Let H be a subgroup of index n. H' is the intersection of all the subgroups that conjugate to H. $G/H'$ is isomorphic to some subgroup of $S_n$ , so that $(G:H')| n!$ , $(H:H')| (n-1)!$ . (The reason will be explained below.) In fact, H' is the union of all the conjugate classes of G that are contained in H, and at the same time the largest normal subgroup of G contained by H.
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This action is transitive. Actually, any transitive G-action can be given in this way.
When we want a group to be transitive, the larger the better, which is quite the contrary of being faithful or free. In the finite case, an action can be n-transitive only if $A_{|X|}^n\big ||G|$ , can be faithful only if $|G|\big||X|!$ , can be free only if $|G|\big||X|$ .
![[Pasted image 20221028231443.png]] Let H' be the largest normal subgroup of G contained in H. Then $[H:H']|(p-1)!$ . Also $[H:H']\big||G|$ . So $[H:H']$ can only be 1. That is, $H=H'\lhd G$ .