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

## 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 $K$.

## 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]