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A104109
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Numbers n such that whenever a group G has a solvable subgroup of index n, then G itself is solvable.
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0
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1, 2, 3, 4, 11, 19, 22, 23, 29, 31, 37, 41, 43, 46, 47, 53, 57, 58, 59, 61, 67, 69, 71, 73, 79, 83, 86, 87, 89, 92, 93, 94, 97, 101, 103, 106, 107, 109, 111, 113, 116, 118
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OFFSET
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1,2
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COMMENTS
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The way the sequence is defined, it is not obvious that it is computable.
The following condition is obviously computable and equivalent to the criterion for membership in this sequence: n is in the sequence iff it is true that a transitive subgroup of the symmetric group S_n is solvable iff it has a solvable point stabilizer.
As for the primes in the sequence, a theorem proved by Burnside guarantees that any unsolvable permutation group on a prime number of points is doubly transitive. The classification of doubly transitive permutation groups has been completed and it can be used to show that the only primes not in this sequence are 7, 13 and the Fermat primes larger than 3. So assuming the standard conjecture that 65537 is the largest Fermat prime, all primes except for these six (5, 7, 13, 17, 257 and 65537) are in this sequence.
If q (not equal to 2 or 3) is a power of a prime, then q+1 is not in this sequence, since PGL(2,q) is an unsolvable group with a solvable subgroup of index q+1 (isomorphic to the general affine group GA(1,q)).
Finally, if n is a nonmember of this sequence, then so is any multiple of n:
Since n is a nonmember of this sequence, there is a group G such that G is unsolvable and G has a solvable subgroup H of index n. Suppose N=d*n is a multiple of n. Then G x C_d (here C_d denotes a cyclic group of order d) is a group which is unsolvable because G is unsolvable. Nevertheless, it has a subgroup of index N isomorphic to H, which is solvable: this is the H x 1 subgroup.
This also means that any divisor of a member of this sequence is in this sequence and this form of the above-proved property is useful in proving membership in the sequence. One nice application of this form of that property is the following:
If n is insipid (that is, n is in A102842) and n/p is in this sequence for all primes p dividing n, then n is in this sequence.
Proof. Since n is insipid, n=1,2,3 or 4 or n>=34. If n=1,2,3 or 4, this is trivial since all of those values of n are in this sequence.
If n>=34, then let G be a group and let H be a solvable subgroup of index n in G.
Claim. H is not maximal in G.
Proof of Claim. Suppose H is maximal in G. Then let N be the core of H in G. G/N is isomorphic to a primitive permutation group on n points. Since n is insipid, this means G/N ~= A_n or S_n. Then restricting the quotient map (the map from G to G/N) to H yields a surjective homomorphism from H to a point stabilizer in S_n or A_n. This point stabilizer is isomorphic to S_(n-1) or A_(n-1). Since n>=34, n-1 >= 33 >= 5 so this image of H under this homomorphism is unsolvable and therefore H is unsolvable for a contradiction. The Claim is proved.
Since H is not maximal in G, there is a subgroup K of G such that G is properly contained in G and properly contains H. Then, by Lagrange's Theorem, [G:K] and [K:H] are proper divisors of n. This means there are (not necessarily distinct) prime factors p,q of n such that [G:K] | n/p and [K:H] | n/q. By our assumption, n/p and n/q are in this sequence. Then the property proved earlier implies that [G:K] and [K:H] are in this sequence. This means that the solvability of H implies the solvability of K, which, in turn, implies the solvability of G and the membership of n in this sequence. The proof is now complete.
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LINKS
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EXAMPLE
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22 is in this sequence, since the only primitive permutation groups on 22 points are M_22, Aut(M_22), A_22 and S_22. All of these have unsolvable point stabilizers.
This shows that a counterexample to 22 being in the sequence will be (by basic properties of the action-on-cosets homomorphism) an unsolvable group G with a solvable _nonmaximal_ subgroup H of index 22. Since H is nonmaximal, there is a subgroup K lying properly between G and H. Then by Lagrange's Theorem, we know [G:K]=2 and [K:H]=11 or vice versa. In either case, since 2 and 11 are both in this sequence, the solvability of H implies the solvability of K which implies the solvability of G. Thus 22 is in the sequence.
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CROSSREFS
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KEYWORD
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more,nonn
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AUTHOR
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STATUS
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approved
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