elements whose orders divide m is the subgroup of G which is the direct product of two cyclic
(iii) p” divides the order of M n Z(P) and the exponent of P/M. We shall call (P, f, ,u) a
of two cyclic Galois extensions need not be a cyclic extension. For instance
Moreover, m divides n, and Gal(E/F) is cyclic and is generated by the automorphism ...
N = LF, where F is a finite abelian extension of K, totally split at all
injective and H is cyclic of order d dividing p−1. It is easy to
Lemma : Suppose FELEK with Klf and LF. Galois.
Let O be an order in a division algebra E = O ⊗ Q of dimension nm2 over Q, where n
is a cyclic subgroup of G with order |H| = m. Thus m divides |G| by the theorem. 2. Since |G| is prime, we may take a = 1 in G, and since the order of a has.
In a finite cyclic group, the order of an element divides the order of the group.