[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 ].

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- November 18th 2008, 08:59 PMaliceinwonderlandSymmetric 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 ]. - November 19th 2008, 06:40 AMThePerfectHacker
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, .

Finally, .