[FOM] Query on alternating groups
joeshipman@aol.com
joeshipman at aol.com
Sun Jan 8 19:13:51 EST 2006
Shipman:
> I am looking for results on the orders (or, equivalently, the
indexes)
> of maximal subgroups of the alternating groups An.
Cohen:
Not quite what you are looking for, but the O'Nan-Scott Theorem can be
seen as classifying the maximal subgroups of the alternating group. A
maximal subgroup is an intransitive direct product, a transitive but
imprimitive wreath product, a primitive wreath product in its product
action, a regular abelian group or an almost simple group - the first
three of specific groups. Cameron's permutation groups book has the
statement of the theorem in this form, but you might also try Dixon and
Mortimer's book of the same title.
I reply:
Thanks very much, Jon, the O'Nan-Scott theorem is exactly what I
needed. There are actually 6 cases:
1) intransitive direct product Sk X Sm where n=k+m; |G|=k!m!
2) transitive imprimitive wreath product Sk wr Sm where n=k*m;
|G|=m!*(k!^m)
3) primitive wreath product Sk wr Sm where n=k^m; |G|=m!*(k!^m)
4) affine group of dimension d over Zp when n=p^d; |G|=(p^d)*(p^d -
1)*(p^d - p)*...*(p^d - p^(d-1))
5) an almost simple group acting on the cosets of a maximal subgroup
(an almost simple group is contained between T and Aut(T) for a
nonabelian simple group T)
6) If n=t^k, where t is the order of a nonabelian simple group T, a
group of order t^(k+1) * (k+1)! * |Out(T)|. The simplest case of this
is a subgroup of S60 of order 7200.
Although this needs checking against the C.O.F.S.G., it appears that
only the first 2 cases are relevant to finding maximal orders, because
the groups in the other cases are so small in comparison that their
orders always divide the orders of groups in the first 2 cases, with
exceptions when n=4 or n=8.
This allows me to calculate finitary implications between "degree
axioms" easily. In what follows, [n] denotes the statement "all
polynomials of degree n have a root", and <a, b, c, ...> denotes the
additive semigroup generated by the positive integers a, b, c, ....
Then the degree axiom [n] entails the axioms [d] for all d properly
dividing n, and is entailed by the combination of the divisor axioms
[d] and [i] where i is ANY element of the semigroup associated with n
below:
[p] for prime p: <p>
[4]: <3,4>
[6]: <6,10,15>
[8]: <8, 15, 28, 35>
[9]: <9, 84, 280>
[10]: <10, 45, 126>
[12]: <12, 66, 220, 495>
[14]: <14, 91, 1716>
[15]: <15, 455, 3003>
[16]: <16, 120, 1820, 6345>
[18]: <18, 153, 816, 24310>
[20]: <20, 190, 4845, 15504>
[21]: <21, 1330, 116280>
[22]: <22, 231, 352716>
[24]: <24, 276, 2024, 10626, 735471>
[25]: <25, 53130, 5194672859376>
[26]: <26, 325, 5200300>
[27]: <27, 2925, 4686825, 75957810500>
The large number in the case n=25 comes from the index in S5 of the
imprimitive group with order (5!)^6 where the blocks {1,2,3,4,5},
{6,7,8,9,10}, {11,12,13,14,15},{16,17,18,19,20}, and {21,22,23,24,25}
are permuted both internally and externally. If there is a polynomial
of degree 25 with that Galois group over some field, I can find you a
polynomial of degree 5194672859376 with no roots, which is why, even
though ([5]&[n])-->[25] for all sufficiently large n, you have to go
very high up!
These calculations are not "guaranteed"; they will be checked with John
Conway when he returns from New Zealand next month....
-- JS
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