Embedding rank of finite groups and quotients

Question

Let $G$ be a finite group, and $n$ a positive integer. It is not hard to check that the following are equivalent:

1. For every $g\in G\setminus\{1\}$ there is a subgroup $H\leq G$ with $|G/H|\leq n$ and $g\notin H$.
2. There is a finite $G$-set $X$ with faithful action such that each orbit has size at most $n$.
3. There is an injective homomorphism $G\to\Sigma_n^r$ for some $r\geq 0$ (where $\Sigma_n$ is the symmetric group).

I'll say that the embedding rank of $G$ (written $\operatorname{erank}(G)$) is the least $n$ such that these conditions hold. (Is there a more established name for this?) Basic facts about embedding rank include the following.

1. If $H\leq G$, it is clear that $\operatorname{erank}(H)\leq\operatorname{erank}(G)$.
2. If $G$ is abelian then we can write it as a product of cyclic groups of prime power order, and then $\operatorname{erank}(G)$ is just the maximum of those orders.
3. Using the same ideas: if $N\lhd G$ and $G/N$ is abelian then $\operatorname{erank}(G/N)\leq\operatorname{erank}(G)$.

My main question: if $N\lhd G$ but $G/N$ is not abelian, is it still true that $\operatorname{erank}(G/N)\leq\operatorname{erank}(G)$? I can prove this if $\operatorname{erank}(G)\leq 3$, but the proof uses many special considerations that are unlikely to generalise much further. It is easy to produce examples where $G$ acts faithfully on a set $X$ but the induced action of $G/N$ on $X/N$ is no longer faithful, so there is no obvious argument based on that construction or anything equivalent to it.

(UPDATE: Assertion based on incorrect computer calculations removed)

Answer 1

The answer is no.

There is a well known family (due to Peter Neumann) of groups $G \le {\rm Sym}(n)$ with normal subgroups $N$ such that $G/N$ does not embed into ${\rm Sym}(n)$, and the same examples are counterexamples for your question.

Let $G$ be a direct product of $k$ copies of the dihedral group of order $8$, and let $N$ be a subgroup of order $2^{k-1}$ of $Z(G)$ such that $G/N$ is an extraspecial group of order $2^{1+2k}$.

Then ${\rm erank}(G) = 4$ (clearly), but ${\rm erank}(G/N) = 2^{1+k}$.

To see this note that, since $Z(G/N)$ of order $2$ is the only minimal normal subgroup of $G/N$, any faithful action of $G/N$ must have an orbit with faithful action. The largest core-free subgroup of $G/N$ is an elementary abelian subgroup of order $2^k$, so the index of this subgroup is the ${\rm erank}$ of $G/N$.