A Family of Groups Containing a Nonabelian Fully Ramified Section
✍ Scribed by Mark L. Lewis
- Publisher
- Elsevier Science
- Year
- 1996
- Tongue
- English
- Weight
- 224 KB
- Volume
- 186
- Category
- Article
- ISSN
- 0021-8693
No coin nor oath required. For personal study only.
✦ Synopsis
Thus, we observe that nrt is a divisor of p a y 1. This implies p b y 1 divides p a y 1, and so, b divides a. Since a divides b, we conclude that a s b which implies that b divides d.
Write k for the Galois field having order p b , and observe that E : k : F. We define N to be the subgroup of M identified with k. We will now e e prove that N is invariant under the action of C. Consider an element e Ž t . n r t n t cgC, and observe that c s c s 1. This implies that c has order dividing nrt, and c t lies in k. Since t is a divisor of e, there exists an e Ž t . a integer a so that e s at, and we note that c s c g E. Given an element y g N , we have y и c s yc e . Because yc e lies in k, it follows that e y и c lies in N . Therefore, N is invariant under the action of C. Since e e k ) 1 and M is irreducible under the action of C, we conclude that e N s M . This implies that b s d as desired. e e Ž . Conversely, suppose that ord p s d. We note that tn, which n r t e Ä e < implies that there is some element c g C such that c / 1. Let B s b 4 b g C , and observe that B is a cyclic group having order nrt. Given an element x g M such that x / 0, we see that x и c s xc e / x. Thus, every e orbit under the action of C in M not containing 0 has cardinality nrt. If e W is a C-invariant subgroup of M , then W is a union of C-orbits. Since W e