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

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. 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] By Fact (1), there is a bijection between $\operatorname {Hom} (L,G)\times \operatorname {Hom} (L,H)$ and $\operatorname {Hom} (L,G\times H)$. Since all groups involved are finite, so are all $\operatorname {Hom}$-sets, so the bijection gives a fact about cardinalities.
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] By Fact (1), there is a bijection between $\operatorname {Hom} (L,G)\times \operatorname {Hom} (L,K)$ and $\operatorname {Hom} (L,G\times K)$. Since all groups involved are finite, so are all $\operatorname {Hom}$-sets, so the bijection gives a fact about cardinalities.
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] Combining the previous three steps, we get $|\operatorname {Hom} (L,G)||\operatorname {Hom} (L,H)|=|\operatorname {Hom} (L,G)||\operatorname {Hom} (L,K)|$. Note now that $|\operatorname {Hom} (L,G)|$ is nonzero since there is at least one homomorphism (the trivial homomorphism) between the groups. So, it can be canceled from the product of numbers (it is critical here that we are working with finite numbers rather than infinite cardinals)
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] Base case for induction on the order of $L$: Case of trivial group is obvious. Inductive step: By Fact (2), we have$|\operatorname {Hom} (L,H)|=\sum _{N\triangleleft L}|\operatorname {IHom} (L/N,H)|$ and $|\operatorname {Hom} (L,K)|=\sum _{N\triangleleft L}|\operatorname {IHom} (L/N,K)|$. By induction on the order, we can assume that for every nontrivial normal subgroup $N$ of $L$, $\operatorname {IHom} (L/N,H)=\operatorname {IHom} (L/N,K)$. Since the left sides of both equations are equal by step (4), we obtain that $\operatorname {IHom} (L,H)=\operatorname {IHom} (L,K)$.
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] Set $L=H$ in step (5). Clearly, $\operatorname {IHom} (H,H)$ is nonempty (it contains the identity map, and all the automorphisms of $H$). Thus, $\operatorname {IHom} (H,K)$ is nonempty, yielding that $H$ is isomorphic to the subgroup of $K$. Similarly, $K$ is isomorphic to a subgroup of $H$.
7	$H$ is isomorphic to $K$		$H,K$ are finite	Step (6)	[SHOW MORE] For finite groups, each being isomorphic to a subgroup of the other forces them to be equal in size, hence forcing the injective homomorphisms to be isomorphisms.