Isomorphism from inner direct product to external.

Question

Prompt me, please, why in this proof of existence of products' isomorphism the one use, in the last equivalence: $\alpha(ac,bd) = \alpha((a,b),(c,d))$ and for what purpose, in this case, had he used permutation of $b$ and $c$ if eventually he permuted them back?

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: $$ \alpha \colon N_1 \times N_2 \to G, \quad \alpha(a,b) = ab $$ 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: $$ a b a^{-1} b^{-1} = a ( b a^{-1} b^{-1} ) \in N_1; \qquad a b a^{-1} b^{-1} = ( a b a^{-1} ) b^{-1} \in N_2 $$ Combining these, we get $$ a b a^{-1} b^{-1} \in N_1 \cap N_2 = \{ e \} $$ Hence we get: $$ ab = ba. $$ Thus, any element of $N_1$ commutes with any elements 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.

(Original images here and here.)

Answer 1

In the line in question he is trying to show that $\alpha$ is a homomophism of groups. That is, if $u, v$ are in the domain of $\alpha$, then $\alpha(uv) = \alpha(u)\alpha(v)$. Or, turning it around, $$\alpha(u)\alpha(v) = \alpha(uv)$$ In this case, the domain is $N_1 \times N_2$. So $u = (a, b)$ and $v = (c,d)$ for arbitrary elements $a, c \in N_1; b, c\in N_2$

The steps are:

1. $\alpha(a,b)\alpha(c,d) = (ab)(cd)$, which is just the definition of $\alpha$ applied twice ($\alpha(a,b) = ab$).
2. $(ab)(cd) = a(bc)d$, an application of associativity.
3. $a(bc)d = a(cb)d$, an application of the previous result that elements of $N_1$ commute with elements of $N_2$.
4. $a(cb)d = (ac)(bc)$, associativity again.
5. $(ac)(bd) = \alpha(ac, bd)$, another application of the definition of $\alpha$
6. $\alpha(ac,bd) = \alpha((a,b)(c,d))$, an application of the definition of multiplication on $N_1 \times N_2$, which is $(a,b)(c,d)=(ac,bd)$.

As for your other question, he doesn't permute them back, so I don't understand what you are asking about. He needed that $bc = cb$ because in order to reach the product in step 6, he needs $a$ multiplied by $c$ and $b$ multiplied by $d$, not $a$ by $b$ and $c$ by $d$ as it starts with.