Hypersubstitutions are mappings which map operation symbols to terms of the corresponding arities. They were introduced as a way of making precise the concept of a hyperidentity and generalizations to M-hyperidentities. A variety in which every identity is satisfied as a hyperidentity is called solid. If every identity is an M-hyperidentity for a subset M of the set of all hypersubstitutions, the variety is called M-solid. There is a Galois connection between monoids of hypersubstitutions and sublattices of the lattice of all varieties of algebras of a given type. Therefore, it is interesting and useful to know how semigroup or monoid properties of monoids of hypersubstitutions transfer under this Galois connection to properties of the corresponding lattices of M-solid varieties. In this paper, we study the order of each hypersubstitution of type (2,1), that is, the order of the cyclic subsemigroup of the monoid of all hypersubstitutions of type (2,1) generated by that hypersubstitution.

1. Preliminaries

Let ℕ denote the set of all positive integers. Let τ = {(f_i, n_i)∣i ∈ I} be a type. Let X = {x₁, x₂, x₃, …} be a countably infinite alphabet of variables such that the sequence of the operation symbols (f_i) _i∈I is disjoint with X, and let X_n = {x₁, x₂, …, x_n} be an n-element alphabet where n ∈ ℕ. Here, f_i is n_i-ary for a natural number n_i ≥ 1. An n-ary (n ≥ 1) term of type τ is inductively defined as follows:

(i)
every variable x_j ∈ X_n is an n-ary term,
(ii)
if are n-ary terms and f_i is an n_i-ary operation symbol, then is an n-ary term.

Let W_τ(X_n) be the smallest set containing x₁, …, x_n and being closed under finite application of (ii). The set of all terms of type τ over the alphabet X is defined as the disjoint union .

Any mapping σ : {f_i : i ∈ I} → W_τ(X) is called a hypersubstitution of type τ if σ(f_i) is an n_i-ary term of type τ for every i ∈ I. Any hypersubstitution σ of type τ can be uniquely extended to a map

on W_τ(X) as follows:

(i)
if t ∈ X,
(ii)
if .

A binary operation ∘_h is defined on the set Hyp(τ) of all hypersubstitutions of type τ by

(1.1)

for all n_i-ary operation symbols f_i. This binary associative operation makes Hyp(τ) into a monoid, with the identity hypersubstitution σ_id which maps every f_i to

as an identity element. For a submonoid M of Hyp(τ) an identity s ≈ t of a variety V of type (τ) is called an M-hyperidentity of V if for every hypersubstitution σ ∈ M, the equation

holds in V. A variety V is called M-solid if every identity of V is an M-hyperidentity of V. If M is a submonoid of Hyp(τ), then the collection of all M-solid varieties of type τ is a complete sublattice of the lattice of all varieties of type τ [1].

Let σ ∈ Hyp(τ) and let 〈σ〉∶ = {σⁿ∣n ∈ ℕ} be the cyclic subsemigroup of Hyp(τ) generated by σ. The order of σ ∈ Hyp(τ) is defined as the order of the semigroup 〈σ〉. If 〈σ〉 is finite, then the order of the hypersubstition σ is finite, otherwise the order of σ is infinite. The hypersubstitution σ is idempotent if and only if the order of σ is 1.

The order of a hypersubstitution of type (2) is 1, 2, or infinite [2]. The order of a hypersubstitution of type (3) is 1, 2, 3, or infinite [3]. The order of a hypersubstitution of type (2,2) is 1, 2, 3, 4, or infinite [4]. We are interested in type (2,1). The main result is as follows

Main Theorem. Any hypersubstitution of type (2,1) has order either infinite or less than or equal to 3.

Throughout this paper, let f and g be the binary operation symbols and the unary operation symbols of type τ = (2,1), respectively. For a binary term a and an unary term b of type τ, the hypersubstitution which maps the operation symbol f to the term a and the operation symbol g to the term b will be denoted by σ_a,b.

For a binary term t ∈ W_(2,1)(X₂), we introduce the following notations:

leftmost(t)—the first variable (from the left) occurring in t,
rightmost(t)—the last variable occurring in t,
var(t)—the set of all variable occurring in t,
op(t)—the total number of all operation symbols occurring in t,
ops(t)—the set of all operation symbols occurring in t,
firstops(t)—the first operation symbol (from the left) occurring in t.

For t ∈ W_(2,1)(X₂), let Lp(t) denote the left path from the root to the leaf which is labelled by the leftmost variable in t and Rp(t) denote the right path from the root to the leaf which is labelled by the rightmost variable in t. The operation symbols occurring in Lp(t) and Rp(t) will be denoted by ops(Lp(t)) and ops(Rp(t)), respectively. If t ∈ W_(2,1)(X₂) such that var(t) = {x₁} or var(t) = {x₂}, we define t¹ = t and

(1.2)

For a and b of the hypersubstitution σ_a,b, we break our analysis into the following six cases, which cover all possibilities for a and b:

(I)
op(a) ≤ 1 and op(b) ≤ 1,
(II)
op(a) > 1 and op(b) > 1,
(III)
op(a) = 1 and op(b) > 1,
(IV)
op(a) > 1 and op(b) = 1,
(V)
op(a) > 1 and op(b) = 0,
(VI)
op(a) = 0 and op(b) > 1.

