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!
 A: Hints: $S_3$ has order $6$. It has a subgroup $N$ of index 2, so $N$ is normal. Then prove that $S_3\not\cong \mathbb{Z}_2\times\mathbb{Z}_3$, which is contrary to your claim.
A: The claim fails even more profoundly. While a finite abelian group always has a subgroup isomorphic to its any quotient, this is not true in general. (As said before, the original claim is still false for abelian groups.)
The smallest counterexample is the quaternion group $Q_8$. Its center is $Z(Q) \cong \mathbb{Z}_2$ and the quotient $Q/Z(Q)$ is isomorphic to the Klein four-group, but the subgroups of order $4$ of $Q_8$ are cyclic.
A: 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.
