Date: Fri, 1 Dec 1995 17:30:05 -0500
To: rusin@math.niu.edu (Dave Rusin)
From: jkelley@postoffice.ptd.net (Jim Kelley)
Subject: Thanks, have a look at this.
>In article <49j6dd$q4o@ns2.ptd.net> you write:
>>I have heard that it is possible to turn a sphere inside out without
>>breaking the surface. Is this possible, and does a video of it exist?
>Yes, and yes, although the transformation takes place in R^4, so a
>2-dimensional movie picture which attempts to create a 3-dimensional
>image in our heads will have to merely convey some of the idea. But indeed,
>there is a function f(theta, phi, t) = (f_1, f_2, f_3, f_4) which
>has the properties that it is continuous, it's really defined on the
>sphere (i.e. f(0,phi,t)=f(2pi,phi,t) for all phi and t, etc.) and
>f(theta,phi,0)=-f(theta,phi,1)=(cos(theta)sin(phi),
sin(theta)sin(phi),cos(phi),0)
>for all theta and phi [here I mean whatever those usual coordinates are for
>the sphere -- sorry I forget the convential trig combinations, but these work.]
>
>A video does exist; as with everything else mathematical a reasonable
>place to begin is with the Amer Math Society (URL something like
>http://www.ams.org)
>
>> Also, does anyone know a general rule for the m such that U/m is
>>cyclic? I could not find a proof or anything in my number theory
>>books.
>
>Do you mean U/m to stand for the multiplicative group of the ring
>Z/mZ ? It's cyclic iff m=P or 2P where P can be 1, 2, or any power
>of any odd prime.
>
>dave
>
>
Dave--
Dang! I thought I could impress my friends with a sphere, but I
don't have the Planck energy yet. That's the result I got for U/m. Do you
know of the standard proof? I figured it out on my own, and I thought it
might be unknown (it was too easy for that). I tried posting a note before,
but when I told that I already knew the answer, nobody replied. It might be
sci.math taboo (of course, lying probably isn't appreciated much either...)
Here's my proof:
Part A) A power of an odd prime is cyclic.
Subpart A) U/p cyclic (this is notation that I learned from
an 88 year old professor, so it might not be up-to-date) Easy proof from
any basic textbook.
Subpart B) U/p^n cyclic ==> U/p^(n+1) cyclic. First, there
are phi(phi(p^n)) generators in the first. Assume no generators of the form
g and g+p exist. Then there is a limit on the number of generators that I
have forgotten. Anyway, it's less than phi(phi(p^n)) ==><== :-). Then
there are two generators of U/p^n that equal g and g+p. Let g+p=r (to save
time typing). The order of these numbers (or any two of the p pairs
congruent to them mod p^n) in U/p^(n+1) is either phi(p^n) or phi (p^(n+1)).
If the second case is true, we have a generator, so let's assume the first
case is true for both. Then g^(phi(p^n))==(congruent to) 1(mod P^(n+1)).
Expand (g+p)^(phi(p^n)) with binomial and set this == 1. Subtracting the
first congruence and all the terms that are divisible by p, we get some
product of (p-1)/2 and a power of g divisible by p. ==><==
Part B) P an odd modulo of a cyclic group==> U/2P is cyclic. Take
any generator, and find a unit congruent to it mod 2P(CRT). Since there is
only one unit in U/2P congruent to a unit in U/P, all powers of it are
distinct until you reach phi(P)=phi(2P). Also, it obviously reaches all
units.
Part C) If m contains two distinct odd prime factors, U/m not
cyclic. From the trivial lemma: g a generator of U/kl==>g a generator of k,
We get U/qr cyclic, where q and r are the primes. These are relatively
prime, so g^k==1 mod q&g^k==1mod r==> g^k==1 mod qr. Also
|U/(qr)|=phi(qr)=phi(q)phi(r). Then, the lcm of phi(q) and
phi(r)
(x mod m, x mod n) from Z/(mn) to (Z/m) x (Z/n), is an isomorphism
of rings if m and n are relatively prime. That is, it's one-to-one and
onto [those are equivalent, in view of the sizes of the sets] and it takes
sums to sums and products to products. In particular, it establishes an
isomorphism of the multiplicative groups U(mn) and U(m) x U(n) (again,
if m and n are relatively prime). Incidentally, a comparison of orders
at this point establishes the multiplicativity of the Euler phi function.
Anyway, you see that U(N) in general is isomorphic to the direct
product of its factors U(P) as p ranges over the distinct maximal prime
power divisors of N. As for cyclicity, you've pretty much burned your
bridges when N is not a prime power, since each Z/m has a group of
order two of units: {+1, -1}. Thus, U(P1) x U(P2) contains a non-cyclic
group of order 4. Your only hope is to have only a single prime factor,
or an extra factor of 2 (since U(2) = {1}). Even powers of two by themselves
are no good since U(8) = {1, 3, 5, 7} is non-cyclic.
In the other direction, your proof that U(P) _is_ a cyclic group is fine.
You might want to think about this in terms of a homomorphism of rings
Z/(p^(r+1))Z --> Z/(p^r)Z which induces a homomorphism of the corresponding
multiplicative groups.
This stuff ought to appear in most abstract algebra books. For example, we
use here a book by Beachy and Blair which has this scattered in the text
and in exercises.
If you enjoy reading about material on this level, you might find a few other
topics of interest to you on my home page -- http://www.math.niu.edu/~rusin
dave