From: aitken@halcyon.com (William E. Aitken)
Subject: Group theory question
Date: 5 Dec 1999 01:57:57 -0800
Newsgroups: sci.math
Keywords: Structure of groups of order pq^2

I'm trying to prove theorems about the structure of finite groups G of
order pq^2, where p \ne q are primes.  Ideally, I'd like some tight
characterization of exactly what the structure of these groups can be.

I use Z_n to represent the non-negative integers less than n, under
mod n addition.  And I use Z_n^* to represent the positive
integers that are both less than n and relatively prime to n,
under mod n multiplication.  Let P be a Sylow p-subgroup of G
and Q a Sylow q-subgroup.

I have made the following headway.

1) if q < p, everything works out nicely.  

2) If q > p, Q is normal in
G.  G is isomorphic to a semi-direct product of Q with P.  If P is
also normal, then this semi-direct product is actually a direct
product.  In particular if p divides neither q-1 nor q+1, P is normal.
Note too that if Z(G) = Q then G/Z(G) is cyclic, so G is Abelian, and
P is normal.  Otherwise, we must consider isomorphisms from P into
Aut(Q).  All that remains is characterization of p-subgroups of
Aut(Q)., and determining which automorphisms of Q yield non-isomorphic
groups G.

3) if q > p, I am fairly sure I know how to characterize the groups of order
pq^2 in which Q is cyclic, P non-normal.   I am missing only the following 

LEMMA: If q is prime then Z_{q^2}^* is cyclic. [I only actually need
it for odd primes, but this is no simplification, since Z_4^* is cyclic 
of order 2].

4) If Q is non-cyclic, the automorphisms of Q can be viewed as 
elements of GL(2, q)

5) When p odd, p | q - 1, Q is non-cyclic I have managed to
convince myself that I need only consider the diagonal
elements of GL(2, q).  These form a group of order (q-1)^2
that is isomorphic to Z_q^* \times Z_q^*.  In particular,
this group is generated by the following 

        (1 0)     (u 0)
        (0 u) and (0 1)

where u denotes a primitive (q-1)th root of unity (mod q).   It's now easy to 
read off the automorphisms that matter:

        (a 0)
        (0 b) where a and b are pth roots of unity (mod q).   Unfortunately,
this yields too many automorphisms: distinct automorphisms can yield 
isomorphic G's.   Clearly if

        (a 0)   (a' 0)^n         (a 0)      (a' 0)
        (0 b) = (0 b')      then (0 b) and  (0 b') yield isomorphic G.  
But this still doesn't identify enough automorphisms.   I'm wondering
what else to try here.   Can anyone suggest anything?

6) When p odd p | q + 1, Q is non-cyclic I am lost.  Certainly GL(2, q) 
has a subgroup of order p (by Sylow), but I don't know how to
characterize it.  Clearly, the relevant automorphisms are NOT
diagonal elements of GL(2, q) since p can not divide q - 1.  Can
someone suggest an avenue of attack.

7) Presumably p = 2 will require us to combine the reasoning of 5 and
6.  Depending on the exact details of our investigation we may have
to do some more tidying up here.  No questions here, at least not
yet.

8) it seems that whatever transpires here, we will be able to fairly
quickly apply our results to groups of order p^2q^2 too.  Again,
its going to depend on fine details of what we do in 5 and 6, but
everything so far is sound.

Incidentally, I have found myself at times wanting something like the
following lemma:

Let q be an odd prime and suppose that n divides q - 1.  Let u
be a primitive nth root of unity (mod q).  Then u + u^2 + ... + u^n = 0. 
Is this (or something similar true)?  I don't think it comes up in
anything I have discussed above, but I think I wanted something like
it when I was thinking about groups of order p^3 a few days ago.  In
that case, my recollection is that all I ended up needing was the
special case in which n = q -1, where the result certainly holds
because the integers modulo q are a field.

-- 
William E. Aitken 			    | Formal verification is the 
email: aitken@halcyon.com                   |  future of computer science ---
Snail: 6124 86th Ave SE Mercer Island WA    | Always has been, always will be.
===============================================================================

From: Robin Chapman <rjc@maths.ex.ac.uk>
Subject: Re: Group theory question
Date: Sun, 05 Dec 1999 21:54:49 GMT
Newsgroups: sci.math
To: aitken@halcyon.com

In article <82dcv5$5om$1@halcyon.com>,
  aitken@halcyon.com (William E. Aitken) wrote:
> I'm trying to prove theorems about the structure of finite groups G of
> order pq^2, where p \ne q are primes.  Ideally, I'd like some tight
> characterization of exactly what the structure of these groups can be.
>
> I use Z_n to represent the non-negative integers less than n, under
> mod n addition.  And I use Z_n^* to represent the positive
> integers that are both less than n and relatively prime to n,
> under mod n multiplication.  Let P be a Sylow p-subgroup of G
> and Q a Sylow q-subgroup.

