Timeline for Characterizations of subgroups of $S_n$ with small centralizer

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Nov 4, 2016 at 21:43	comment	added	YCor		So the centralizer $C$ is abelian iff $\max_i n_i\le 2$, all $M_i$ are abelian, and for all $i$, $n_i=2$ implies $M_i=\{1\}$. That $C$ is cyclic is even more constraining: for instance it implies that $n_i=2$ for at most one value of $i$, and that the $M_i$ have pairwise coprime order (and odd order in case $\max n_i=2$).
Nov 4, 2016 at 21:40	comment	added	YCor		In the non-transitive case the good point of view is to view $X=\{1,\dots,n\}$ as an $H$-set. Then it decomposes as disjoint union $\bigsqcup n_iX_i$ where $X_i$ are non-isomorphic transitive $H$-sets and $n_i$ is a disjoint union of copies of $X_i$ (this consists in partitioning $X$ according to conjugacy classes of stabilizers, and then into orbits). Then the centralizer is the product over all $i$ of the permutational wreath product $M_i\wr S_{n_i}$, where $M_i$ is the automorphism group of the $H$-set $X_i$ (that is, the centralizer of $H$ in the symmetry group of $X_i$).
Nov 4, 2016 at 21:34	comment	added	YCor		If $H$ is transitive, then it has trivial centralizer in $S_n$ iff $H_1$ is its own normalizer in $H$, where $H_1$ is the stabilizer of $1\in\{1,\dots,n\}$. More generally if $H$ is transitive, the centralizer is isomorphic to $N/H_1$ where $N$ is the normalizer of $H_1$ in $H$.
Nov 4, 2016 at 20:30	history	asked	stupid_question_bot	CC BY-SA 3.0