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