Equivalence of internal and external direct product

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 $$N_1N_2 = G$$.

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, $$N_1N_2 = G$$.

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 $$(e,b)$$.