login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A340511 Numbers k such that there exists a group of order k which has no subgroup of order d, for some divisor d of k. 5

%I #47 Dec 09 2021 06:03:14

%S 12,24,36,48,56,60,72,75,80,84,96,108,112,120,132,144,150,156,160,168,

%T 180,192,196,200,204,216,225,228,240,252,264,276,280,288,294,300,312,

%U 320,324,336,348,351,360,363,372,375,384,392,396,400,405,408,420

%N Numbers k such that there exists a group of order k which has no subgroup of order d, for some divisor d of k.

%C Suggested by the fact that the converse to Lagrange's theorem does not hold. These numbers might be called "Non-Converse Lagrange Theorem Orders".

%C A subsequence of A066085. The first difference between them is that 224 is missing from the present sequence (see MacHale-Manning, 2016). The sequence of terms of A066085 not in the present sequence is infinite, and begins 224, 2464, ... [This sequence is now A341048. - _Bernard Schott_, Feb 15 2021]

%C From _Jianing Song_, Dec 06 2021: (Start)

%C If k is a term, gcd(k,m) = 1, then k*m is again a term. Proof: If G is a group of order k without a subgroup of order k', then G X C_m has no subgroup of order k'*m. Suppose that it has, let G' be that subgroup. For every (a,b) in G', let m_0 be a multiple of m congruent to 1 modulo k, then (a,b)^(m_0) = (a,1) in G'; let k_0 be a multiple of k congruent to 1 modulo m, then (a,b)^(k_0) = (1,b) in G'. This shows that G' itself is of the form H X C_{m'}, where H is a subgroup of G and m' divides m. We have |H|*m' = k'*m, so |H| = k' and m' = m, contradicting with our assumption that G has no subgroup of order k'.

%C On the other hand, if gcd(k,m) > 1, then k*m need not be a term, as 56 is here but 224 is missing. In fact, N has a proper divisor here but N itself is not in this sequence if and only if N is in A341048. For the "only if" part, if N = k*m is a CLT order and k is a NCLT order, then k is a NSS order. Since every multiple of a NSS order is a NSS order, N is a NSS order, so by definition N is in A341048. The "if" part follows from MacHale-Manning, 2016, Corollary 13, Page 5.

%C Conjecture: If k = p^a*q^b, where p, q are primes, q !== 1 (mod p), b >= ord(q,p), then k is a term of this sequence, unless k is an NSS-CLT order of the form described in MacHale-Manning, 2016, Theorem 8, Page 5. Here ord(q,p) is the multiplicative order of q modulo p. Moreover, if k satisfies this condition, it seems that for each NCLT group of order k, the missing orders of subgroups are of the form p^a'*q^b' where either a' = a or b' = b, and a' = a if p == 1 (mod q) or a < ord(p,q). (End)

%H Des MacHale and J. Manning, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Manning/manning5.html">Converse Lagrange Theorem Orders and Supersolvable Orders</a>, Journal of Integer Sequences, 2016, Vol. 19, #16.8.7.

%e 12 belongs to this sequence because there is a group of order 12 (A_4) which has no subgroup of order 6, despite the fact that 6 divides 12.

%Y Cf. A066085, A340517, A341048.

%K nonn,more

%O 1,1

%A _Des MacHale_, Jan 24 2021

%E a(35)-a(53) from _Bernard Schott_, Feb 15 2021

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified March 28 16:58 EDT 2024. Contains 371254 sequences. (Running on oeis4.)