the variety ωB_{6}, and since B_{6} is locally finite, we are done.

Proposition 6. Let charK ≠ 2, 3. Then a basis of identities of Reg_{K}S_{4} consists of ω_{24}^{∗}, (2.10)*–(2.13)*, plus the identities

x^{12} − 1,  (6.4)
[[[x_{1}, x_{2}], [x_{3}, x_{4}]], [[x_{5}, x_{6}], [x_{7}, x_{8}]]] − 1.  (6.5)

P r o o f. We know that ω_{24} and (2.10)–(2.13) form a basis of d-identities of S_{4}. Since S_{4} is a 3-solvable group of exponent 12, the identities (6.4) and (6.5) are valid in Reg_{K}S_{4}. Moreover, these two identities define the variety of representations ω𝔙, where 𝔙 is the variety of 3-solvable groups of exponent 12. Since 𝔙 is locally finite and char K ∤ 12, the result follows. □

The argument above generalizes to arbitrary solvable groups. Indeed, let G be a finite solvable group such that char K ∤ |G|, and suppose that δ_{1}, δ_{2}, …, δ_{n} is a (weak) basis of d-identities of G. To find a basis of identities of Reg_{K}G, we take all δ_{i}^{∗} and add to them

x^{e} − 1 and v_{s} − 1  (6.6)

where e = exp G, s is the solvability class of G, and v_{s} ∈ F is the identity of solvability of class s. Then the variety of group representations defined by (6.6) is locally finite and ordinary, and we are done.

For example, if G = D_{2m}, then the identities (6.6) are x^{2m} −1 (or x^{m} −1 if m is even) and [[x_{1}, x_{2}], [x_{3}, x_{4}]] − 1.

If the group in question is not solvable, there are other ways to obtain identities that guarantee that the corresponding variety is locally finite and ordinary. Let

s_{n}(x_{1}, …, x_{n}) = ∑_{σ∈S_{n}}(−1)^{σ}x_{σ(1)} … x_{σ(n)}

be the standard polynomial of degree n.

Lemma 8. If G is a group of order n and exponent e, then x^{e} − 1 and s_{n+1} are identities of Reg_{K}G. Furthermore, the variety of group representations defined by these two identities is locally finite (and ordinary if charK ∤ e).

P r o o f. Since s_{n+1} ≡ 0 in any algebra of dimension n, the first statement is obvious. To prove the second, it is enough to show that if ρ : G ⟶ Aut_{K}V is a faithful representation satisfying x^{e} − 1 and s_{n+1}, then G is locally finite. We may assume that G ⊂ End_{K}V, and then s_{n+1}(g_{1}, …, g_{n+1}) = 0 for any g ∈ G. Let A be the subalgebra of End_{K}V generated by G. Since s_{n+1}(x_{1}, …, x_{n+1}) is a multilinear polynomial, it is clear that s_{n+1}(x_{1}, …, x_{n+1}) ≡ 0 is identically true on A. Hence G is a periodic subgroup of a P I-ring A. It is known that such a group is locally finite (see, for example, [Pr]). □

Thus, if we know a basis (or at least a weak basis) of d-identities of a group G, the procedure of finding a basis of identities of Reg_{K}G becomes automatic. For example, combining Theorem 3, Lemma 8 and Fact 2, we obtain

Proposition 7. Let char K ≠ 2, 3, 5. Then the identities (4.1)*–(4.7)* plus x^{60} − 1 and s_{361} form a basis of identities of Reg_{K}A_{6}. □

References

[B] Baker, K.A., Equational axioms for classes of lattices, Bull. Amer. Math. Soc. 77 (1971), 97 –102.

[ES] Eilenberg, S. and Schützenberger, M.P., On pseudovarieties, Adv. Math. 19 (1976), 413–418.

[G] Grätzer, G., Universal Algebra, 2nd Ed., Springer, 1979.

[P] Plotkin, B.I., Varieties in representations of finite groups. Locally stable varieties. Matrix groups and varieties of representations, Uspekhi Mat. Nauk 34 (1979), no. 4, 65–95.

[PK] Plotkin, B.I. and Kushkuley, A.H., Identities of regular representations of groups, preprint.

[PV] Plotkin, B.I. and Vovsi, S.M., Varieties of Group Representations: General Theory, Connections, and Applications, Zinātne, Riga, 1983.

[Pr] Procesi, C., The Burnside problem, J. Algebra 4 (1966), 421–425.

[V1] Vovsi, S.M., On critical representations of groups, Latv. Mat. Yezhegodnik 20 (1976), 141–159.

[V2] Vovsi, S.M., Topics in Varieties of Group Representations, Cambr. Univ. Press, 1991.

[VN] Vovsi, S.M., and Nguyen Hung Shon, Identities of stable-by-finite representations of groups, Mat. Sb. 132 (1987), 578–591.

School of Mathematics, Institute for Advanced Study, Princeton, NJ 08540 E-mail address: vovsi@math.ias.edu, vovsi@math.rutgers.edu