 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/2/a/ii/solution

∣ G ∣ = p^{a} m

, where

p

p ∤ m

p^{a}

, the largest power of

p

dividing

∣ G ∣

The three Sylow theorems assert:

Existence: a Sylow p-subgroup exists, and every subgroup of prime power order a power of $p$ is contained in one.
Conjugacy: any two Sylow p-subgroups are conjugate in $G$ .
Counting: their number $n_{p}$ satisfies $n_{p} ∣ m$ and $n_{p} \equiv 1 (mod p)$ .

In particular, a unique Sylow p-subgroup is a normal subgroup, since all its conjugates must equal it.

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