> 6) When p odd p | q + 1, Q is non-cyclic I am lost.  Certainly GL(2,q)
> has a subgroup of order p (by Sylow), but I don't know how to
> characterize it.  Clearly, the relevant automorphisms are NOT
> diagonal elements of GL(2, q) since p can not divide q - 1.  Can
> someone suggest an avenue of attack.

The field F_q of q elements has a quadratic extension F_{q^2}
of q^2 elements. The multiplicative group of this field
is cyclic of order q^2-1 = (q-1)(q+1). If p divides q+1, then
there is an element t of multiplicative order p in F_{q^2}.
Let X^2 + aX + b be the minimum polynomial of t over F_q. Then
its companion matrix
( 0  1)
(-b -a)
has order p.

> Incidentally, I have found myself at times wanting something like the
> following lemma:
>
> Let q be an odd prime and suppose that n divides q - 1.  Let u
> be a primitive nth root of unity (mod q).  Then u + u^2 + ... + u^n = 0.
> Is this (or something similar true)?

Yes. It's an arithmetic progression summing to u(1-u^n)/(1-u) = 0.
--
Robin Chapman
 http://www.maths.ex.ac.uk/~rjc/rjc.html
  "`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'"
    Greg Egan, _Teranesia_


Sent via Deja.com http://www.deja.com/
Before you buy.
==============================================================================

From: mareg@mimosa.csv.warwick.ac.uk ()
Subject: Re: Group theory question
Date: 6 Dec 1999 00:20:42 GMT
Newsgroups: sci.math

In article <82dcv5$5om$1@halcyon.com>,
	aitken@halcyon.com (William E. Aitken) writes:
>I'm trying to prove theorems about the structure of finite groups G of
>order pq^2, where p \ne q are primes.  Ideally, I'd like some tight
>characterization of exactly what the structure of these groups can be.
>

Here are a few quick hints

>I use Z_n to represent the non-negative integers less than n, under
>mod n addition.  And I use Z_n^* to represent the positive
>integers that are both less than n and relatively prime to n,
>under mod n multiplication.  Let P be a Sylow p-subgroup of G
>and Q a Sylow q-subgroup.
>
>I have made the following headway.
>
>1) if q < p, everything works out nicely.  
>2) If q > p, Q is normal in
>G.  G is isomorphic to a semi-direct product of Q with P.  If P is
>also normal, then this semi-direct product is actually a direct
>product.  In particular if p divides neither q-1 nor q+1, P is normal.
>Note too that if Z(G) = Q then G/Z(G) is cyclic, so G is Abelian, and
>P is normal.  Otherwise, we must consider isomorphisms from P into
>Aut(Q).  All that remains is characterization of p-subgroups of
>Aut(Q)., and determining which automorphisms of Q yield non-isomorphic
>groups G.
>
>3) if q > p, I am fairly sure I know how to characterize the groups of order
>pq^2 in which Q is cyclic, P non-normal.   I am missing only the following 
>
>LEMMA: If q is prime then Z_{q^2}^* is cyclic. [I only actually need
>it for odd primes, but this is no simplification, since Z_4^* is cyclic 
>of order 2].

The element (1+q) in Z_{q^2} generates the cyclic subgroup
Q = {1+rq | 0 <= r < q } of Z_{q^2}^* and |Q| = q. 
It is easy to see that Z_{q^2}^*/Q is the same as Z_q^*, which is cyclic of
order q-1, and now it follows that Z_{q^2}^* is cyclic of order q(q-1),
because q and q-1 are coprime.

More generally, it turns out that Z_{q^r}^* is cyclic for q odd, but not
when q=2 and r>=3.

>4) If Q is non-cyclic, the automorphisms of Q can be viewed as 
>elements of GL(2, q)
>
>5) When p odd, p | q - 1, Q is non-cyclic I have managed to
>convince myself that I need only consider the diagonal
>elements of GL(2, q).  These form a group of order (q-1)^2
>that is isomorphic to Z_q^* \times Z_q^*.  In particular,
>this group is generated by the following 
>
>        (1 0)     (u 0)
>        (0 u) and (0 1)
>
>where u denotes a primitive (q-1)th root of unity (mod q).   It's now easy to 
>read off the automorphisms that matter:
>
>        (a 0)
>        (0 b) where a and b are pth roots of unity (mod q).   Unfortunately,
>this yields too many automorphisms: distinct automorphisms can yield 
>isomorphic G's.   Clearly if
>
>        (a 0)   (a' 0)^n         (a 0)      (a' 0)
>        (0 b) = (0 b')      then (0 b) and  (0 b') yield isomorphic G.  
>But this still doesn't identify enough automorphisms.   I'm wondering
>what else to try here.   Can anyone suggest anything?