In case (I), we have term a ∈ {x₁, x₂, f(x₁, x₁), f(x₁, x₂), f(x₂, x₁), f(x₂, x₂), g(x₁), g(x₂)} and b ∈ {x₁, g(x₁), f(x₁, x₁)}. For σ_a,b we have twenty-four cases, for which the orders can be verified by simple calculations. The case a = g(x₂) and b = f(x₁, x₁) gives order 3, six other cases give order 2, and the remaining 17 cases give order 1.

2. Case II: op(a) > 1 and op(b) > 1

In this section, we consider the order of a hypersubstitution σ_a,b where op(a) > 1 and op(b) > 1. We consider three subcases of Case II:

(II-1) var(a) = {x₁, x₂}, var(b) = {x₁},
(II-2) var(a) = {x₁}, var(b) = {x₁},
(II-3) var(a) = {x₂}, var(b) = {x₁}.

The following formula for the operation symbol count of the compound term s(t₁, …, t_n) for some s, t₁, …, t_n ∈ W_τ(X) was proved in [5]:

(2.1)

where vb_j(s) = the number of occurrences of variable x_j in the term s.

Using the facts (see [5]) that for all t ∈ W_τ(X) if σ ∈ Hyp(τ) is regular, that is, var(σ(f_i)) = X_n, for all i ∈ I, and the formula above, we obtain the following theorem.

Theorem 2.1. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁). If op(a) > 1, op(b) > 1, var(a) = {x₁, x₂}, var(b) = {x₁}, then the order of σ_a,b is infinite.

Proof. Since σ_a,b is regular, by induction we obtain . Then, the order of σ_a,b is infinite.

The following lemmas are easy to prove.

Lemma 2.2. Let a, t ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) ≥ 1, op(b) ≥ 1. If t ∉ X₂, then is not a variable for all n ∈ ℕ.

Lemma 2.3. Let a, t ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁). If var(t) = {x_i} for some i ∈ {1,2}, then for all n ∈ ℕ.

Lemma 2.4. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) ≥ 1, op(b) ≥ 1, and a = f(a₁, a₂), b = g(b₁) for some a₁, a₂ ∈ W_(2,1)(X₂), b₁ ∈ W_(2,1)(X₁). Then, the following hold:

(i)
if var(a) = {x₁} and a₁ ≠ x₁, then the order of σ_a,b is infinite;
(ii)
if var(b) = {x₁} and b₁ ≠ x₁, then the order of σ_a,b is infinite.

Proof. (i) Assume var(a) = {x₁} and a₁ ≠ x₁. Then, a₁ ∉ X₂. Let n ∈ ℕ. By Lemma 2.2, . Since var(a) = {x₁}, by Lemma 2.3, . Now,

(2.2)

This shows that

for all n ∈ ℕ. Hence, the order of σ_a,b is infinite.

(ii) Assume var(b) = {x₁} and b₁ ≠ x₁, then b₁ ∉ X₁. Let n ∈ ℕ. By Lemma 2.2, . Since var(b) = {x₁}, by Lemma 2.3 for all n ∈ ℕ. Then,

(2.3)

Therefore,

for all n ∈ ℕ. Hence we have the claim.

Throughout the rest of this paper, we assume that when a is not a variable term it has the form f(a₁, a₂) or g(a₁), for some terms a₁, a₂ and that when b is not a variable term it has the form f(b₁, b₂) or g(b₁), for some terms b₁, b₂.

In Case (II-2), we have var(a) = {x₁}, var(b) = {x₁}. We consider the following four subcases:

(2.1) firstops(a) = f, firstops(b) = f;
(2.2) firstops(a) = g, firstops(b) = g;
(2.3) firstops(a) = f, firstops(b) = g;
(2.4) firstops(a) = g, firstops(b) = f.

In Case (2.1), we can separate into four subcases:

(2.1.1) a₁ ∉ X₂;
(2.1.2) a₁ = x₁, b₁ = x₁;
(2.1.3) a₁ = x₁, b₁ ∉ X₁, ops(Lp(b)) = {f};
(2.1.4) a₁ = x₁, b₁ ∉ X₁, ops(Lp(b))≠{f}.

