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%I #25 Sep 20 2021 17:41:19
%S 24,48,60,72,96,120,144,160,168,180,192,216,240,288,300,320,324,336,
%T 360,384,432,480,504,540,576,600,640,648,660,672,720,768,784,800,840,
%U 864,896,900,960,972,1008,1053,1080,1092,1152,1176,1200,1280,1296,1320,1344,1440
%N List of numbers k > 1 such that there exists a group of order k without nontrivial normal Sylow subgroups.
%C Equivalently, numbers k > 1 such that there exists a group of order k with Sylow number > 1 for every prime dividing k.
%C The corresponding numbers of groups of order k without nontrivial normal Sylow subgroups are: 1, 4, 1, 4, 17, 3, 17, 1, 1, 1, 86, 18, 8, 90, 1, 5, 1, 3, 6, 536, 80, 27, ...
%C Note that if G has no normal Sylow p-subgroups, p divides |G|, then G X C_p also has no normal Sylow p-subgroups. That is to say, if k is in this sequence and p divides k, then k*p is also in this sequence. In particular, every number of the form 24 * 2^a * 3^b * 5^c * 7^d with nonnegative a,b,c,d is here.
%C The "primitive" terms (the terms not of the form k*p where k is a previous term and p divides k) are 24, 60, 160, 168, 324, 660, 784, 840, 896, 1053, ...
%C Includes A001034 as a subsequence, by the definition of non-cyclic simple groups. If q is not a prime power (not in A246655) and A338757(q) > 0, then q is here, as guaranteed by Schur-Zassenhaus theorem.
%C If k = p^e * q is here, p, q distinct primes, then q == 1 (mod p) and e >= 1+ord(p,q), where ord(p,q) is the multiplicative order of p modulo q. Proof: Let G be a group of order k without nontrivial normal Sylow subgroups. Let n_p (respectively n_q) be the number of Sylow p-subgroups (respectively q-subgroups), then n_p, n_q > 1. By Sylow's 3rd theorem, we have n_p == 1 (mod p), n_p | q; n_q == 1 (mod q), n_q | p^e. It is possible only if q == 1 (mod p) and e >= ord(p,q).
%C If e = ord(p,q), then we must have n_q = p^e. The Sylow q-subgroups have order q, which is a prime, so the pairwise intersections must be trivial, i.e., there are p^e * (q-1) = k - p^e elements in G of order q. The remaining p^e elements are just enough to make a unique Sylow p-subgroup, so n_p = 1, which is a contradiction. Hence, e >= 1+ord(p,q).
%C The terms of the form p^e * q where e = 1+ord(p,q), q == 1 (mod p) are 24 = 2^3 * 3, 160 = 2^5 * 5, 1053 = 3^4 * 13 and so on. Note that q == 1 (mod p) and e >= 1+ord(p,q) are only necessary but not sufficient: 112 = 2^4 * 7 satisfies 7 == 1 (mod 2) and 4 >= 1+ord(2,7), but 112 is not here. Similarly, 19375 = 5^4 * 31 satisfies 31 == 1 (mod 5) and 4 >= 1+ord(5,31), but 19375 is not here.
%H Jianing Song, <a href="/A338853/b338853.txt">Table of n, a(n) for n = 1..66</a>
%H The Group Properties Wiki, <a href="https://groupprops.subwiki.org/wiki/Sylow_subgroup">Sylow subgroup</a>
%H The Group Properties Wiki, <a href="https://groupprops.subwiki.org/wiki/Schur-Zassenhaus_theorem">Schur-Zassenhaus theorem</a>
%H <a href="/index/Gre#groups">Index entries for sequences related to groups</a>
%e All the normal subgroups of S_4 (symmetric group of degree 4, order 24) are the trivial group, the Klein four-group (order 4), A_4 (alternating group of degree 4, order 12) and S_4 itself. None of these is a Sylow 2-subgroup or a Sylow 3-subgroup. So 24 is a term.
%e All the normal subgroups of SmallGroup(1053,51) are the trivial group, C_3 X C_3 X C_3 (order 27), SmallGroup(351,12) and SmallGroup(1053,51) itself. None of these is a Sylow 3-subgroup or a Sylow 13-subgroup. So 1053 is a term. In fact, 1053 is the smallest odd term. [As a result, every number of the form 1053 * 3^a * 13^b with nonnegative a,b is a term, showing that there are infinitely many odd terms in this sequence. What is the smallest odd term not of this form? - _Jianing Song_, Sep 08 2021]
%o (GAP)
%o HasNoSylow := function(G)
%o local c, l, i;
%o c := FactInt(Size(G))[1];
%o l := Length(c);
%o if c[1] = c[l] then # |G| is 1 or a prime power
%o return false;
%o else
%o for i in [1..l] do
%o if IsNormal(G, SylowSubgroup(G, c[i])) then
%o return false;
%o fi;
%o od;
%o return true;
%o fi;
%o end;
%o IsA338853 := function(n)
%o local c, l, i;
%o c := FactInt(n)[1];
%o l := Length(c);
%o if c[1] = c[l] then # |G| is 1 or a prime power
%o return false;
%o else
%o i := NumberSmallGroups(n);
%o while i > 0 do
%o if(HasNoSylow(SmallGroup(n,i))) then
%o return true;
%o fi;
%o i := i-1;
%o od;
%o return false;
%o fi;
%o end;
%Y Cf. A001034, A246655, A338757.
%K nonn
%O 1,1
%A _Jianing Song_, Nov 12 2020
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