Internal weak product of groups (in the chapter of direct product and direct sum of groups) In the book of algebra by Hungerford in the course of proof to the theorem that if,
i) $G = \bigl\langle \bigcup_{i \in I} N_i \bigl\rangle$, where $\{N_i \mid i\in I\}$ is a family of normal subgroups of $G$;
ii) for each $k \in I$, $N_k \cap \bigl\langle \bigcup_{i \neq k} N_{i} \bigl\rangle = \langle e \rangle$,
then $G \simeq \prod_{i \in I}^w N_i$.
We noticed that it is written that the map $\phi \colon \prod_{i \in I}^w N_i \rightarrow G$, given by $\phi(\{a_i\}) = \prod_{i \in I_0} a_i$ where $I_0$ is the finite set $\{i \in I \mid a_i \neq e\}$ is a homomorphism. But I failed to frame to show that it is a homomorphism. Please help me in framing the argument in an ‘easy to understand’ way. Thanks in advance.
 A: Let $\{a_i\}, \{b_i\} \in \prod_{i \in I}^w N_i$. Then $$\phi(\{a_i\}\{b_i\}) = \phi(\{a_ib_i\}) = \prod_{i \in I} a_ib_i$$ where the right hand has $a_i, b_j = e$ for cofinitely many $i,j \in I$. On the other hand,
$$\phi(\{a_i\})\phi(\{b_i\}) = \left( \prod_{i \in I} a_i \right)\left( \prod_{i \in I} b_i \right)$$ and thus, what we need to show is that $\prod_{i \in I} a_ib_i = \left( \prod_{i \in I} a_i \right)\left( \prod_{i \in I} b_i \right)$. If $I_0 \subset I$ is the set of indices $i \in I$ for which $a_i \neq e$ or $b_i \neq e$, then, we have that $I_0$ is finite of cardinality, say $n$, and after reindexing, we can write
\begin{align}
a_1b_1a_2b_2\cdots a_nb_n = (a_1a_2\cdots a_n)(b_1b_2\cdots b_n)
\end{align}
We show that terms can be shifted around to get the left hand side from the right hand side. Suppose $b_i \in N_i$ for some $1 \leq i < n$. Then
$$ a_ia_{i+1}\cdots a_nb_i = a_ib_ia_{i+1} \cdots a_n$$
This is true because for each $j > i$, $N_j \cap N_i = e$ by assumption and hence $a_jb_i = b_ia_j$ (by a theorem in Hungerford). So, to obtain the "interleaving" product, we just apply this commutativity to each of the $b_i$ one-by-one until the $a_j$ with $j = i$ is directly to the left of $b_i$.
