Dietzfelbinger M. Primality testing in polynomial time ... - tiera.ru

4.2 Cyclic Groups 65
(c) Since m is a divisor of jzd for 0 ≤ j < d,wehave(a jz ) d = e for all
elements a jz ∈ H. Conversely,ifb d = e, forb = a i ,thena id = e, and hence
m is a divisor of id = im/z. Thisimpliesthati/z is an integer, and hence
that a i ∈ H, by(b). ⊓⊔
We consider a converse of part (c) of Lemma 4.2.11.
Lemma 4.2.12. Assume G = 〈a〉 is a cyclic group of size m and s ≥ 0 is
arbitrary. Then
Hs = {a ∈ G | a s = e}
is a subgroup of G with gcd(m, s) elements.
(In particular, every divisor s of m gives rise to the subgroup Hs = {a ∈ G |
a s = e} of size s.)
Proof. It is a simple consequence of the subgroup criterion Lemma 4.1.6 that
Hs is indeed a subgroup of G. Which elements a i ,0≤ i