About Weak-Direct Products being Normal Subgroups of the Direct Product The theorem (Algebra, T.Hungerford, p.60) clearly states that:

If $ \{ G_i \}_{i \in I} $ is a family of groups, then $ \prod_{i \in I} {^{w}G_i} $ is a normal subgroup of $ \prod_{i \in I} G_i $.
NOTE that $ \prod ^{w} $ represents a weak-direct product (a temporary notation in the book to highlight the difference), and $ \prod $ is the direct product as usual.

I miss the point when I try the passage from being subgroup to normal subgroup.
(I also insist on my weak-product being external by preference, as internal semi-products imply by definition normality of subgroups.)
 A: You misquoted the book (unless you are using a different edition than what I use).  The term is weak direct products, not semidirect products as you said.  Semidirect products are completely different beasts and I am not sure if the term is even defined for an arbitrary number of groups.
Weak direct products, or what some group theorists call direct sums, are defined as follows.  If you have a family $(G_i)_{i\in I}$ of groups $G_i$ where $i$ ranges over an index set $I$, then the weak direct product $G:=\prod_{i\in I}^w\,G_i$ in Hungerford's notation (sometimes written as my preferred notation $\bigoplus\limits_{i\in I}\,G_i$ or in Wikipedia's notation $\sum\limits_{i\in I}\,G_i$) consists of tuples $(g_i)_{i\in I}$ such that each $g_i\in G_i$ for all $i\in I$, and that $g_i$ is the identity of $G_i$ for all but finitely many $i\in I$.  The multiplication rule in $G$ is given by pointwise multiplication.
When $I$ is a finite set (or when $G_i$ is trivial for all but finitely many $i\in I$), then of course, the weak direct product $G=\bigoplus\limits_{i\in I}\,G_i$ and the usual direct product $\tilde{G}:=\prod\limits_{i\in I}\,G_i$ coincide.  However, in general, $G$ is a normal subgroup of $\tilde{G}$, which is always proper when there are infinitely many $i\in I$ such that $G_i$ is nontrivial.  
Normality of the inclusion $G\leq \tilde{G}$ is quite easy to see.  Let $g:=(g_i)_{i\in I}\in G$ and $h:=(h_i)_{i\in I}\in\tilde{G}$.  Then,
$$hgh^{-1}=\left(h_ig_ih_i^{-1}\right)_{i\in I}\,.$$
Note that $g_i$ is the idenitity for all but finitely many $i\in I$.  Therefore,
$$h_ig_ih_i^{-1}=h_ih_i^{-1}=1$$
for such $i$.  Thus, $hgh^{-1}$ is in $G$.  Therefore, $G\trianglelefteq \tilde{G}$.
