A Hall divisor of an integer n is a divisor d of n such that d and n/d are coprime. The number of Hall divisor of n is $2^k$, where k is the number of distinct prime divisors of n. Now a Hall subgroup of G is a subgroup whose order is a Hall divisor of the order of G. Also we know:
Any Sylow subgroup of a group is a Hall subgroup.
If $G = A_{5}$, the only simple group of order 60, then 15 and 20 are Hall divisors of the order of G, but G has no subgroups of these orders.
Let $n_{p}$ be the number of Sylow p-subgroups of G. Then the following hold:
$n_{p}$ divides m, which is the index of the Sylow p-subgroup in G. $n_{p} ≡ 1$ mod p. $n_{p} = |G : N_{G}(P)|$, where P is any Sylow p-subgroup of G and $N_{G}$ denotes the normalizer.
Now what do we can say about the number of the Hall subgroups in a group?