Number of elements of prime order for multiplicity-free prime divisor equals Sylow number times one less than the prime

Statement
Suppose $$p$$ is a prime number, $$m$$ is a natural number relatively prime to $$p$$, and $$G$$ is a group of order $$pm$$. Let $$n_p$$ be the $$p$$-fact about::Sylow number of $$G$$. Then, the number of elements of order $$p$$ in $$G$$ is:

$$n_p(p-1)$$.

Related facts

 * Order is product of three distinct primes implies normal Sylow subgroup
 * Order is product of Mersenne prime and one more implies normal Sylow subgroup

Related survey articles

 * Combinatorial enumeration technique: This technique is used to prove that groups of certain orders cannot be simple, because the possible combinations of Sylow numbers give rise to too many elements. In other cases, it may help rule out certain combinations of Sylow numbers.