If the kernel is complete, the extension splits over its centralizer

By [L2], the extension defines a homomorphism $\omega :Q\to \mathrm{Out}(N)$. Since $\mathrm{Out}(N)=1$ by [L1], this homomorphism is trivial. Therefore for every $g\in E$, conjugation by $g$ on $i(N)$ is an inner automorphism.

If $x\in i(N)\cap C_E(i(N))$, then $x$ commutes with every element of $i(N)$. Under the identification of $i(N)$ with $N$, this says that $x$ lies in $Z(N)$, which is trivial by [L1]. So $i(N)\cap C_E(i(N))=1$.

Fix $g\in E$. By step 1.1 there exists $\gamma \in N$ such that $gi(n)g^{-1}=i(\gamma n{\gamma }^{-1})$ for all $n\in N$. Then $i(\gamma {)}^{-1}g$ commutes with every element of $i(N)$, so $i(\gamma {)}^{-1}g\in C_E(i(N))$. Hence every $g\in E$ lies in $i(N)C_E(i(N))$, and therefore $E=i(N)C_E(i(N))$.

Because every element of $C_E(i(N))$ commutes with every element of $i(N)$ by [L3], the two subgroups centralize one another. Together with steps 2.1 and 1.2, this makes $E$ the internal direct product of $i(N)$ and $C_E(i(N))$. In particular $C_E(i(N))$ is a complement to $i(N)$, so [L4] gives a split extension.