Let p be a prime number and write the order of G as pn s (with p not dividing s). We define a Sylow p-subgroup of G to be a subgroup of G which has order pn. Then:
1. If H is a subgroup of G with order pm for some m, then H is contained in some Sylow p-subgroup of G.
2. All Sylow p-subgroups of G are conjugate to each other, i.e. if H1 and H2 are Sylow p-subgroups of G, then there exists an element g in G with g-1H1g = H2.
3. The number np of Sylow p-subgroups of G is equal to [G:NG(H)], where H is any Sylow p-subgroup of G and NG(H) is the normalizer? of H in G. Furthermore, np divides s, and is congruent to 1 (mod p).
This last statement implies that G contains at least one subgroup of order pn.