Direct product is cancellative for finite algebras in any variety with zero

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Statement

Suppose \mathcal{V} is a variety of algebras that is a variety with zero. Suppose G,H,K are finite algebras in \mathcal{V}. Further, suppose that the algebras G \times H and G \times K are isomorphic as algebras in \mathcal{V}. Then, H is isomorphic to K.

Related facts

Facts used

  1. Homomorphism set to direct product is Cartesian product of homomorphism sets: If A,B,C are algebras, then there is a natural bijection:
    • \operatorname{Hom}(A,B) \times \operatorname{Hom}(A,C) \leftrightarrow \operatorname{Hom}(A,B \times C).
    • The bijection is defined as: (\alpha,\beta) \mapsto (g \mapsto (\alpha(g),\beta(g)).
  2. Homomorphism set is disjoint union of injective homomorphism sets: For algebras A and B, let \operatorname{Hom}(A,B) denotes the set of homomorphisms from A to B, and \operatorname{IHom}(A,B) denote the set of injective homomorphisms from A to B. Then we have:

\operatorname{Hom}(A,B) = \bigsqcup_{~} \operatorname{IHom}(A/~, B).

Here ~ varies over the set of all possible congruences on the algebra A.

Proof

Given: Finite algebras G, H, K such that G \times H \cong G \times K.

To prove: H \cong K.

Proof: Let L be an arbitrary finite algebra in \mathcal{V}. Note that the trivial homomorphism refers to the map that sends every element to the zero element.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 |\operatorname{Hom}(L,G \times H)| = |\operatorname{Hom}(L,G)||\operatorname{Hom}(L,H)| as an equality of finite numbers. Fact (1) G,H are finite [SHOW MORE]
2 |\operatorname{Hom}(L,G \times K)| = |\operatorname{Hom}(L,G)||\operatorname{Hom}(L,K)| as an equality of finite numbers. Fact (1) G,K are finite [SHOW MORE]
3 |\operatorname{Hom}(L,G \times H)| = |\operatorname{Hom}(L,G \times K)| as an equality of finite numbers. G \times H \cong G \times K The number of homomorphisms to an algebra depends only on its isomorphism type.
4 |\operatorname{Hom}(L,H)| = |\operatorname{Hom}(L,K)| as an equality of finite numbers. Steps (1),(2),(3) [SHOW MORE]
5 For any finite algebra L, the number of injective homomorphisms from L to H equals the number of injective homomorphisms from L to K. We show this by induction on the order of L. In other words, |\operatorname{IHom}(L,H)| = |\operatorname{IHom}(L,K)| Fact (2) Step (4) [SHOW MORE]
6 H is isomorphic to a subalgebra of K and K is isomorphic to a subalgebra of H H,K are finite Step (5) [SHOW MORE]
7 H is isomorphic to K H,K are finite Step (6) [SHOW MORE]