From: mareg@csv.warwick.ac.uk (Dr D F Holt) Newsgroups: sci.math Subject: Re: When every group of order n is abelian (Was: example) Date: 7 Oct 1997 10:00:41 +0100 In article <61bpmr\$sv\$1@nntp.Stanford.EDU>, hwatheod@leland.Stanford.EDU (theodore hwa) writes: >Einar Andreas R|dland (einara@ulrik.uio.no) wrote: >: >: Also, if o(G)=p^2, G is abelian. >: > >This is slightly off the subject of the original thread but it brings up >the question of when every group of order n is abelian. We know that > >1) There's a non-abelian group of order p^3 for every prime p: > the group of 3 by 3 upper triangular matrices having 1's on the main >diagonal over the field Z_p. > >2) If p and q are primes such that p divides q-1 then there's a >non-abelian semidirect product of Z_p by Z_q. > >Therefore certainly 2 necessary conditions for n are that it be cube-free >and for every two primes p,q dividing n, p cannot divide q-1. Is this >sufficient? > >A _Monthly_ article from a few years ago (June-July 1992 ?! I know it was >a summer issue) proved that if "cube-free" is replaced by "square-free" >then the above is sufficient for every group of order n to be cyclic. (The >condition was stated as gcd(n,phi(n))=1, which is the same thing.) No, your condition is not quite sufficient. You need the extra condition that, whenever p and q are distinct primes with q^2 and p dividing n, then p does not divide q+1. Then it is sufficient. For example, there does exist a nonabelian group of order 75, because 3 divides 5+1, and so 3 divides the order of the automorphism group of Z_5 x Z_5, and you can construct the group as a semidirect product. e.g. G = < x,y,z | x^5=y^5=z^3=1, xy=yx, z^-1xz=y, z^-1yz=x^-1y^-1 >. In general, the order of the automorphism group of Z_p x Z_p is (p-1)p(p+1) and of Z_{p^2} is p(p-1). Using those facts, it is not difficult to prove by induction that every group of order n is abelian, whenever, for distict primes p,q dividing n: (i) p^3 does not divide n, (ii) p does not divide q-1 (iii) if q^2 divides n, then p does not divide q+1. Sketch of proof (using Odd Order Theorem). Since groups of order p and p^2 are well-known to be abelian, we can assume n is divisible by at least two primes, and then (ii) implies that n is odd. Hence G is soluble by the Odd Order Theorem. Thus G has a proper nontrivial normal subgroup N, and by induction N and G/N are abelian. By choosing N minimal, it has order q or q^2. In fact, if q^2 divides N, then we can take N to have order q^2 in any case. Then N has a complement H in G, and by the above, |H| is relatively prime to |Aut(N)|, which implies H centralizes N, so G is abelian. Can anyone find a proof that does not invoke the Odd Order Theorem? Derek Holt. ============================================================================== From: mareg@csv.warwick.ac.uk (Dr D F Holt) Newsgroups: sci.math Subject: Re: When every group of order n is abelian (Was: example) Date: 8 Oct 1997 12:28:13 +0100 In article <61ecmb\$sft\$1@nntp.Stanford.EDU>, hwatheod@leland.Stanford.EDU (theodore hwa) writes: >Dr D F Holt (mareg@csv.warwick.ac.uk) wrote: >: >1) There's a non-abelian group of order p^3 for every prime p: >: > the group of 3 by 3 upper triangular matrices having 1's on the main >: >diagonal over the field Z_p. >: > >: >2) If p and q are primes such that p divides q-1 then there's a >: >non-abelian semidirect product of Z_p by Z_q. >: > >: >: No, your condition is not quite sufficient. You need the extra condition >: that, whenever p and q are distinct primes with q^2 and p dividing n, >: then p does not divide q+1. Then it is sufficient. >: >: For example, there does exist a nonabelian group of order 75, because >: 3 divides 5+1, and so 3 divides the order of the automorphism group >: of Z_5 x Z_5, and you can construct the group as a semidirect product. >: >: e.g. G = < x,y,z | x^5=y^5=z^3=1, xy=yx, z^-1xz=y, z^-1yz=x^-1y^-1 >. >: >: In general, the order of the automorphism