Group extension that doesn't realize a coupling Let $E$ be an extension of $N$ by $G$:
$$N \hookrightarrow E \twoheadrightarrow G$$
If $N$ is abelian, then $E$ uniquely defines an action of $G$ on $N$. More generally, it defines a unique class $\chi$ on:
$$\text{Out}(N) = \text{Aut}(N)/\text{Inn}(N)$$
We call the pair $(G, \chi)$ a coupling of $G$ to $N$. Robinson says:

[...] principal aims of the theory of group extensions may be summarized as follows:
(i) to decide which couplings of $G$ to $N$ give rise to an extension of $N$ by $G$;

Unfortunately, I'm failing to find a counter example, a coupling of $G$ to $N$ that does not gives rise to an extension of $N$ by $G$. So far, I've looked only at finite, abelian groups $N$. Can someone point me to such counter example? 

I know one example of this, but it might not be the smallest.
Let $N = {\rm SL}(2,9)$, which is isomorphic to a double cover $2.A_6$ of $A_6$. Then ${\rm Out}(N) \cong C_2 \times C_2$. Let $G = {\rm Out}(N)$ with $\chi$ the identity map. Then there is no extension $E$ that induces this coupling.
In the ATLAS of Finite Simple groups, if you look under $A_6$, you will find that there is no extension of the form $2.A_6.2_3 = 2.M_{10}$.
