By V. A. Krechmar

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Show that A is one-generated (but it is not necessarily a cyclic group). 6. 3. Prove that q(x, y, z) = fn (x, z, y) is a Mal’cev term for any n–step nilpotent algebra in U. 7. Let G and H be Abelian groups and let A be their disjoint union. Define a ternary operation p(x, y, z) on A as follows. If x, y and z all lie in one of the groups, let p(x, y, z) = x − y + z. Otherwise two of x, y and z lie in one group and the other lies in the other group. In this case p(x, y, z) is the one that is alone.

Hence [θ, θ] ≤ [γ, α ∨ β] = [γ, α] ∨ [γ, β] ≤ (γ ∧ α) ∨ (γ ∧ β). 5 x (γ ∧ α) ∨ (γ ∧ β) d(x, z, z). So (7) x (γ ∧ α) ∨ (γ ∧ β) d(x, z, z) β d(x, y, z) α d(x, x, z) = z. Notice that this shows that every modular variety satisfies the congruence identity (α◦β)∧γ ≤ ((γ ∧α)∨(γ ∧β))◦β ◦α. Now we apply this to F = FV(x, y, z) with α = Cg(x, y), β = Cg(y, z), and γ = Cg(x, z). By (7) there are terms q0 (x, y, z) = x, q1 (x, y, z), . . , qn (x, y, z) = d(x, z, z) such that, in F, qi (x, y, z) β qi+1 (x, y, z) if i is even, qi (x, y, z) α qi+1 (x, y, z) if i is odd, and qi (x, y, z) γ qi+1 (x, y, z) for i = 0, .

If A ∈ U and x, y, b, c ∈ A the for all n fn (p(x, b, c), b, c) (1]n x. and p(fn (y, b, c), b, c) (1]n y. Proof. We prove the second relation: the proof of the first is similar but easier. Induct on n. Since (1]0 = 1 the result is trivial for n = 0. Let y ′ = p(fn (y, b, c), b, c). By induction y ′ (1]n y. In the algebra A/(1]n+1, (1]n is contained in the center. 7 (since b = p(b, y, y) (1]n p(b, y, y ′)) p(fn+1 (y, b, c), b, c) = p((p(b, p(b, y, y ′), fn (y, b, c)), b, c) = p(p(b, p(b, y, y ′), fn (y, b, c)), p(b, b, b), p(b, b, c)) ≡ p(p(b, b, b), p((p(b, y, y ′), b, b), p(fn (y, b, c), b, c)) (mod (1]n+1 ) = p(b, p(b, y, y ′), y ′).

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