Let $N$, $Q$ be groups and there is group homomorphism $$\Phi: Q \to Out(N)= Aut(N)/Inn(N),$$ we called such $\Phi$ to be coupling. For every extension of $N$ by $Q$ there exist a coupling.
What about the converse? That is, if we have a coupling $\Phi$, does there always exist an extension of $N$ by $Q$ that induce the coupling $\Phi$?
I know converse is also true, if $N$ is abelian group. What about for non-abelian case?
Please give a counterexample.
A counterexample is $N = 2.A_6 \cong {\rm SL}(2,9)$, the double cover of the alternating group $A_6$.
The automorphism group of $N$ is isomorphic to $A_6.2^2 \cong {\rm P \Gamma L}(2,9)$. The three extensions $A_6.2$ are isomorphic to $S_6$, ${\rm PGL}(2,9)$ and a group that is often denoted by $M_{10}$, the point stabilizer in the Mathieu group $M_{11}$. (Another description is that $M_{10}$ is the extension of ${\rm PSL}(2,9)$ by hte product of a diagonal and a field automorphism.)
There is no associated group with structure $2.M_{10}$ (the ATLAS of Finite Groups is a good source for this type of information).
So if we let $Q=C_2$ and $\Phi$ the homomorphism with image the outer automorphism inducing $M_{10}$, then there is no associated coupling.