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 $G$ is a group, and $N_{1},N_{2}$ are normal subgroups. We say that $G$ is an internal direct product of $N_{1}$ and $N_{2}$ if $N_{1}\cap N_{2}$ is trivial, and $N_{1}N_{2}=G$.

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 $G$ that is the internal direct product of normal subgroups $N_{1},N_{2}$, we can identify $G$ naturally with the external direct product $N_{1}\times N_{2}$.

Proof: Consider the map:

$\alpha :N_{1}\times N_{2}\to G,\qquad \alpha (a,b)=ab$

We first must verify that $\alpha$ is a group homomorphism. For this, we first observe that if $a\in N_{1},b\in N_{2}$, then:

$aba^{-1}b^{-1}=a(ba^{-1}b^{-1})\in N_{1};\qquad aba^{-1}b^{-1}=(aba^{-1})b^{-1}\in N_{2}$

Combining these, we get:

$aba^{-1}b^{-1}\in N_{1}\cap N_{2}=\{e\}$

Hence, we get:

$ab=ba$.

Thus, any element of $N_{1}$ commutes with any element of $N_{2}$. Thus, if we pick two pairs $(a,b),(c,d)\in N_{1}\times N_{2}$, then:

$\alpha (a,b)\alpha (c,d)=abcd=a(bc)d=a(cb)d=(ac)(bd)=\alpha (ac,bd)=\alpha ((a,b)(c,d))$

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

Next we note that:

1. $\alpha$ is injective because $\alpha (a,b)=e\implies ab=e\implies b=a^{-1}\implies a,b\in N_{1}\cap N_{2}=\{e\}$.
2. $\alpha$ is surjective because by assumption, $N_{1}N_{2}=G$.

Thus, $\alpha$ defines an isomorphism from $N_{1}\times N_{2}$ to $G$.

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

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

1. The subgroup isomorphic to $A$ is $A\times \{e\}$ (i.e., elements of the form $(a,e)$).
2. The subgroup isomorphic to $B$ is $\{e\}\times B$ (i.e., elements of the form $(e,b)$).

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

1. Both the subgroups are normal: This is a direct check using the coordinate-wise multiplication.
2. The subgroups intersect only at $(e,e)$: This is set-theoretically obvious.
3. The product of the subgroups is $A\times B$: Any element $(a,b)$ can be written as the product of $(a,e)$ and $(e,b)$.