Theorem 2.5. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₁} = var(b), and a = f(a₁, a₂), b = f(b₁, b₂) for some a₁, a₂ ∈ W_(2,1)(X₂), b₁, b₂ ∈ W_(2,1)(X₁). Then, the following hold:

(i)
if a and b satisfy (2.1.1) or (2.1.4), then σ_a,b has infinite order;
(ii)
if a and b satisfy (2.1.2) or (2.1.3), then the order of σ_a,b is less than or equal to 3.

Proof. (i) Assume a and b satisfy (2.1.1). By Lemma 2.2, we have for all n ∈ ℕ. Since , by Lemma 2.3 for all n ∈ ℕ. Therefore,

(2.4)

This shows that

for all n ∈ ℕ. Hence, σ_a,b has infinite order.

Assume a and b satisfy (2.1.4). Since , for all n ∈ ℕ. Since g ∈ ops(Lp(b)), we have for all n ∈ ℕ. Hence, the order of σ_a,b is infinite.

(ii) Assume a and b satisfy (2.1.2). Since , a₁ = x₁ and firstops(a) = f, we obtain . This gives

(2.5)

Since b₁ = x₁,

and firstops(b) = f, we have

. So,

(2.6)

Hence,

. This shows that the order of σ_a,b is less than or equal to 2.

Assume a and b satisfy (2.1.3). Because of (2.1.2), we have , which implies . Since ops(Lp(b)) = {f}, we have

(2.7)

for some

with m = op(Lp(b)) and

. It follows that

. Since

, we obtain

(2.8)

Thus,

. Hence, the order of σ_a,b is less than or equal to 3.

In Case (2.2), we have firstops(a) = g and firstops(b) = g. Since op(a) > 1 and op(b) > 1, a₁ ∉ X₂ and b₁ ∉ X₁. Then, we obtain the following.

Theorem 2.6. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₁}, a = g(a₁) and b = g(b₁) for some a₁ ∈ W_(2,1)(X₂), b₁ ∈ W_(2,1)(X₁). Then, σ_a,b has infinite order.

Proof. Assume that a = g(a₁), b = g(b₁), op(a) > 1, op(b) > 1, var(a) = {x₁}. Then, a₁ ∉ X₂, b₁ ∉ X₁. It can be proved, as in the proof of Theorem 2.5(i), that the order of σ_a,b is infinite.

In Case (2.3), we have firstops(a) = f and firstops(b) = g. Since op(b) > 1, b₁ ∉ X₁. Then, we obtain the following.

Theorem 2.7. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₁}, a = f(a₁, a₂) and b = g(b₁) for some a₁, a₂ ∈ W_(2,1)(X₂), b₁ ∈ W_(2,1)(X₁)∖X₁. Then, σ_a,b has infinite order.

Proof. By op(b) > 1, b₁ ∉ X₁. Since firstops(b) = g, var(b) = {x₁} and Lemma 2.4(ii), we have σ_a,b has infinite order.

In Case (2.4), we have firstops(a) = g and firstops(b) = f. Since op(a) > 1 and op(b) > 1, a₁ ∉ X₂ and b₁ ∉ X₁ or b₂ ∉ X₁. We consider the following three subcases:

(2.4.1) b₁ ∉ X₁;
(2.4.2) b₁ = x₁, op(a₁) = {g};
(2.4.3) b₁ = x₁, op(a₁)≠{g}.

Theorem 2.8. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₁}, var(b) = {x₁}, and a = g(a₁), b = f(b₁, b₂) for some a₁ ∈ W_(2,1)(X₂), b₁, b₂ ∈ W_(2,1)(X₁). Then, the following hold:

(i)
if a and b satisfy (2.4.1) or (2.4.3), then σ_a,b has infinite order;
(ii)
if a and b satisfy (2.4.2), then the order of σ_a,b is less than or equal to 2.

Proof. (i) Assume that a and b satisfy (2.4.1). By Lemma 2.2, we have for all n ∈ ℕ. Since , by Lemma 2.3 for all n ∈ ℕ. Therefore,

(2.9)

This shows that

for all n ∈ ℕ. Hence, σ_a,b has infinite order.

Assume that a and b satisfy (2.4.3). This case can be proved as in (2.4.1).

(ii) Assume that a and b satisfy (2.4.2). Then, and . We have and . Hence, the order of σ_a,b is less than or equal to 3.

In Case (II-3), we have var(a) = {x₂}, var(b) = {x₁}. We consider the following four subcases:

(3.1) firstops(a) = f, firstops(b) = f;
(3.2) firstops(a) = g, firstops(b) = g;
(3.3) firstops(a) = f, firstops(b) = g;
(3.4) firstops(a) = g, firstops(b) = f.

In Case (3.1), we can separate into four subcases:

(3.1.1) a₂ ∉ X₂;
(3.1.2) a₂ = x₂, b₂ = x₁;
(3.1.3) a₂ = x₂, b₂ ∉ X₁, ops(Rp(b)) = {f};
(3.1.4) a₂ = x₂, b₂ ∉ X₁, ops(Rp(b))≠{f}.

