Let $G$ be a group and $H$ a subgroup such that there is a unique (non-trivial) intermediate subgroup $K$ (i.e. $H < S < G$ implies $S=K$).

Question: Is it true that $HgK=KgH$, $\forall g \in G$ ?

Experiment (GAP) : It's true if $[G:H] \le 30$ and $\vert G/H_G \vert \le 10000$.
($H_G$ is the normal core of $H$ in $G$)


1 Answer 1


There are counterexamples with $G = {\rm PSL}(2,11)$, $H$ cyclic of order $6$ and $K$ dihedral of order $12$.

I find it difficult to understand why you would expect such a property to be true. The reason why it holds in many small examples is that $HgK=KgK=KgH$.

  • $\begingroup$ Thank you! $\vert G/H_G \vert \le 10000$, it's not so small examples. Do you suspect it's true if $[G:H] \le 30$ without restriction on the order of $G/H_G$ ? If yes, is there a proof without computation ? $\endgroup$ Mar 21, 2014 at 14:56
  • $\begingroup$ Since there are databases of permutation groups of degree up to $30$, it could be checked by computer for $|G:H| \le 30$. I find it very unlikely that there would be acomputer-free proof. $\endgroup$
    – Derek Holt
    Mar 21, 2014 at 16:54
  • $\begingroup$ Yes, I've used the databases of transitive permutation groups, but beyond order $10000$, the verification begins to be long for my computer. I think I can't check this with my own computer in a reasonable time. $\endgroup$ Mar 21, 2014 at 17:04
  • 2
    $\begingroup$ I've checked it already! The unique intermediate condition is equivalent to there being a unique system of blocks of imprimitivity, which is easy to check. For such groups, take $H$ to be a point stabilizer, and $K$ the stabilizer of a block containing that point. I found that for all such groups of degree up to 30, the number of double cosets $KgK$ was equal to the number of double cosets $HgK$, so we always get $HgK = KgH = KgK$. $\endgroup$
    – Derek Holt
    Mar 21, 2014 at 17:19
  • $\begingroup$ It's a quite strange phenomenon that this property is true for all the transitive groups of degree up to $30$, but not in general. Perhaps your counterexample of degree $110$ is the first. I've also found a counterexample of degree $182$ with $G=PSL(2,13)$, $K=D_{12}$ and $H=S_3$. $\endgroup$ Mar 22, 2014 at 0:08

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