Is $ G \cong G/N \times N$? If G is a finite group and N is a normal subgroup in G , then can we say G $\cong$ G/N $\times$ N always? Is it true for like normal nilpotent or normal solvable or any such special classes.   I couldn't help but ask for it. What if G is infinite? Please tell me if you can, the concept I am skipping here. If you can avoid using exact sequences, It will be much clearer to me. I have seen an answer which still is not so clear to me. Thanks!

I've intended to ask a similar question. Fortunately, the community has avoided a duplicate. Moreover, I've had more information to better my proposition, which is written below.
The proposition: The necessary and sufficient conditions of group $G$ and its normal subgroup $N$ (regardless of commutativity and finiteness) to satisfy $G \sim N \times G / N$ is that:
$i) \; \; N \le Z(G)$ (center of group $G$), and
$ii)$ There is a way to choose representative elements for classes of $G/N$ so that those elements would make up a subgroup of $G$.
I think I proved this, but I'm not so sure.
