Direct product is cancellative for finite groups
Suppose are finite groups, such that:
where denotes the external direct product. (Note that the isomorphism need not
Then, we have:
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.
- 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:
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]|