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.


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).


No. With a question like that, I could immediately guess the answer is no, and then try a computer search to look for an example. I would only start thinking about it if I didn't find an example quickly. In this case you don't have to look too far.

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 $S_7$.

  • $\begingroup$ Thanks! I was on the wrong track because I thought the statement might well be true. I also had some GAP code to test the statement for alternating groups, but my code takes very long on the alternating group of degree 7, probably because it isn't very efficient, which is why I didn't find the counterexample. $\endgroup$ – Vipul Naik Jul 21 '11 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.