Direct product is cancellative for finite groups

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 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 $$G^m \cong H^m$$, and $$G$$ satisfies both the ascending and descending chain conditions on normal subgroups, then $$G \cong H$$.
 * 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 $$G,H,K$$ are finite groups such that $$G \rtimes H \cong G \rtimes K$$, that does not imply that $$H$$ is isomorphic to $$K$$.
 * Every group of given order is a permutable complement for symmetric groups: Any group of order $$n$$ is a permutable complement to $$\operatorname{Sym}(n-1)$$ inside $$\operatorname{Sym}(n)$$. 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.

Other related facts

 * Series-equivalent characteristic subgroups may be distinct

Facts used

 * 1) uses::Homomorphism set to direct product is Cartesian product of homomorphism sets: If $$A,B,C$$ are groups, then there is a natural bijection:
 * 2) * $$\operatorname{Hom}(A,B) \times \operatorname{Hom}(A,C) \leftrightarrow \operatorname{Hom}(A,B \times C)$$.
 * 3) * The bijection is defined as: $$(\alpha,\beta) \mapsto (g \mapsto (\alpha(g),\beta(g))$$.
 * 4) uses::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.