Suppose *n* is a natural number, at least 3. Is the following true?


*Any two subgroups of the alternating group $A\_n$ that are conjugate inside the symmetric group $S\_n$ (With the natural embedding of $A\_n$ in $S\_n$) are also conjugate inside $A\_n$?*


Background: Elements of $A\_n$ that are conjugate in $S\_n$ need not be conjugate in $A\_n$ (there is a criterion for splitting based on the cycle decomposition -- if the cycle type comprises distinct odd cycle sizes, then the conjugacy class splits). However, it turns out that cyclic subgroups of $A\_n$ that are conjugate in $S\_n$ must be conjugate in $A\_n$, because for any two elements of the same cycle type, we can always find a power of one element that is conjugate to the other element (with a little combinatorial manipulation). It's not immediately clear, though, how this can be extended to an arbitrary subgroup.



Subgroups of alternating group conjugate in symmetric group conjugate in alternating group?
What you need is an example of a subgroup $G\subset A\_n$ such that the normalizer of $G$ in $S\_n$ is contained in $A\_n$. If every automorphism of $G$ is inner, then it will be enough if the centralizer of $G$ in $S\_n$ is contained in $A\_n$. How about $n=8$ and $G$ the diagonal copy of $S\_4$ in $S\_4\times S\_4\subset S\_8$, if you know what I mean?


EDIT: If you replace this $G$ by its normal subgroup of order $4$, you get another example. This one is minimal with regard to the order of $G$ (though not with regard to $n$, as Derek's Holt's examples show).


