# Direct product is cancellative for finite groups

From Groupprops

## Contents

## 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:

- Direct product is cancellative for finite algebras in any variety with zero: The proof outlined here generalizes to finite algebras in any variety with zero.

### Stronger facts for groups

- Krull-Remak-Schmidt theorem: This states something stronger -- there is an analogue of unique factorization for finite groups into directly indecomposable groups. In fact, the result holds for a somewhat more general class of groups: groups that satisfy both the ascending and descending chain conditions on subgroups. This has two important corollaries:
- Corollary of Krull-Remak-Schmidt theorem for cancellation of powers: This states that if , and satisfies both the ascending and descending chain conditions on normal subgroups, then .
- Corollary of Krull-Remak-Schmidt theorem for cancellation of direct factors: This states the cancellation result for the larger collection of groups, namely those satisfying the ascending and descending chain conditions on normal subgroups.

### Facts for other kinds of products

- Retract not implies normal complements are isomorphic: This basically shows that the analogue of the direct product, namely, the semidirect product, is
*not*right-cancellative for finite groups (it actually shows something stronger). - Semidirect product is not left-cancellative for finite groups: If are finite groups such that , that does
*not*imply that is isomorphic to . - Every group of given order is a permutable complement for symmetric groups: Any group of order is a permutable complement to inside . Thus, if two subgroups are permutable complements to the same subgroup in the whole group, the
*only*conclusion we can draw is that they gave the same order.

## Facts used

- 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: .

- 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] |