Theorem 2.9. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₂}, var(b) = {x₁}, and a = f(a₁, a₂), b = f(b₁, b₂) for some a₁, a₂ ∈ W_(2,1)(X₂), b₁, b₂ ∈ W_(2,1)(X₁). Then, the following hold:

(i)
if a and b satisfy (3.1.1) or (3.1.4), then σ_a,b has infinite order;
(ii)
if a and b satisfy (3.1.2) or (2.1.3), then the order of σ_a,b is less than or equal to 2.

Proof. (i) Assume that a and b satisfy (3.1.1). By Lemma 2.2, we have for all n in ℕ. Since , by Lemma 2.3 for all n ∈ ℕ. Therefore,

(2.10)

This shows that

for all n ∈ ℕ. Hence, σ_a,b has infinite order.

Assume that a and b satisfy (3.1.4). Then, for all n ∈ ℕ. Since and b₂ ∉ X₁, we have

(2.11)

This implies that

for all n ∈ ℕ. Hence, the order of σ_a,b is infinite.

(ii) Assume that a and b satisfy (3.1.2). Then, we get , or , . In the first, the order of σ_a,b is equal to 2. For the latter, σ_a,b is idempotent.

Assume that a and b satisfy (3.1.3). Then, , , which can be proved the same way as the first case of (ii).

In Case (3.2), we have firstops(a) = g and firstops(b) = g. Since op(a) > 1 and op(b) > 1, a₁ ∉ X₂ and b₁ ∉ X₁. Then, we obtain the following.

Theorem 2.10. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₂}, a = g(a₁) and b = g(b₁) for some a₁ ∈ W_(2,1)(X₂)∖X₂, b₁ ∈ W_(2,1)(X₁)∖X₁. Then σ_a,b has infinite order.

Proof. Since var(a) = {x₂} and b₁ ∉ X₁, we have

(2.12)

Hence, the order of σ_a,b is infinite.

In Case (3.3), we have firstops(a) = f and firstops(b) = g. Since op(b) > 1, b₁ ∉ X₁. Then, we obtain the following.

Theorem 2.11. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₂}, a = f(a₁, a₂) and b = g(b₁) for some a₁, a₂ ∈ W_(2,1)(X₂), b₁ ∈ W_(2,1)(X₁)∖X₁. Then σ_a,b has infinite order.

Proof. Since b₁ ∉ X₁ firstops(b) = g and Lemma 2.4(ii), we have σ_a,b has infinite order.

In Case (3.4), we have firstops(a) = g and firstops(b) = f. Since op(a) > 1 and op(b) > 1, a₁ ∉ X₂ and b₁ ∉ X₁. We have the following result.

Theorem 2.12. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) > 1, var(a) = {x₂}, var(b) = {x₁}, and a = g(a₁), b = f(b₁, b₂) for some a₁ ∈ W_(2,1)(X₂), b₁, b₂ ∈ W_(2,1)(X₁). Then, σ_a,b has infinite order.

Proof. Since op(a) > 1, a₁ ∉ X₂. We have, by Lemma 2.2, for all n in ℕ. Since , by Lemma 2.3 for all n ∈ ℕ. We obtain

(2.13)

Hence, σ_a,b has infinite order.

3. Case III: op(a) = 1 and op(b) > 1

We consider three subcases:

(III-1) var(a) = X₂, var(b) = {x₁};
(III-2) var(a) = {x₁}, var(b) = {x₁};
(III-3) var(a) = {x₂}, var(b) = {x₁}.

In Case (III-1), we separate into the following subcases:

(1)
a = f(x₁, x₂), firstops(b) = f;
(2)
a = f(x₁, x₂), firstops(b) = g;
(3)
a = f(x₂, x₁), firstops(b) = f;
(4)
a = f(x₂, x₁), firstops(b) = g.

We have the following results. Subcases (2) and (4) give infinite order, by Lemma 2.4(ii). The next proposition deal with subcases (1) and (3).

Proposition 3.1. Let b = f(b₁, b₂) for some b₁, b₂ ∈ W_(2,1)(X₁) be such that var(b) = {x₁} and op(b) > 1. Then, the following hold:

(i)
if a = f(x₁, x₂), then the order of σ_a,b is equal to 1 or is infinite;
(ii)
if a = f(x₂, x₁), then the order of σ_a,b is less than or equal to 3 or infinite.

Proof. (i) Assume that a = f(x₁, x₂). Then, σ_a,b[a] = a. We consider the following.

(i′) If ops(b) = {f}, then σ_a,b[b] = b. We get that σ_a,b is idempotent.

(i′′) If g ∈ ops(b), suppose that g ∈ ops(b₁). Let k ∈ ℕ. Then .

