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A finite group $G$ with a "large" (comparatively to $G$) core-free subgroup $P$ is "relatively highly compressible"$^\dagger$, meaning that the "comprimibility factor": $$c(G):=\frac{|G|-\mu(G)}{|G|}$$ ($\mu(G)$ is the minimal faithful permutation degree of $G$) has $\frac{|G|-[G:P]}{|G|}$ as lower bound, which by itself is already "relatively close to $1$".

But I'm afraid that this approach is too naive to get cases where really $c(G)\lesssim 1$. How can we do better, and to what extent can we push $c(G)$ close to $1$?

$^\dagger$As opposedly to the "stiff" Klein 4-group, cyclic groups of prime power order and generalized quaternion groups, for which $c(G)=0$.

Kan't
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    I find the question a bit too vague/open-ended to know whether this qualifies as an answer, but the obvious candidates for "good" groups (whatever that means) would probably be the symmetric groups $S_n$. Clearly $c(S_n) = 1 - 1/(n-1)!$, so you can get arbitrarily close to $c(G) = 1$. Also, $S_n$ contains $S_{n-1}$ as a core-free subgroup of index $n$, so perhaps your approach wasn't that naive after all? – sTertooy Sep 27 '24 at 22:36
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    Just to echo the above comment, once you fix a particular value of $\mu(G)$, you have a bound on $|G|$, so taking the largest possible (the symmetric group $S_{\mu(G)}$) is the best you can do when trying to get close to $1$. – Steve D Sep 27 '24 at 23:27

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