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A338853
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List of numbers k > 1 such that there exists a group of order k without nontrivial normal Sylow subgroups.
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2
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24, 48, 60, 72, 96, 120, 144, 160, 168, 180, 192, 216, 240, 288, 300, 320, 324, 336, 360, 384, 432, 480, 504, 540, 576, 600, 640, 648, 660, 672, 720, 768, 784, 800, 840, 864, 896, 900, 960, 972, 1008, 1053, 1080, 1092, 1152, 1176, 1200, 1280, 1296, 1320, 1344, 1440
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OFFSET
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1,1
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COMMENTS
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Equivalently, numbers k > 1 such that there exists a group of order k with Sylow number > 1 for every prime dividing k.
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, ...
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.
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, ...
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.
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).
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).
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.
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LINKS
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EXAMPLE
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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.
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]
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PROG
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(GAP)
HasNoSylow := function(G)
local c, l, i;
c := FactInt(Size(G))[1];
l := Length(c);
if c[1] = c[l] then # |G| is 1 or a prime power
return false;
else
for i in [1..l] do
if IsNormal(G, SylowSubgroup(G, c[i])) then
return false;
fi;
od;
return true;
fi;
end;
IsA338853 := function(n)
local c, l, i;
c := FactInt(n)[1];
l := Length(c);
if c[1] = c[l] then # |G| is 1 or a prime power
return false;
else
i := NumberSmallGroups(n);
while i > 0 do
if(HasNoSylow(SmallGroup(n, i))) then
return true;
fi;
i := i-1;
od;
return false;
fi;
end;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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