Now,

(3.1)

A similar argument works for g ∈ ops(b₂). This shows that the order of σ_a,b is infinite.

(ii) If ops(b) = {f}, then and . This gives . Then, the order of σ_a,b is less than or equal to 3. If g ∈ ops(b), then it can be proved as in (i′′) that the order is infinite.

In Case (III-2), we have var(a) = {x₁}, var(b) = {x₁}. We consider the following subcases:

(3.2.1) a = f(x₁, x₁), firstops(b) = f;
- (3.2.1.1) b₁ = x₁;
- (3.2.1.2) b₁ ≠ x₁, g ∈ ops(Lp(b));
- (3.2.1.3) b₁ ≠ x₁, g ∉ ops(Lp(b));
(3.2.2) a = g(x₁), firstops(b) = g, then b₁ ∉ X₁;
(3.2.3) a = f(x₁, x₁), firstops(b) = g, then b₁ ∉ X₁;
(3.2.4) a = g(x₁), firstops(b) = f;
- (3.2.4.1) b₁ = x₁;
- (3.2.4.2) b₁ ≠ x₁.

Proposition 3.2. Let a = f(x₁, x₁), firstops(b) = f, op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (3.2.1.1) or (3.2.1.3), then the order of σ_a,b is less than or equal to 2;
(ii)
if b satisfies (3.2.1.2), then the order of σ_a,b is infinite.

Proof. (i) If a and b satisfy (3.2.1.1), then . Assume that a and b satisfy (3.2.1.3). Since a = f(x₁, x₁) and , we have . Clearly, . Then, . Hence, the order of σ_a,b is less than or equal to 2.

(ii) Assume that a and b satisfy (3.2.1.2). Since a = f(x₁, x₁), firstops(b) = f and g ∈ ops(Lp(b)), it follows that for some v, v^′ ∈ W_(2,1)(X₁). For n ∈ ℕ, we have

(3.2)

Hence, the order of σ_a,b is infinite.

Proposition 3.3. Let a = g(x₁) or a = f(x₁, x₁), firstops(b) = g, op(b) > 1, and var(b) = {x₁}. Then, the order of σ_a,b is infinite.

Proof. Assume that a = g(x₁). Since firstops(b) = g, a = g(x₁), . For any natural number n, makes σ_a,b have infinite order.

Assume that a = f(x₁, x₁). Since firstops(b) = g, a = f(x₁, x₁), . For n ∈ ℕ, a similar proof to those above gives . Hence, the order of σ_a,b is infinite.

Proposition 3.4. Let a = g(x₁), firstops(b) = f, op(b) > 1 and var(b) = {x₁}. Then the following hold:

(i)
if b satisfies (3.2.4.1), then the order of σ_a,b is equal to 2;
(ii)
if b satisfies (3.2.4.2), then the order of σ_a,b is infinite.

Proof. (i) If a and b satisfy (3.2.4.1), then and . Hence, the order of σ_a,b is equal to 2.

(ii) Assume that a and b satisfy (3.2.4.2). Since a = g(x₁), firstops(b) = f, it follows that . For n ∈ ℕ. As in the proof of Proposition 3.3, we have . Hence, the order of σ_a,b is infinite.

In Case (III-3), we have var(a) = {x₂}, var(b) = {x₁}. We consider the following subcases:

(3.3.1) a = f(x₂, x₂), firstops(b) = f;
- (3.3.1.1) b₂ = x₁;
- (3.3.1.2) b₂ ≠ x₁;
(3.3.2) a = g(x₂), firstops(b) = g.
(3.3.3) a = f(x₂, x₂), firstops(b) = g.
(3.3.4) a = g(x₂), firstops(b) = f;
- (3.3.4.1) b₂ = x₁;
- (3.3.4.2) b₂ ≠ x₁.

Proposition 3.5. Let a = f(x₂, x₂), firstops(b) = f, op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (3.3.1.1), then the order of σ_a,b is equal to 2;
(ii)
if b satisfies (3.3.1.2), then the order of σ_a,b is infinite.

Proof. (i) If a and b satisfy (3.3.1.1), then and . Hence the order of σ_a,b is equal to 2.

(ii) This can be proved in the same way as Proposition 3.2 (ii).

Proposition 3.6. Let a = g(x₂) or a = f(x₂, x₂), firstops(b) = g, op(b) > 1, and var(b) = {x₁}. Then, the order of σ_a,b is infinite.

Proof. Assume that a = g(x₂). Since firstops(b) = g, a = g(x₂), . The same argument works as in Proposition 3.3. Hence, the order of σ_a,b is infinite. The case that a = f(x₂, x₂) can be proved similarly.

Proposition 3.7. Let a = g(x₂), firstops(b) = f, op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (3.3.4.1), then the order of σ_a,b is equal to 2;
(ii)
if b satisfies (3.3.4.2), then the order of σ_a,b is infinite.

