For a normal Subgroup $N$ of $G$, is there an Isomorphism from $G$ to $N\times (G/N)$? For a normal Subgroup $N$ of $G$, is there an Isomorphism from $G$ to $N\times (G/N)$ (the Product with the Quotient)?
[What I thought about: The number of elements is the same, so one could establish a bijective Map. But it is difficult for me to come up with a Homomorphism. One could map $g$ to $(n,gN)$ in $N\times (G/N)$ and if $g\in N$, one could choose $n=g$, otherwise I thought about taking one representative from each coset $g_1N, g_2N, \cdots, g_nN$ and define the map as $\varphi(g\in g_iN):=(g_i^{-1}g,g_iN)$. But then
$\phi(g\in g_iN)\phi(g'\in g_jN)=(g_i^{-1}g,g_iN)(g_j^{-1}g',g_jN)=(g_i^{-1}gg_j^{-1}g',g_ig_jN)$
which does not seem to be a Homomorphism.]
Thanks for all answers.
 A: No, take $G=Z$ and $N=2Z$. $Z$ is not isomorphic to $2Z\times Z/(2Z)$ since it does not have element of order 2.
If you assume $G$ is finite, let $G=S_n, n>5$ the symmetric group, and $A_n\subset S_n$ the alternating subgroup the kernel of the signature. $S_n/A_n=S_2$, but $S_n$ is not isomorphic to $A_n\times S_2$, since $A_n$ is the only normal proper subgroup of $S_n$.
http://planetmath.org/normalsubgroupsofthesymmetricgroups
A: Here's an example of a small finite group, for which you can check exhaustively that no isomorphism exists:
Take $G:=\Bbb{Z}/4\Bbb{Z}$ and $N:=2\Bbb{Z}/4\Bbb{Z}$ its only nontrivial subgroup. Then $G/N\cong\Bbb{Z}/2\Bbb{Z}$, but there is no isomorphism
$$G\ \stackrel{\sim}{\longrightarrow}\ N\times(G/N),$$
because there is no element of order $4$ on the right.
A: No. This is not true at all. Note that if what you think were true every solvable group would be abelian.  
For a specific example take $S_3$ and the subgroup $A_3$ which is cyclic of order $3$. The quotient is cyclic of order $2$. So the product is abelian, in fact cyclic of order $6$. But $S_3$ is non-abelian. 
It is not true for commutative groups either. It even does not  work with a group of smallest cardinality that has non-trivial subgroups: the cyclic group of order $4$ has a subgroup of order $2$, the quotient is also of order $2$, so the product is the Klein four group, not the cyclic group of order $4$.
There are some classes of groups where what you ask for is true (e.g., elementary abelian groups, divisible abelian groups), but in general it's not true at all. 
