How to prove that
in $A_n$ (Alternating group), the subgroup of second smallest index has index $n \choose 2$ if $n\ge 9$ ?
I know how to prove it for the smallest index, but for the second smallest I don't know how to prove it.
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1$\begingroup$ Cross posted : math.stackexchange.com/questions/2341526/… $\endgroup$– verretCommented Jul 2, 2017 at 19:52
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$\begingroup$ Why? I had already given a sketch of the proof there... $\endgroup$– verretCommented Jul 3, 2017 at 7:12
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2 Answers
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Although some fiddling will be necessary to get a precise result, one can understand problems of this type without using powerful modern results in group theory.
For large enough $n$, there are a lot of primes strictly between $n/2$ and $n-2$. Any subgroup whose index is at most $n^2$ will have order divisible by at least one of these primes, call it $p$, and thus contain a $p$-cycle. C. Jordan showed long ago that a proper subgroup of $A_n$ containing a $p$-cycle with $p<n-2$ cannot be primitive (see for example Theorem 3.3E in Dixon and Mortimer's ``Permutation Groups"). The indices of the intransitive and imprimitive maximal subgroups of $A_n$are straightforward to compute.
answered Jul 2, 2017 at 18:16
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$\begingroup$ That actually is not what the theorem says :) It just says that a primitive subgroup with a $p$-cycle will be will be "small". $\endgroup$ Commented Jul 2, 2017 at 18:43
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$\begingroup$ Are we talking about Theorem 3.3E in Dixon-Mortimer? This says that if $G \leq Sym(\Omega)$ is primitive and contains a $p$-cycle then either $G$ contains $Alt(\Omega)$ or $|\Omega| \leq p+2$. Letting $n=|\Omega|$, it follows that no primitive proper subgroup of $A_n$ contains a $p$-cycle unless $n \leq p+2$, right? An example of this phenomenon is the more well-known result that says a primitive subgroup of $S_n$ containing a transposition must be all of $S_n$. $\endgroup$ Commented Jul 2, 2017 at 19:24
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$\begingroup$ Or maybe I misunderstood your objection, and you were referring to the theorem in the post. In that case, the point is that every proper primitive proper subgroup is ``small", and therefore, when considering subgroups of small index, one need only consider indices of intransitive and imprimitive maximal subgroups. $\endgroup$ Commented Jul 2, 2017 at 19:38
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$\begingroup$ No, I was referring to 3.3E, and you are right, your statement in the post follows from the statement in the comment (which is, in fact, the statement of the theorem), but not exactly the same. Just picking nits. $\endgroup$ Commented Jul 2, 2017 at 21:42
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$\begingroup$ I would have thought you'd be better off using Bochert's result which bounds all the primitive subgroups of $S_n$ directly without any need to consider primes... The result is described in Wielandt's book (of which I have an e-copy if you want one) and dates back to the 1800s I think. $\endgroup$ Commented Jul 4, 2017 at 14:54
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This should follow from the O'Nan-Scott theorem, as described in the accepted answer to this question: maximal subgroups of finite simple groups
A completely explicit classification is given by Liebeck, Praeger, Saxl, using the methodology suggested above.
Liebeck, Martin W.; Praeger, Cheryl E.; Saxl, Jan, A classification of the maximal subgroups of the finite alternating and symmetric groups, J. Algebra 111, 365-383 (1987). ZBL0632.20011.
answered Jul 2, 2017 at 16:48
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$\begingroup$ The problem with using O'Nan-Scott is that you need a bound on the size of an almost simple group that embeds primitively into $A_n$. As far as I know if you want to avoid using CFSG, then the best bounds work for ANY primitive group and knowing that the group is almost simple isn't much of an advantage (so O'Nan--Scott hasn't really helped). $\endgroup$ Commented Jul 4, 2017 at 14:57