This is probably the trickiest case.
What we are doing is equivalent to finding the conjugacy classes of subgroups
of order p in GL(2,q). As you say, each such subgroup is conjugate to one
generated by a diagonal matrix, which I will write as [a,b] for simplicity
of notation.
There is one class [a,1] and another [a,a].
The others have the form [a,b], where a and b are unequal and not 1.
We can fix a to be a fixed element of order p in Z_q^*.
There is one further conjugacy to be considered - [a,b]  is conjugate to
[b,a].  This conjugates [a,a^-1] to itself, for the remaining values of b,
the elements are paired off by this conjugation.
So, in total, we get 1 + 1 + 1 + (p-3)/2 isomorphism types of nonabelian
groups of this type, coming from
[a,1], [a,a], [a,a^-1], and the (p-3)/2 paired off  [a,b] s.

>6) When p odd p | q + 1, Q is non-cyclic I am lost.  Certainly GL(2, q) 
>has a subgroup of order p (by Sylow), but I don't know how to
>characterize it.  Clearly, the relevant automorphisms are NOT
>diagonal elements of GL(2, q) since p can not divide q - 1.  Can
>someone suggest an avenue of attack.

The easiest way to imagine the element of order q+1 is to note that
multiplication by a multiplicative generator of the field GF(q^2) of
order q^2 induces an element of order q^2-1 in GL(2,q). In particular,
we have an element of order q+1, which tells us that a Sylow p-subgroup
of GL(2,q) is cyclic. Hence, by Sylow, all subgroups of GL(2,q) of order
p are conjugatye, and there is just one class of nonabelian groups coming
from  this case.

>7) Presumably p = 2 will require us to combine the reasoning of 5 and
>6.  Depending on the exact details of our investigation we may have
>to do some more tidying up here.  No questions here, at least not
>yet.

No, an elemnt of order 2 in GL(2,q) is conjugate to a diagonal matrix, so
this is just a rather easy instance of 5) - there are just two
possibilities, [a,1] and [a,a].

Hope this helps.
Derek Holt.
==============================================================================

From: kramsay@aol.commangled (Keith Ramsay)
Subject: Re: Group theory question
Date: 06 Dec 1999 05:01:24 GMT
Newsgroups: sci.math

In article <82dcv5$5om$1@halcyon.com>,
aitken@halcyon.com (William E. Aitken) writes:
|LEMMA: If q is prime then Z_{q^2}^* is cyclic. [I only actually need
|it for odd primes, but this is no simplification, since Z_4^* is cyclic 
|of order 2].

This is standard fare of number theory textbooks.

You already know Z_q^* is cyclic. So take a primitive root e mod q.
The order of e in Z_{q^2}^* divides the order of Z_{q^2}^*, i.e.,
q(q-1), and is a multiple of q-1, its order in Z_q^*. The only possible
orders of e in Z_{q^2}^* are then q-1 and q(q-1). So e generates all
of Z_{q^2}^*, unless e^{q-1} is congruent to 1 modulo q^2. If so, then
replace e with e+q. We know e+q is congruent to e mod q, but
(e+q)^{q-1} is not congruent to e^{q-1} mod q^2, as we can see by
expanding (e+q)^{q-1} as a binomial, with all but the terms e^{q-1}
and (q-1)e^{q-2}q divisible by q^2.

|6) When p odd p | q + 1, Q is non-cyclic I am lost.  Certainly GL(2, q) 
|has a subgroup of order p (by Sylow), but I don't know how to
|characterize it.  Clearly, the relevant automorphisms are NOT
|diagonal elements of GL(2, q) since p can not divide q - 1.  Can
|someone suggest an avenue of attack.

Identify Z_q + Z_q with the additive group of a finite field F of order
q^2. (There is only one up to isomorphism.) The multiplicative group
of F acts on the additive group by automorphisms, and has order q^2-1
=(q-1)(q+1). There are q+1-st roots of unity in F, including among
them p-th roots of 1, which act as required.

|Let q be an odd prime and suppose that n divides q - 1.  Let u
|be a primitive nth root of unity (mod q).  Then u + u^2 + ... + u^n = 0. 
|Is this (or something similar true)?

You are thinking of this in the finite field of order q, but it's
generally true. We would more often write 1 instead of u^n. It follows
from  the factorization u^n-1=(u-1)(u^{n-1}+...+1)=0, and u-1<>0.

Keith Ramsay