If $N_1 \times N_2 \times ... \times N_p$ is a normal subgroup of $G_1 \times G_2 \times ... \times G_p$, is it true that $N_i \unlhd G_i$ for each value of $i$?
Conjugating an arbitrary $(n_1, n_2, ..., n_p) \in N_1 \times N_2 \times ... \times N_p$ by $(g,1, ..., 1) \in G_1 \times G_2 \times ... \times G_p$ gives $(gn_1g^{-1}, n_2, ..., n_p)$, which must be an element of $N_1 \times N_2 \times ... \times N_p$ by normality of $N_1 \times N_2 \times ... \times N_p$ in $G_1 \times G_2 \times ... \times G_p$.
So $gn_1g^{-1} \in N_1$ for an arbitrary $g \in G_1$ and $n_1 \in N_1$, in other words $N_1$ is closed under conjugation from $G_1$ hence $N_1 \unlhd G_1$, and since we can show this for any $N_i$ and $G_i$, each component of the normal subgroup direct product must be normal to the corresponding component of the group direct product. Is this the correct reasoning?