# 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:

### 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.

## 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]