$N\unlhd G$ with $N\cong S_3$ then there exists $H\leq G$ with $G=N\times H$ Given $G$ a finite group and $N\unlhd G$ with $N\cong S_3$ then  Question is to prove that :
there exists $H\leq G$ such that  $G=N\times H$
I was thinking as $N\unlhd G$ then, $G/N$ would be a group and letting $G/N=H$ one can write $G=N\times H$
But then, I realized that $\{1,−1\}\unlhd Q_8$ and $Q_8/\{1,−1\}\cong V_4$ but $V_4\nleq Q_8$
so what all i have been thinking is nonsense...
But then i am not getting any other way to proceed further..
Question is how do we construct $H\leq G$ given a $N\unlhd G$ 
I have not used that $N\cong S_3$ i am not sure where to use this :O
please provide some hints to work out this.
Thank you
 A: This is because $S_{3}$ is a complete group, that is, it has trivial centre, and all automorphisms are inner. The property you name is shared by all complete groups.
Let us review the proof. If $N$ is complete, and it is normal in $G$, consider $C_{G}(N)$. Because $N$ has trivial centre, one has $C_{G}(N) \cap N = \{ 1 \}$, and obviously $C_{G}(N)$ and $N$ commute elementwise.
Now if $g \in G$, the map
$$
N \to N, \qquad n \mapsto g^{-1} n g
$$
is an automorphism of $N$, and thus it is inner, as $N$ is complete, 
that is, there is $x \in N$ such that 
$$
g^{-1} n g = x^{-1} n x
$$
for all $n \in N$, so that $g x^{-1} \in C_{G}(N)$, or $g \in C_{G}(N) N$.
It follows that $G = C_{G}(N) N = C_{G}(N) \times N$.
To complete ;-) the picture, let us as check that $S_{3}$ is indeed complete. It is clearly centreless. Moreover, an automorphism of $S_{3}$ is completely ;-) determined by its action on the three elements of order $2$. But the $6$ inner automorphisms already induce all possible permutations of these three elements.
