Erdos-Turan theorem for normal subgroups?

Question

Suppose $G$ is a finite group and $N \triangleleft G$. Let’s define the relative commuting fraction as $$cf(G, N) := \frac{|\{(g, h) \in G \times H| [g, h] = e\}|}{|G||H|}$$

Does there exist such $\epsilon > 0$, that for all $G$ and $H$ if $cf(G, N) > 1 - \epsilon$, then $N \leq Z(G)$?

For the particular case when $N = G$ we have the following fact:

Erdos-Turan theorem

If $cf(G, G) > \frac{5}{8}$, then $G$ is abelian.

However, the method that is used to prove it can not be extended onto this general case…

It may still be useful to get this inequality:

$$cf(G, N) \leq \frac{|Z(G) \cap N|}{|N|} + \frac{1}{2}$$

$$cf(G, N) = \frac{1}{|G||H|}(|G||Z(G)\cap N| + \Sigma_{g \in N} C_G(g)) \leq \frac{1}{|G||H|}(|G||Z(G)\cap N| + \frac{|N||G|}{2}) = \frac{|Z(G) \cap N|}{|N|} + \frac{1}{2}$$

But I do not know how to proceed further...

Answer 1

You've done most the work finding the above bound, but you made one subtle error.

The sum should be over $g\in N\setminus(Z(G)\cap N)$.

If you follow this through you get $$cf(G,N)\le\frac{|Z(G)\cap N|}{2|N|}+\frac{1}{2}$$

So if $cf(g,N)>3/4$ then $\frac{|Z(G)\cap N|}{|N|}>\frac{1}{2}$ and $Z(G)\cap N=N$