Inductive Step when working with Quotient Groups, Proof about normal subgroups of special groups I have a question on the following proof of the characterisation of normal subgroups of groups which are direct products of non-abelian simple groups:

Theorem: Let $G = G_1 \times \cdots \times G_n$, where $G_1, \ldots, G_n$ are non-abelian and simple, and let $N \unlhd G$. Then there exists some set $J \subseteq \{1,\ldots, n\}$ such that
  $$
 N = \times_{j \in J} G_j \quad \mbox{ and } \quad G_k \cap N = 1 \mbox{ for } k \notin J.
$$

The proof goes by induction on $n$. I just show the relevant part:

Suppose $k \in \{1,\ldots, n\}$ with $N\cap G_k \ne 1$. Then $G_k \le N$ because $G_k$ is simple., let $\overline G/G_k$, then
  $$
 \overline G = \times_{i\ne k} G_i/G_k
$$
  with $G_i / G_k \cong G_i$ for $i \ne k$. By induction hypothesis there exists $J' \subseteq \{1, \ldots, n \}$ with $k \notin J'$ such that
  $$
 \overline N = \times_{j\in J'} G_j/G_k \quad \mbox{ with } \quad
 G_i/G_k \cap \overline N = 1 \mbox{ for } i \notin j'.
$$
  But then for $J := J' \cup \{ k \}$ we have
  $$
 N = \times_{j\in J} G_j \quad \mbox{ with } \quad N\cap G_i = 1 \mbox{ for } i \notin J.
$$

The last step I do not understand, the logic is something like ''because of 
$$
 G/N = G'/N
$$
we can conclude $G = G'$'', but as I found out this logic does not hold in general, see the Extension problem. So whats going on here?
(remark on notation: do not know how to do the big $\times$...)
 A: You want to use the following result:
If $G$ is any group, $K$ is a direct factor of $G$, and $H$ is a subgroup of $G$ containing $K$, then $K$ is a direct factor of $H$.
In your case, you would use this to conclude that $G_k$ is a direct factor of $N$, and the inductive hypothesis tells you that the remaining factors (if any) are some (possibly empty) collection of the remaining $G_i$.
EDIT:
Proof of the claimed result:
Suppose $G=K\times L$ and $K\subseteq H\subseteq G$.  For any $h\in H$ there exists $a\in K,b\in L$ with $h=ab$.  Since $K\subseteq H$ we have $a^{-1}h = b\in H$.  Setting $T=\{b\in L \ : \ ab\in H \mbox{ for some } a\in K\}$, we may conclude that $T$ is a subgroup of $H$ (and $L$, in fact) such that $H=KT$, $K\cap T=\{1_G\}$, and $[K,T]=\{1_G\}$.  It follows that $H$ is the direct product of $K$ and $T$.  
Note that you may use slightly different notation depending on how you opt to write out direct products like this.  Namely you may write $h=(a,b)$ instead, in which case we'd then say $(a^{-1},1)h = (1,b)\in H$, etc. This is just a technical distinction between whether we interpret $G$, at the level of sets, as the actual cartesian product of the sets $K$ and $L$, or if we simply use the $\times$ to denote the direct product conditions without necessarily implying a Cartesian product of the underlying sets.  The latter is the more general situation, and is what I was using.
In your case, we would thus have that $N=G_k\times L$ for some normal subgroup $L$ of $\times_{i\neq k} G_i$.  Apply the inductive hypothesis to $L$.  In this way we actually avoid the quotients entirely, thus avoiding your concerns about the extension problem.  Though, see my comment about a result of Ayoub which shows the problem doesn't exist for direct products of certain groups.
