# Subgroups of alternating group conjugate in symmetric group conjugate in alternating group?

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.

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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).

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Thanks! That example definitely works. –  Vipul Naik Jul 21 '11 at 16:53
The subgroups $\langle (1,2)(3,4), (1,2,3)(5,6,7) \rangle$ and $\langle (1,2)(3,4), (1,2,3)(5,7,6) \rangle$ of $A_7$ are isomorphic to $A_4$ and are conjugate in $S_7$ (by (6,7), for example) but not in $A_7$.
$A_7$ also has two classes of subgroups isomorphic to the simple group $L_2(7)$ which are fused in $S_7$.