Let $G$ be a simple group and $H$ a proper subgroup such that there is a unique intermediate subgroup $K$ (i.e. $H < S < G$ implies $S=K$).
Question: Is it possible that $[G:K]=[K:H]$ ?
2026-03-25 06:27:59.1774420079
Is there a simple group and a proper subgroup with a unique and equidistant intermediate?
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$G=M_{11}$, $K={\rm PSL}(2,11)$, $H = 11:5$, $|G:K|=|K:H|=12$.