Is there always a way to break a group?

Question

Let $G$ be a finite group ( some finite representation is given ) and Let $N$ be a normal subgroup of group $G$. I know that given a $N$ and $G/N$ one can't get back a $G$ ( always ). My question is opposite given a $G$, Is it possible to find a $N$ which is a normal subgroup of $G$ ( suppose there always exists a normal subgroup $N$ in $G$) such that one can get back $G$ from $G/N$ and $N$ only?

Please note that $N$ should be non-trivial and $G$ is indecomposable means we can't write $G$ as a direct product of non-trivial subgroups.

Answer 1

As pointed out by Don Sir, you can write it in trivial way..
but in general if group is abelian then by fundamental theorem of finite abelian group you can decompose that into product of normal.

If you want just in 2 factor then apart from triviality this is not possible,

to give example see.
$G=Z_{49}$ Now $Z_7$ is only possible normal subgroup ofcourse nontrivial then you get that $Z_7\times Z_7$ , which are not isomorphic to original group,
Some groups can be decomposed using semidirect product..
Example $D_n$ in that A= is normal group. B=< s > ,
$D_n\cong A\rtimes B$

Note that not every group can be decomposed into semi direct product . Example $Q_8$.

Answer 2

Unless you allow $N=1$ or $N=G$, the answer is no. There will always be the extension $N\times G/N$ but you said the group is indecomposable, so this will not be isomorphic to $G$.