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Dummit and Foote, p. 204

They suppose that $G$ is simple with a subgroup of index $k = p$ or $p+1$ (for a prime $p$), and embed $G$ into $S_k$ by the action on the cosets of the subgroup. Then they say

"Since now Sylow $p$-subgroups of $S_k$ are precisely the groups generated by a $p$-cycle, and distinct Sylow $p$-subgroups intersect in the identity"

Am I correct in assuming that these statements follow because the Sylow $p$-subgroups of $S_k$ must be cyclic of order $p$? They calculate the number of Sylow $p$-subgroups of $S_k$ by counting (number of $p$-cycles)/(number of $p$-cycles in a Sylow $p$-subgroup). They calculate the number of $p$-cycles in a Sylow $p$-subgroup to be $p(p-1)$, which I don't see.

  • 2
    You left out the important fact that $k=p$ or $p+1$.2011-05-22
  • 0
    What do you mean, "the number of $p$-cycles in a Sylow $p$-subgroup"?2011-05-22

2 Answers 2