Direct product is cancellative for finite groups

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Statement

Suppose G, H, K are finite groups, such that:

G \times H \cong G \times K

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

Then, we have:

H \cong K.

Related facts

Related facts for other algebraic structures

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

Stronger facts for groups

Facts for other kinds of products

Other related facts

Facts used

  1. Homomorphism set to direct product is Cartesian product of homomorphism sets: If A,B,C are groups, 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 groups 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_{N \triangleleft A} \operatorname{IHom}(A/N, B).

Proof

Given: Finite groups G, H, K such that G \times H \cong G \times K.

To prove: H \cong K.

Proof: Let L be an arbitrary finite group.

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 a group 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 group 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 subgroup of K and K is isomorphic to a subgroup of H H,K are finite Step (5) [SHOW MORE]
7 H is isomorphic to K H,K are finite Step (6) [SHOW MORE]