Proof. (i) If a and b satisfy (3.3.4.1), then and . Hence, the order of σ_a,b is equal to 2.

(ii) Assume a and b satisfy (3.3.4.2), Since b = g(x₂), firstops(b) = f, b₂ ∉ X₂. By Lemma 2.2, we have . Then,

(3.3)

This shows that,

for all a ∈ W_(2,1)(X₁)∖X₁. Consequently,

for all k ∈ ℕ. Then, order of σ_a,b is infinite.

4. Case IV: op(a) > 1 and op(b) = 1

We consider three subcases:

(IV-1) var(a) = X₂, var(b) = {x₁};
(IV-2) var(a) = {x₁}, var(b) = {x₁};
(IV-3) var(a) = {x₂}, var(b) = {x₁}.

In Case (IV-1), we separate into four cases:

(4.1.1) b = f(x₁, x₁), firstops(a) = f;
(4.1.2) b = f(x₁, x₁), firstops(a) = g;
(4.1.3) b = g(x₁), firstops(a) = f;
(4.1.4) b = g(x₁), firstops(a) = g.

Theorem 4.1. Let a ∈ W_(2,1)(X₂), b ∈ W_(2,1)(X₁) be such that op(a) > 1, op(b) = 1, var(a) = {x₁, x₂}, var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (4.1.1) or (4.1.2) or (4.1.3), then the order of σ_a,b is infinite;
(ii)
if b satisfies (4.1.4), then the order of σ_a,b is equal to 1 or is infinite.

Proof. (i) If a and b satisfy (4.1.1), then we let a = f(a₁, a₂) for some a₁, a₂ ∈ W_(2,1)(X₂). Since op(a) > 1, we may assume that a₁ ∉ X₂. By Lemma 2.2, we have . For any natural number n, makes σ_a,b have infinite order. Cases (4.1.2) and (4.1.3) can be proved similarly.

(ii) Assume that a and b satisfy (4.1.4). If ops(a₁) = {g}, then and . Hence, σ_a,b is idempotent. The case that op(a₁)≠{g} can be proved as in (i).

In Case (IV-2), we have var(a) = {x₁}, var(b) = {x₁}. If a = g(a₁) for some term a₁, then op(a) > 1 means that a₁ is not variable.

Proposition 4.2. Let b = f(x₁, x₁) or b = g(x₁), op(a) > 1, a = g(a₁) for some a₁ and var(a) = {x₁} = var(b). Then, the order of σ_a,b is infinite.

Proof. Assume op(a) > 1, a = g(a₁) for some a₁ and var(a) = {x₁} = var(b). Since a₁ ∉ X₂. By Lemma 2.2 we have . It can be proved, as in the proof of Theorem 4.1(i), that the order of σ_a,b is infinite.

In Case (IV-3), we have var(a) = {x₂}, var(b) = {x₁}. If firstops(a) = g, then a₂ ∉ X₂. We consider the following cases:

(4.3.1) a₂ ∉ X₂, b = f(x₁, x₁);
(4.3.2) a₂ ∉ X₂, b = g(x₁).

Proposition 4.3. Let b = f(x₁, x₁), op(a) > 1, and var(a) = {x₂}, var(b) = {x₁}. If a and b satisfy (4.3.1), then the order of σ_a,b is infinite.

Proof. The case (4.3.1) can be verified as in Proposition 4.2.

Proposition 4.4. Let b = g(x₁), op(a) > 1, and var(a) = {x₂}, var(b) = {x₁}. If a and b satisfy (4.3.2), then the order of σ_a,b is equal to 1 or is infinite.

Proof. Assume that a and b satisfy (4.3.2). If b = g(x₁), firstops(a) = g, and ops(a₂) = {g}, then and . Hence, σ_a,b is idempotent. Assume that a₂ ∉ X₂. This case can be proved as in Proposition 4.3.

5. Case V: op(a) > 1 and op(b) = 0

In this case, we have a = f(a₁, a₂) or a = g(a₁), b = x₁. We consider three subcases:

(V-1) var(a) = X₂, b = x₁;
(V-2) var(a) = {x₁}, b = x₁;
(V-3) var(a) = {x₂}, b = x₁.

In Case (V-1), we separate into the following subcases:

(5.1.1) firstops(a) = f, a₁ ∉ X₂, ops(Lp(a))≠{g};
(5.1.2) firstops(a) = f, a₁ ∉ X₂, ops(Lp(a)) = {g};
(5.1.3) firstops(a) = f, a₁ ∈ X₂, ops(Rp(a))≠{g};
(5.1.4) firstops(a) = f, a₁ ∈ X₂, ops(Rp(a)) = {g};
(5.1.5) firstops(a) = g, ops(a₁) = {g};
(5.1.6) firstops(a) = g, ops(a₁)≠{g}.

