If we let $G$ be a group with $n$ subgroups $N_i$ such that
$\prod_{i=1}^n N_i= G$
$N_i \cap N_j = \{e\}$ for all $i \ne j$ s.t. $1 \le i < j \le n$
$N_i \unlhd G$ for any $1 \le i \le n$
then $G$ is not necessarily an internal direct product of the $N_i$, for there exist counter-examples that show $G$ could still not be isomorphic to the external direct product $N_1 \times \ldots \times N_n$.
But I'm curious if the converse is true: that is, if $G$ is an internal direct product of $N_1, \ldots, N_n$, then does this imply either that (i) $N_i \cap N_j = \{e\}$ for all $1 \le i < j \le n$ or (ii) any $N_i \unlhd G$?
NOTE: I'm assuming that if $G = \prod_{i=1}^n N_i$ and each $g \in G$ has a unique representation of form $h_1 \cdot \ldots \cdot h_n$ s.t. $h_i \in N_i$, then $\prod_{i=1}^n N_i$ forms an internal direct product of $G$.