Group Extensions and Couplings In this post, I will be referencing this paper. Here is proposition 1.2 from said paper:

Proposition 1.2 --- Let $G \neq \{1\}$ be a group that decomposes as an extension:
$$1 \longrightarrow K \longrightarrow G \stackrel{\pi}{\longrightarrow} Q \longrightarrow 1$$
and $\Theta : Q \to \text{Out}(K)$ the associated coupling. A sufficient condition for $G$ to be ICC is (i) $FC_{G}(K) = \{1\}$, and (ii) $\Theta$ restricted to $FC(Q)$ is injective.

I am guessing that a coupling is a group homomorphism, but perhaps it satisfies further conditions. So I guess my first question is, what exactly is a coupling? Second, is this author saying that are taking a group $G$ which decomposes as an extension and comes with associated coupling, or do all group extensions have an associated coupling? If all group extensions come with a coupling, how does one define/construct it?
 A: Suppose that we have a group extension
$1 \to K \to G \xrightarrow{\pi} Q \to 1$
Wlog suppose that $K$ is a subgroup of $G$ to simplify notation.
To define a homomorphism $Q \to \mathrm{Out}(K)$, we can do the following: take $q \in Q$. Then take any lift $q'$ of $q$ under $\pi$. Because $K$ is normal in $G$, conjugation by $q'$ gives an automorphism of $K$. We claim that the image of this automorphism in $\mathrm{Out}(K)$ is independent from the choice of lift $q'$ so that this construction is well-defined. Suppose that $q''$ is another lift so that $\pi(q'')=\pi(q')$. Then we have $\pi(q''q'^{-1})=1$ so by exactness we have $a:=q''q'^{-1} \in K$ now if $C_g:K \to K$ denotes conjugation by $g \in G$, we have $C_{q''}(k)=q''kq''^{-1}=aq'kq'^{-1}a^{-1}=C_a(C_{q'}(k))$ for all $k \in K$ showing that the automorphisms $C_{q'}$ and $C_{q''}$ differ only by the inner automorphism $C_a$, thus the class in $\mathrm{Out}(K)$ is well-defined as claimed.
To see that this is actually a homomorphism just use that if $q'$ is a lift of $q \in Q$ and $p'$ is a lift of $p \in Q$, then $q'p'$ is a lift of $qp$.