Proposition 5.1. Let a = f(a₁, a₂), op(a) > 1, op(b) = 0, var(a) = {x₁, x₂}, and b = x₁. Then, the following hold:

(i)
if a satisfies (5.1.1) or (5.1.3), then the order of σ_a,b is infinite;
(ii)
if a satisfies (5.1.2) or (5.1.4), then the order of σ_a,b is less than or equal to 2.

Proof. (i) Assume a and b satisfy (5.1.1). Then, firstops(a) = f, var(a) = {x₁, x₂}, and f ∈ ops(Lp(a)). Then, . For any natural number n, makes σ_a,b have infinite order. A similar argument works for (5.1.3).

(ii) Assume a and b satisfy (5.1.2). Then, firstops(a) = f, var(a) = {x₁, x₂}, and ops(Lp(a)) = {g}. If leftmost(a₁) = x₁, then and . Hence, σ_a,b is idempotent. If leftmost(a₁) = x₂, then and . Hence, the order of σ_a,b is equal to 2.

Assume a and b satisfy (5.1.4). Then, firstops(a) = f, var(a) = {x₁, x₂} and ops(Rp(a)) = {g}. If rightmost(a₂) = x₂, then and . Hence, σ_a,b is idempotent. If leftmost(a₂) = x₁, then and . Hence, the order of σ_a,b is equal to 2.

Proposition 5.2. Let a = g(a₁), op(a) > 1, op(b) = 0, var(a) = {x₁, x₂}, and b = x₁. If a satisfies (5.1.5) or (5.1.6), then, the order of σ_a,b is less than or equal to 2.

Proof. Assume that a and b satisfy (5.1.5). Then, firstops(a₁) = g and ops(a₁) = {g}. If leftmost(a₁) = x₁, then and . Hence, the order of σ_a,b is equal to 2. If leftmost(a₁) = x₂, then and . Again, the order of σ_a,b is equal to 2.

Assume that a and b satisfy (5.1.6). Then firstops(a₁) = f and f ∈ ops(a₁). If leftmost(a₁) = x₁, then and . Hence, σ_a,b is idempotent. If leftmost(a₁) = x₂, then and . Hence, the order of σ_a,b is equal to 2.

In Case (V-2), we have var(a) = {x₁}, var(b) = {x₁}. We separate this case into following subcases:

(5.2.1) firstops(a) = f, a₁ = x₁;
(5.2.2) firstops(a) = f, a₁ ∉ X₂, ops(Lp(a)) = {g};
(5.2.3) firstops(a) = f, a₁ ∉ X₂, ops(Lp(a))≠{g};
(5.2.4) firstops(a) = g, ops(a₁) = {g};
(5.2.5) firstops(a) = g, ops(a₁)≠{g}.

Proposition 5.3. Let a = f(a₁, a₂), op(a) > 1, op(b) = 0, var(a) = {x₁}, and b = x₁. Then, the following hold:

(i)
if a satisfies (5.2.1) or (5.2.2), then the order of σ_a,b is equal to 1;
(ii)
if a satisfies (5.2.3), then the order of σ_a,b is infinite.

Proof. (i) Assume that a and b satisfy (5.2.1). Since firstops(a₁) = f and var(a) = {x₁}, we have . Since b = x₁, . Hence, σ_a,b is idempotent.

Assume a and b satisfy (5.2.2). Since ops(a₁) = g and var(a) = {x₁}, we have and . Hence, σ_a,b is idempotent.

(ii) Assume a and b satisfy (5.2.3). Since firstops(a₁) = f and var(a) = {x₁}, we have for all k ∈ ℕ. For any natural number n, makes σ_a,b have infinite order.

Proposition 5.4. Let a = g(a₁), op(a) > 1, op(b) = 0, var(a) = {x₁}, and b = x₁. Then, the following hold:

(i)
if a satisfies (5.2.4), then the order of σ_a,b is equal to 2;
(ii)
if a satisfies (5.2.5), then the order of σ_a,b is equal to 1 or is infinite.

Proof. (i) Assume a and b satisfy (5.2.4). Then, . Hence, the order of σ_a,b is equal to 2.

(ii) Assume a and b satisfy (5.2.5). If there is only one f ∈ ops(a₁), then and . Hence, σ_a,b is idempotent.

Assume that the symbol f occurs more than twice in term a₁. Then, for all k ∈ ℕ. For any natural number n, makes σ_a,b have infinite order.

In Case (V-3), we have var(a) = {x₂}, b = x₁. We separate into the following subcases:

(5.3.1) firstops(a) = f, a₂ = x₂;
(5.3.2) firstops(a) = f, a₂ ∉ X₂, ops(a₂) = {g};
(5.3.3) firstops(a) = f, a₂ ∉ X₂, ops(a₂)≠{g};
(5.3.4) firstops(a) = g, ops(a₁) = {g};
(5.3.5) firstops(a) = g, ops(a₁)≠{g}.

