Number theory, geometry and algebra - Dynamics-approx.jku.at

The same proof demonstrates the following fact:
Proposition 20 Let G be a group with |G| = m, so that d|m ⇒ G has at
most d elements with x d = e. Then G is cyclic.
Corollar 4 Let K be a finite field. Then K ∗ is cyclic.
For the equation x d −e = 0 (which is of degree d) has at most d solutions.
Lemma 3 Let n = p α . then n has a primite root.
Proof. Let g be a primitive root (mod p). We show that there exists x, so
that h = g +px is a primitive root (mod p α ).
We know that g p−1 = 1+py for some y ∈ Z. Hence
h p−1 = (g +px) p−1 (65)
( ) p−1
= g p−1 +p(p−1)xg p−2 +p 2 x 2 g p−3 +... (66)
2
( ) p−1
= 1+py +p(p−1)xg p−2 +p 2 x 2 pg p−3 +... (67)
2
= 1+pz (68)
where z = y + (p − 1)xg p−2 (mod p). The coefficient of x is relatively
prime to p and so we can choose x so that z = 1 (mod p). We claim that for
this choice h is a primitive root. i.e. ord(h) = φ(p α ) = p α−1 (p−1). For put
d = ord h, so that d | p α−1 (p−1). Since h is a primitive root (modp), then
p−1 | d, and so d = p k (p−1), where k < α. p is odd and so
with ¯z and p relatively prime. Hence
(1+pz) pk = 1+p k+1¯z
1 = h d = h pk (p−1) = (1+pz) pk = 1+p k+1¯z(modp α ).
This implies that α = k +1, as was to be proved.
Corollar 5 m = 2p α has a primitive root.
For φ(2p α ) = φ(2)φ(p α ) = φ(p α ). Let g be a primitive root (mod p α ). One of
the two, g or g+p α , is odd and so is an element of Z ∗ m and thus a primitive
root.
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