# Embedding rank of finite groups and quotients

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)

• I hope I haven't misunderstood your question, because according to my answer below $G=D_8 \times D_8$ is a counterexample of order $64$. Commented Aug 12 at 21:53
• @DerekHolt Indeed, there was a stupid typo in my code. The minimal counterexample if $G=\langle a,b|a^4=b^4=aba^{-1}b=1\rangle$ of order $16$ and rank $4$: the quotient by $\langle a^2b^2\rangle$ is $Q_8$ with rank $8$. Commented Aug 13 at 7:59

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 $$G/N$$.