# Equivalence of internal and external direct product

This article gives a proof/explanation of the equivalence of multiple definitions for the term direct product

View a complete list of pages giving proofs of equivalence of definitions

## The definitions we have to prove as equivalent

### Internal direct product

Suppose is a group, and are normal subgroups. We say that is an internal direct product of and if is trivial, and .

### External direct product

## What we need to prove

### Every internal direct product is naturally isomorphic to the external direct product

We need to prove that given a group that is the internal direct product of normal subgroups , we can identify naturally with the external direct product .

*Proof*: Consider the map:

We first must verify that is a group homomorphism. For this, we first observe that if , then:

Combining these, we get:

Hence, we get:

.

Thus, any element of commutes with any element of . Thus, if we pick two pairs , then:

In the intermediate step, we use that and commute. Thus, is a group homomorphism. It's clear that preserves the identity element and inverses, too.

Next we note that:

- is injective because .
- is surjective because by assumption, .

Thus, defines an isomorphism from to .

### Every external direct product is naturally realized as an internal direct product

We need to show that if and are abstract groups, then we can find subgroups in isomorphic to and respectively, whose internal direct product is . Indeed:

- The subgroup isomorphic to is (i.e., elements of the form ).
- The subgroup isomorphic to is (i.e., elements of the form ).

Let's now check the conditions for an internal direct product:

- Both the subgroups are normal: This is a direct check using the coordinate-wise multiplication.
- The subgroups intersect only at : This is set-theoretically obvious.
- The product of the subgroups is : Any element can be written as the product of and .