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Science Forum Index » Mathematics Forum » Proof that the mult groups of nonzero elements of a finite f
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| Van Jacques |
 Posted: Sat Dec 27, 2003 2:50 pm |
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I did a search on sci.math and see that the fact that the
nonzero elements of a finite field F*_q are a cyclic groups
has been discussed several times. If you don't want to read the
basic arguments again, skip this post.
I thought I could clarify the proof, at least to myself.
I don't know if this will help anyone else.
(I learned this in Birkhoff and Maclane, and thought I would never forget
their approach, but I have. I don't have a copy or access
to one, so if anyone does and will take the time and trouble, ...)
I think it goes something like this. A field F_q has q-1 nonzero
elements. Write q-1 = p_1^e_1 ... p_r^e_r.
(Note: The following is not restricted to p != 2. e.g., for q = 2^3 = 8,
F*_8 has q-1=7 elements, so is iso. to Z_7. Z*_8 is not
the multiplicative group of the nonzero elements of a field.
It has 4 elements, and isn't cyclic. q = 3^2, q-1 = 2^3
gives F*_9 = Z_8 is cyclic. But Z*_(2^e) is not cyclic for e > 2--see below.)
Let p = some p_i, e = e_i. There are p^(e-1) distinct solns. of
x^(p^(e-1)) - 1 = 0. These have orders <= p^(e-1), so there are
[p^(e-1)]*(p-1) with order p^e. Since this is true for
every p_i, and for abelian groups, if the orders of a and b
are relatively prime, |ab| = |a||b| (where |a| = order of a),
and there are elements of order q-1 = p_1^e_1 ... p_r^e_r so the
group is cyclic.
For example, if q = 3^2, q-1 = 2^3 = p^e, and p^(e-1) = 2^2 = 4.
There are 4 solns. of x^4 - 1 = 0, with orders <= 4, so there
are 4*(2-1) = 4 solns. with order 8, so F*_9 is cyclic.
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(Note: I use Z_n to denote both Z/nZ with + as group multiplication,
and as C_n = <x> = {x^i|i in Z_n}, with xy = x*y as group multiplication.)
I also like the following proof:
any finite abelian group is isomorphic to
G = Z_n_1 x Z_n_2 ... x Z_n_k ; where n_i divides n_(i+1).
(= usually means "isomorphic to" when applied to groups below.)
The order of G is n = n_1*n_2*...*n_k.
If G has at most m elements x such that |x| | m
for all m|n, then G is cyclic.
(e.g., consider G = Z_2 x Z_4. It has 4 elements of order <= 2,
so is not cyclic, while Z_8 has only 2 elements of order <= 2,
4 with order <= 4, and 8 with order <= 8, so Z_8 is cyclic.
Finally, if F is a field, F* the nonzero elements, the elements of
order dividing m are solutions of x^m - 1 = 0, which has at most
m solns., so F* is cyclic.
[Z_p is the splitting field for f(x) = x^p - x, i.e., the solns. of
f(x) = 0 are the elements of Z_p.]
=============
I also wanted to say something on the multiplicative groups Z*_n,
which are isomorphic the the automorphisms A(Z_n).
Let n = p_1^e_1 ... p_r^e_r. The algebraic form of the
Chinese remainder theorem gives the isomorphism
Z_n = Z_[p_1^(e_1)] x ... x Z_[p_r^(e_r)]
which induces the isomorphism
Z*_n = Z*_[p_1^(e_1)] x ... x Z*_[p_r^(e_r)]
Again, let p = p_i, e = e_i. Z*_p^e has p^(e-1)(p-1) elements.
p and p-1 are relatively prime if p != 2 (here is where
the special case of p = 2 comes in (I think), so assume p is odd).
This implies that
Z*_p^e is isomorphic to A x B, where |A| = p-1, |B| = p^(e-1).
The map Z*_p^e --> Z*_p induces an isomorphism from A to Z*_p,
which is cyclic of order phi(p) = p-1.
Using the binomial theorem shows that (p+1)^[p^(e-1)] = 1,
so p+1 has order
p^(e-1) = |B| in Z*_p^e, and p+1 generates B, which is also cyclic.
If x generates Z*_p, x(p+1) generates G:
G = Z*_p^e = Z_[p^(e-1)(p-1)] = A x B = Z_(p-1) x Z_p^(e-1),
Any element of G can be written
x^i*(p+1)^j ; i in Z_(p-1), j in Z_p^(e-1). This is for p odd.
For p = 2, Z*_2 = (1) ; Z*_4 = Z_2 ; Z*_2^e = Z_2 x Z_2^(e-2).
I don't have the proof for p = 2 done, so I will spare you that,
but I would like to see it.
Van |
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