# Direct product is cancellative for finite groups

## Statement

Suppose  are finite groups, such that:



where  denotes the external direct product. (Note that the isomorphism need not

Then, we have:

.

## Related facts

### Related facts for other algebraic structures

The statement is true for finite algebras in any variety of algebras:

## Facts used

1. Homomorphism set to direct product is Cartesian product of homomorphism sets: If  are groups, then there is a natural bijection:
• .
• The bijection is defined as: .
2. Homomorphism set is disjoint union of injective homomorphism sets: For groups  and , let  denotes the set of homomorphisms from  to , and  denote the set of injective homomorphisms from  to . Then we have:

.

## Proof

Given: Finite groups  such that .

To prove: .

Proof: Let  be an arbitrary finite group.

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 a group depends only on its isomorphism type.
4  as an equality of finite numbers. Steps (1),(2),(3) [SHOW MORE]
5 For any finite group , 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 subgroup of  and  is isomorphic to a subgroup of   are finite Step (5) [SHOW MORE]
7  is isomorphic to   are finite Step (6) [SHOW MORE]