For a subset $X$ of $S\_n$ (the symmetric group of degree $n$) define $C(X)$ to be the union of all conjugates of elements of $X.$


**Question 1** What is that called?


**Question 2** Over all transitive subgroups $H$ of $S\_n,$ which one has the *smallest* $C(H)?$



transitive subgroups of $S_n$
Since nothing else came up, I'm posting this as an answer.


For $n$ a power of a prime $p$, the answer is any group of exponent $p$, acting regularly.


Indeed, if $n$ is a power of $p$, then a Sylow $p$-subgroup $P$ of $H$ is transitive, and any element of order $p$ in the center of $P$ will be fixed-point-free, so $H$ always contains an element of cycle type $(p,\ldots,p)$. In a group of exponent $p$ acting regularly, every non-trivial element has this cycle type, so this is best possible.


For other small values of $n$, I obtained the following answers computationally:


* $n=6$ : dihedral group, acting regularly.
* $n=10$ : dihedral group, acting regularly.
* $n=12$ : $A\_4$, acting regularly.
* $n=14$ : dihedral group, acting regularly (edging out by less than 0.1% a group isomorphic to $AGL(1,8)$).
* $n=15$ : $A\_5$, acting on the cosets of a Sylow $2$-subgroup.
* $n=18$ : generalised dihedral group over an elementary abelian group, acting regularly.
* $n=20$ : a group of the form $C\_2^4:C\_5$ (edging out another group by about 0.01%).


I'm not sure why there are so many extremely close calls. I'm guessing it has to do with the wide range of size of conjugacy classes in $S\_n$ (so two groups might contain the same large classes, but differ in some small classes).


Perhaps some other cases can be done, like $n$ twice a prime but, to me, the general problem seems hard, even intractable. 


