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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?

Henrique Augusto Souza
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    It's not clear to me exactly what $\chi$ is, but I'll assume you intend it to be an outer action $G \to \text{Out}(N)$. If $N$ is abelian then $\text{Out}(N) = \text{Aut}(N)$, so any such outer action is a genuine action, meaning every coupling gives rise to an extension (namely the semidirect product). So counterexamples necessarily have $N$ nonabelian. – Qiaochu Yuan May 30 '16 at 04:04

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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 $2.A_6.2_3 = 2.M_{10}$.

Derek Holt
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