Proposition 5.5. Let a = f(a₁, a₂), op(a) > 1, op(b) = 0, var(a) = {x₂}, and b = x₁. Then, the following hold:

(i)
if a satisfies (5.3.1) or (5.3.2), then the order of σ_a,b is equal to 1;
(ii)
if a satisfies (5.3.3), then the order of σ_a,b is infinite.

Proof. This can be proved similarly to the proof of Proposition 5.3.

Proposition 5.6. Let a = g(a₁), op(a) > 1, op(b) = 0, var(a) = {x₂}, and b = x₁. Then, the following hold:

(i)
if a satisfies (5.3.4), then the order of σ_a,b is equal to 2;
(ii)
if a satisfies (5.3.5), then the order of σ_a,b is equal to 1 or is infinite.

Proof. This can be proved similarly to the proof of Proposition 5.4.

6. Case VI: op(a) = 0 and op(b) > 1

In this case, we consider the following cases:

(6.1) a = x₁;
- (6.1.1) firstops(b) = f, b₁ = x₁;
- (6.1.2) firstops(b) = f, b₁ ∉ X₁, ops(Lp(b)) = {f};
- (6.1.3) firstops(b) = f, b₁ ∉ X₁, ops(Lp(b))≠{f};
- (6.1.4) firstops(b) = g, ops(b₁) = {f};
- (6.1.5) firstops(b) = g, ops(b₁)≠{f};
(6.2) a = x₂;
- (6.2.1) firstops(b) = f, b₂ = x₁;
- (6.2.2) firstops(b) = f, b₂ ∉ X₁, ops(Rp(b)) = {f};
- (6.2.3) firstops(b) = f, b₂ ∉ X₁, ops(Rp(b))≠{f};
- (6.2.4) firstops(b) = g, ops(b₁) = {f};
- (6.2.5) firstops(b) = g, ops(b₁)≠{f}.

Proposition 6.1. Let a = x₁, b = f(b₁, b₂), op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (6.1.1) or (6.1.2), then the order of σ_a,b is equal to 2;
(ii)
if b satisfies (6.1.3), then the order of σ_a,b is equal to 1 or is infinite.

Proof. (i) If a and b satisfy (6.1.1) or (6.1.2), then and . Hence, the order of σ_a,b is equal to 2.

(ii) Assume a and b satisfy (6.1.3). If there is only one occurrence of g in b₁, then and . Hence, σ_a,b is idempotent.

Assume that g occurs more than twice in b₁. Then, for all k ∈ ℕ. Then,

(6.1)

This shows that the order of σ_a,b is infinite.

Proposition 6.2. Let a = x₁, b = g(b₁), op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (6.1.4), then the order of σ_a,b is equal to 1;
(ii)
if b satisfies (6.1.5), then the order of σ_a,b is infinite.

Proof. (i) If a and b satisfy (6.1.4), then and . Hence, σ_a,b is idempotent.

(ii) Assume a and b satisfy (6.1.5). Then, for all k ∈ ℕ. Then,

(6.2)

This shows that the order of σ_a,b is infinite.

Proposition 6.3. Let a = x₂, b = f(b₁, b₂), op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (6.2.1) or (6.2.2), then the order of σ_a,b is equal to 2;
(ii)
if b satisfies (6.2.3), then the order of σ_a,b is equal to 1 or is infinite.

Proof. (i) If a and b satisfy (6.2.1) or (6.2.2), then and . Hence the order of σ_a,b is equal to 2.

(ii) This case can be proved similarly to the proof of Proposition 6.1(ii).

Proposition 6.4. Let a = x₂, b = g(b₁), op(b) > 1, and var(b) = {x₁}. Then, the following hold:

(i)
if b satisfies (6.2.4), then the order of σ_a,b is equal to 1;
(ii)
if b satisfies (6.2.5), then the order of σ_a,b is infinite.

Proof. The proofs follow those of Proposition 6.2 (i) and (ii), respectively.

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The Order of Hypersubstitutions of Type (2,1)

Abstract

1. Preliminaries

2. Case II: op(a) > 1 and op(b) > 1

3. Case III: op(a) = 1 and op(b) > 1

4. Case IV: op(a) > 1 and op(b) = 1

5. Case V: op(a) > 1 and op(b) = 0

6. Case VI: op(a) = 0 and op(b) > 1

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The Order of Hypersubstitutions of Type (2,1)

Abstract

1. Preliminaries

2. Case II: op(a) > 1 and op(b) > 1

3. Case III: op(a) = 1 and op(b) > 1

4. Case IV: op(a) > 1 and op(b) = 1

5. Case V: op(a) > 1 and op(b) = 0

6. Case VI: op(a) = 0 and op(b) > 1

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