multiplicative group of residues mod p
Anders Gorst-Rasmussen
I'm looking for a short proof of the fact, that the group (Z/pZ)* is cyclic.
I've seen several methods based on the theory of finite fields, more
specifically the following theorem:
"Let K be a field with q elements. Then for any divisor d of q-1 there are
exactly phi(d) elements of order d in the unit group K*"
Does there exist an "easy way" to prove this without theory of fields etc.
(I'm only interested in the group-theoretic result, nothing more)...
Jyrki Lahtonen
Since the group in question is not just any group, but a group of
units of a field, you shouldn't necessarily be surprised that
group theory alone won't do.
The least amount of field theory you can get away with is the fact
that in a field a polynomial of degree n can have at most n zeros.
(Ok, may be somebody will show a way of getting away with even less,
whatever that may mean.)
This suffices (together with the "purely group theoretic" structure
theorem of finitely generated abelian groups), as the following
argument shows (I apologize, if you have seen this standard proof):
The group Z_p^* has p-1 elements and it is abelian. Hence
it is isomorphic to a direct product of one or more cyclic groups.
Moreover the orders k_1, k_2, ... k_t of the t cyclic factor
groups can be chosen to divide each other in sequence (k_1 divides
k_2 divides k_3 ... k_t divides p-1). So all the elements
of Z_p^* are of order k_t. Thus the polynomial
p(x)= x^(k_t)-1
of degree k_t has p-1 zeros in the field Z_p. Therefore k_t=p-1.
Therefore there is only one non-trivial factor group in the
above direct product representation. Therefore Z_p^* is cyclic.
This is easy enough for most people.
Cheers,
Jyrki Lahtonen
Robin Chapman
news:3CC806D9...@utu.fi...
>
> > "Let K be a field with q elements. Then for any divisor d of q-1 there are
> > exactly phi(d) elements of order d in the unit group K*"
> >
> > Does there exist an "easy way" to prove this without theory of fields etc.
> > (I'm only interested in the group-theoretic result, nothing more)...
>
> Since the group in question is not just any group, but a group of
> units of a field, you shouldn't necessarily be surprised that
> group theory alone won't do.
>
> The least amount of field theory you can get away with is the fact
> that in a field a polynomial of degree n can have at most n zeros.
All the proofs I know need this.
> This suffices (together with the "purely group theoretic" structure
> theorem of finitely generated abelian groups), as the following
> argument shows:
Well Gauss in the Disquisitiones din't need this structure
theorem (rather harder than it appears). Basically the Gauss
argument shows that for n |(q-1) there are exactly phi(n)
elements of order exactly n.
Robin Chapman
--
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Bill Dubuque
|
| I'm looking for a short proof that the group (Z/pZ)* is cyclic.
Here is a proof that avoids invoking the high-powered structure theorem
for finite abelian groups (as in the below quoted proof posted by Jyrki).
THEOREM A subgroup G of the multiplicative group of a field is cyclic.
Proof: The proposition below yields, with m = maxord(G) = expt(G),
that X^m = 1 has #G roots. Since a poly over a field satisfies
#roots(P) <= deg(P) we have #G <= m. But m <= #G because the
maxorder m <= #G via g^#G = 1 for all g in G (Lagrange's theorem).
So m = #G = maxord(G) => G has an elt of order #G, so G is cyclic.
PROPOSITION maxord(G) = expt(G) for all finite abelian groups G,
i.e max { ord(g) : g in G } = min { n>0 : g^n = 1 for all g in G }
Proof: The lemma below shows the set S = { ord(g) : g in G } is
a set of naturals closed under lcm, hence every elt s in S is a
divisor of the max elt m (else lcm(s,m) > m), so m = expt(G).
LEMMA Let G be a finite abelian group containing elts a and b
of order d and e resp. Then G contains an element of order
lcm(d,e) of the form a^m b^n for some naturals m, n.
Proof (sketch): Reduce to the easy case where gcd(d,e) = 1
as follows: choose m,n so that a^m, b^n have coprime orders
d/m, e/n resp, with de/mn = lcm(d,e) ...
Note that below quoted proof may also be viewed as being based
upon the equality maxorder(G) = expt(G) (= k_t below).
-Bill Dubuque
| I've seen several methods based on the theory of finite fields, more
| specifically the following theorem:
|
| "Let K be a field with q elements. Then for any divisor d of q-1 there are
| exactly phi(d) elements of order d in the unit group K*"
|
| Does there exist an "easy way" to prove this without theory of fields etc.
| (I'm only interested in the group-theoretic result, nothing more)...
Bob Kolker
Bill Dubuque wrote:
>
>
> THEOREM A subgroup G of the multiplicative group of a field is cyclic.
I think I am missing something here. Let H(p) = { p^k | k an integer } H
is closed under multiplication and p^0 is the unit. For an infinite
field this is not cyclic.
Have I made a mistake?
Bob Kolker
>
Arturo Magidin
The theorem should state that G is supposed to be finite. That is, a
finite subgroup of the multiplicative group of a field is cyclic.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Bob Kolker
Arturo Magidin wrote:
>
> The theorem should state that G is supposed to be finite. That is, a
> finite subgroup of the multiplicative group of a field is cyclic.
Yup. Thank you.
Bob Kolker
Bill Dubuque
>
> THEOREM A subgroup G of the multiplicative group of a field is cyclic.
"A *finite* subgroup G of the ..." was intended, as is clear from the proof.