In article <c5a11011.0312181839.b0ee414@posting.google.com>,
jhsntru@yahoo.com (JHS) wrote:
You can also find the proof in many textbooks, for example Ireland and
Rosen.
Of course, the correct statement is that (Z/p^nZ)^* is cyclic for all
ODD primes. It is not cyclic for even prime(s). Indeed, one has
(Z/2^nZ)^* = C(2) x C(2^{n-2}) for n >= 2,
where C(k) is a cyclic group of order k.
Going back to odd primes, people frequently take the limit as n goes
to infinity to get the p-adic integers Z_p. The unit group Z_p^* of
Z_p is topologically cyclic, i.e., there is an element g whose powers
are dense in Z_p^*.
One final comment. Let U be the subgroup of (Z/p^nZ)^* that consists
of elements c that are congruent to 1 modulo p. Then not only is U
cyclic, there is a natural isomorphism provided by the logarithm map.
The idea is to map
U ----> pZ/p^nZ, 1+p*t ---> log(1+p*t),
where the logarithm is defined using the Taylor series and is reduced
modulo p^n. This also works for the p-adic integers. (And when using
p=2, you need to take the units that are congruent to 1 modulo 8 to
make it work.)
JHS
Hubert <hubert.quatreville@wanadoo.fr> wrote in message
news:<3FE182D0.7000009@wanadoo.fr>...
Thanks a lot, that works.
For example
http://pma106.math.cuhk.edu.hk/Mat3080/MAT3080.htm
Robin Chapman a écrit:
Hubert wrote:
If p is a prime number, it is well know that (Z/pZ)* (the group of
unities of Z/pZ))is a cyclic group.
But it's also true that (Z/p^nZ)* is also a cyclic group when p<>2.
I've looked for a proof with google baut I cannot find the correct key
word.
Yes. The buzzword is "primitive root".
This post confused the hell out of me till I realized that JHS is not
JSH.
