Recently, My Professor taught us about group extension. It is the following: A group $G$ is an extension of $Q$ by $N$ if we have the following short exact sequence: $1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1$.
After learning this definition I observed that for $n \geq 3$, the symmetric group on $n$ letters $S_{n}$ is an extension of $\mathbb{Z_{2}}$ by the alternating group on $n$ letters $A_{n}$, that is, we have an exact sequence: $1 \rightarrow A_{n} \rightarrow S_{n} \rightarrow \mathbb{Z_{2}} \rightarrow 1$. From here, I came up with the following question that I am completely stuck at.
If $G$ is any nontrivial group, does there exist a group $H$ that is an extension of a group $Q$ by $G$ such that $N \unlhd H$ implies $N \subseteq G$? (Here, in this question, I have assumed that $H \neq G$.)
Please help me.
This is not always the case, for instance $G=S_n$ for $n\geq 7$ does not admit such an extension, and generally no complete group has this property:
Suppose for a contradiction that $G$ is a complete group that also arises as a unique maximal normal subgroup of some group $H$. As $H$ acts on $G$ by conjugation, we have a homomorphism $\phi:H \to \mathrm{Aut}(G)$. Since every automorphism of $G$ is inner, this map cannot be injective. Hence, its kernel $C_H(G)$, the centralizer of $G$ in $H$ is nontrivial. By assumption, it must therefore be contained in $G$ already and thus coincide with its center $Z(G)$. But this is a contradiction since $G$ is also centerless.
Both of these assumptions ($Z(G)=1$ and $\mathrm{Aut}(G)=\mathrm{Inn}(G)$) are evidently essential for this argument to work: The alternating groups $A_n$ are centerless but sit inside $S_n$, while $G=C_2$ has no outer automorphisms and sits inside $C_4$ as unique maximal subgroup.
I do not know for which groups there is such an extension in general.