Direct product is cancellative for finite algebras in any variety with zero
- Direct product is cancellative for finite groups: Essentially the same proof, tailored to groups (the key specific detail for groups is that the congruences can be described using the normal subgroups that function as their kernels).
- Homomorphism set to direct product is Cartesian product of homomorphism sets: If are algebras, then there is a natural bijection:
- The bijection is defined as: .
- Homomorphism set is disjoint union of injective homomorphism sets: For algebras and , let denotes the set of homomorphisms from to , and denote the set of injective homomorphisms from to . Then we have:
Here varies over the set of all possible congruences on the algebra .
Given: Finite algebras such that .
To prove: .
Proof: Let be an arbitrary finite algebra in . 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||as an equality of finite numbers.||Fact (1)||are finite||[SHOW MORE]|
|2||as an equality of finite numbers.||Fact (1)||are finite||[SHOW MORE]|
|3||as an equality of finite numbers.||The number of homomorphisms to an algebra depends only on its isomorphism type.|
|4||as an equality of finite numbers.||Steps (1),(2),(3)||[SHOW MORE]|
|5||For any finite algebra , the number of injective homomorphisms from to equals the number of injective homomorphisms from to . We show this by induction on the order of . In other words,||Fact (2)||Step (4)||[SHOW MORE]|
|6||is isomorphic to a subalgebra of and is isomorphic to a subalgebra of||are finite||Step (5)||[SHOW MORE]|
|7||is isomorphic to||are finite||Step (6)||[SHOW MORE]|