group of Z_p x Z_p is (p-1)p(p+1) > >I get p(p+1)(p-1)^2 when I count them. The images of (1,0) and (0,1) have >to be linearly independent, so that gives p^2-1 choices (any non-zero) for >(1,0) and p^2-p choices for (0,1) (anything that isn't a multiple of the >first choice.) (p^2-1)(p^2-p)=p(p+1)(p-1)^2. But that doesn't affect the >argument below. Yes - you are perfectly correct. Sorry, I wrote it in too much of a hurry. Incidentally, (p-1)p(p+1) is the order of SL(2,p), the group of automorphisms having determinant 1. > >: and of Z_{p^2} is p(p-1). Using those facts, it is not difficult to prove >: by induction that every group of order n is abelian, whenever, for distict >: primes p,q dividing n: >: (i) p^3 does not divide n, >: (ii) p does not divide q-1 >: (iii) if q^2 divides n, then p does not divide q+1. >: >: Sketch of proof (using Odd Order Theorem). Since groups of order p and p^2 >: are well-known to be abelian, we can assume n is divisible by at least two >: primes, and then (ii) implies that n is odd. Hence G is soluble by the >: Odd Order Theorem. Thus G has a proper nontrivial normal subgroup N, and >: by induction N and G/N are abelian. > >ok so far... > >: By choosing N minimal, it has order >: q or q^2. In fact, if q^2 divides N, then we can take N to have order >: q^2 in any case. > >I don't follow these last 2 sentences. If the order of N had 2 prime >factors, then ... ? I said it was a sketch! Remember N is abelian, which means that it is a direct product of its Sylow subgroups - i.e. its primary components - e.g. if |N| = 12, N = N1 x N2 with N1 and N2 having orders 3 and 4. Each of the Ni is characteristic in N (i.e. fixed by all automorphisms of N), so must be normal in G. Hence minimality of N implies |N| is a power of a prime. The next statement is slightly trickier. If |N|=p, and p^2 divides |G|, then by a similar argument to above, G/N has a normal subgroup M/N of order p, but then M is normal in G and has order p^2. So, in any case, if P^2 divides |G| and p divides |N| then we can choose |N|=p^2. > >: Then N has a complement H in G, and by the above, |H| >: is relatively prime to |Aut(N)|, which implies H centralizes N, so G >: is abelian. > >What is "complement" in this context? > A complement H of N in G is a subgroup satisfying HN=G and H intersect N = {1}. Its existence follows, for example from the Schur-Zassenhaus Theorem, which says that whenever N is normal in G (finite), and |N| and |G/N| are relatively prime, then N has a complement H in G. This complement H is then abelian (it is isomorphic to G/N) and since |H| and |Aut(N)| are relatively prime, H must centralize N, and then we have G = N x H is abelian. Incidentally Avrinoam Mann has pointed out to me that one can avoid using the Odd Order Theorem (which is an extremely difficult result) by using the less well-known but more elementary theorem known as Burnside's Transfer Theorem. This says that whenever G has a Sylow p-subgroup P such that P is central in its normalizer, then G has a normal subgroup N of order |G|/|P|. This condition applies to any Sylow p-subgroup of G, which implies the existence of N, which proves that G is not simple. Then solvability of G follows easily by induction on |G|. Derek Holt. ============================================================================== Date: Wed, 08 Oct 1997 02:41:15 -0600 From: rjc@maths.ex.ac.uk Subject: Re: When every group of order n is abelian (Was: example) Newsgroups: sci.math In article <61ctnp\$5jo@crocus.csv.warwick.ac.uk>, mareg@csv.warwick.ac.uk (Dr D F Holt) wrote: > > In article <61bpmr\$sv\$1@nntp.Stanford.EDU>, > hwatheod@leland.Stanford.EDU (theodore hwa) writes: > >Einar Andreas R|dland (einara@ulrik.uio.no) wrote: > >: > >: Also, if o(G)=p^2, G is abelian. > >: > > > >This is slightly off the subject of the original thread but it brings up > >the question of when every group of order n is abelian. We know that > > > >1) There's a