Suppose finite directly indecomposable group $G$ has $\frac {|[G, G]|}{|G|} < \alpha$ and $\frac {|A|} {|G|} > \beta$, where $A \lhd G$ abelian. Are there some nontrivial bounds on $\alpha, \beta$ which imply that $G$ is solvable (maybe of particular class) which do not come from trivial observations based on commutativity measure?
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$\begingroup$ Let $G = A \times S$ where $A$ is a finite Abelian group and $S$ a finite perfect group. Then $\frac{\vert [G, G] \vert}{\vert G \vert} = \frac{1}{\vert A \vert}$ and $\frac{\vert A \vert}{\vert G \vert} =\frac{1}{\vert S \vert}$. The group $G$ is not solvable but $\alpha$ can be arbitrarily small while $\beta$ can be as large as $\frac{1}{61}$. $\endgroup$– Luc GuyotCommented Aug 16, 2018 at 16:41
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$\begingroup$ Oh well, I had in mind that group is supposed to be directly indecomposable. That was too obvious for me, so I forgot to mention this property explicitly. $\endgroup$– Denis TCommented Aug 16, 2018 at 17:05
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$\begingroup$ Let $G = B \wr S$ with $B$ finite Abelian, $S$ finite and perfect, and set $A=B^S$. $\endgroup$– Luc GuyotCommented Aug 16, 2018 at 17:52
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$\begingroup$ In the latter example, we have $\frac{\vert [G, G] \vert }{\vert G \vert} = \frac{1}{\vert B \vert}$ because $G/[G, G] \simeq B$ while $\frac{\vert A \vert}{\vert G \vert } = \frac{1}{\vert S \vert}$. $\endgroup$– Luc GuyotCommented Aug 16, 2018 at 17:59
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$\begingroup$ I got your point. But, for example, group with average size of conjugacy class less than $40/3$ is either solvable or $A_5$ times abelian. I guess that some results of same kind are possible here — general statement with few exceptions coming from small perfect groups. $\endgroup$– Denis TCommented Aug 16, 2018 at 18:57
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