Symmetric group of P
[Dummit p132, Q34]
Prove that if p is a prime and P is a subgroup of of order p, then .
[Argue that every conjugate of P contains exactly p-1 p-cycles and use the formula for the number of p-cycles to compute the index of in ].
Let be a subgroup of of order . It means that is generated by a -cycle. Since is prime and it means that is a Sylow -subgroup. Consequently, if is any other subgroup of order then has to be congugate to i.e. . Let be all the Sylow -subgroups. By above each one is conjugate to . However, if is a -cycle then is a -cycle. Furthermore, any -cycle has order and must be contained in one of the Sylow -subgroups. Therefore, contain all the -cycles among them. Since these are distinct subgroups of order (prime) it means for . Each contains -cycles (because we are overlooking the identity). Thus, in total we have -cycles where , as above, is the number of conjugate subgroups to . But we also know that is the index of the normalizer of in , in less fancy language, . Therefore, . It remains to compute . By above we said that there are -cycles, we also know by counting that there are cycles. Thus, .
Originally Posted by aliceinwonderland