non-abelian group of order p^3 for every prime p: > > the group of 3 by 3 upper triangular matrices having 1's on the main > >diagonal over the field Z_p. > > > >2) If p and q are primes such that p divides q-1 then there's a > >non-abelian semidirect product of Z_p by Z_q. > > > >Therefore certainly 2 necessary conditions for n are that it be cube-free > >and for every two primes p,q dividing n, p cannot divide q-1. Is this > >sufficient? > > > >A _Monthly_ article from a few years ago (June-July 1992 ?! I know it was > >a summer issue) proved that if "cube-free" is replaced by "square-free" > >then the above is sufficient for every group of order n to be cyclic. (The > >condition was stated as gcd(n,phi(n))=1, which is the same thing.) > > No, your condition is not quite sufficient. You need the extra condition > that, whenever p and q are distinct primes with q^2 and p dividing n, > then p does not divide q+1. Then it is sufficient. > > For example, there does exist a nonabelian group of order 75, because > 3 divides 5+1, and so 3 divides the order of the automorphism group > of Z_5 x Z_5, and you can construct the group as a semidirect product. > > e.g. G = < x,y,z | x^5=y^5=z^3=1, xy=yx, z^-1xz=y, z^-1yz=x^-1y^-1 >. > > In general, the order of the automorphism group of Z_p x Z_p is (p-1)p(p+1) > and of Z_{p^2} is p(p-1). Using those facts, it is not difficult to prove > by induction that every group of order n is abelian, whenever, for distict > primes p,q dividing n: > (i) p^3 does not divide n, > (ii) p does not divide q-1 > (iii) if q^2 divides n, then p does not divide q+1. > > Sketch of proof (using Odd Order Theorem). Since groups of order p and p^2 > are well-known to be abelian, we can assume n is divisible by at least two > primes, and then (ii) implies that n is odd. Hence G is soluble by the > Odd Order Theorem. Thus G has a proper nontrivial normal subgroup N, and > by induction N and G/N are abelian. By choosing N minimal, it has order > q or q^2. In fact, if q^2 divides N, then we can take N to have order > q^2 in any case. Then N has a complement H in G, and by the above, |H| > is relatively prime to |Aut(N)|, which implies H centralizes N, so G > is abelian. > > Can anyone find a proof that does not invoke the Odd Order Theorem? This is largely based on the exercises in Dummit and Foote, Abstract Algebra, Prentice-Hall, 1991. Say n has property A if it satisfies the three conditions (i), (ii) and (iii). I claim that if n has property A then every group of order n is abelian. Let G be a minimal counterexample of order n. As each factor of n has property A, then each proper subgroup and quotient of G is abelian. Suppose that G is simple. If M and N are distinct maximal subgroups of G, then M cap N is normal in both, so also normal in = G. Thus M cap N is trivial. Let M be a maximal subgroup of index r. The normalizer of M in G is M, so M has r conjugates, each pair of which meet trivially. The conjugates of M contain n-r non-identity elements. As r divides n, this is more than half, but not all of the non-identity elements of G. There must be a maximal subgroup N not conjugate to M. This is impossible, as N's conjugates also cover more than half of the non-identity elements of G. So G isn't simple. Let H be a non-trivial normal subgroup of G. Then H and G/H are abelian. Now G/H acts on H by conjugation. But condition A implies that |G/H| and |Aut(H)| are coprime. Thus G/H acts trivially on H, and so G is abelian. One can extend these ideas to a characterization of the n for which all groups of order n are nilpotent. The condition here is that if p and q are prime factors of n with p^k dividing n, then q cannot divide p^k-1. Robin Chapman "256 256 256. Department of Mathematics O hel, ol rite; 256; whot's University of Exeter, EX4 4QE, UK 12 tyms 256? Bugird if I no. rjc@maths.exeter.ac.uk 2 dificult 2 work out." http://www.maths.ex.ac.uk/~rjc/rjc.html Iain M. Banks - Feersum Endjinn