Discussion
Let G be a group of order \(n\), \(p\) a prime such that \(p \mid n\). Then :
Answer: C
The correct statement is that the number of Sylow \(p\)-subgroups equals the index of N(P) in G.
The Sylow \(p\)-subgroups are all conjugate (second Sylow theorem). G acts on them by conjugation transitively, and the stabiliser of P is its normaliser N(P). By the orbit-stabiliser theorem, the number of Sylow \(p\)-subgroups is \(|G| / |N(P)| = [G : N(P)]\).
The others fail in general. In \(S_{3}\) the Sylow 2-subgroups are not normal (A). A Sylow 2-subgroup of \(S_{4}\) is the dihedral group of order 8, which is not abelian (B). And the index of P itself is usually larger than the number of Sylow subgroups: in \(S_{3}\), \([G : P] = 3\) for \(p = 2\), which happens to match, but for \(p = 3\), \([G : P] = 2\) while there is only one Sylow 3-subgroup (D).
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