Sylow satisfies intermediate subgroup condition

Statement
A Sylow subgroup of a finite group is also a Sylow subgroup in any intermediate subgroup.

Sylow subgroup
A subgroup $$P$$ of a group $$G$$ is a Sylow subgroup if $$P$$ is a group of prime power order and the order of $$P$$ is relatively prime to the index of $$P$$.

$$P$$ is termed a $$p$$-Sylow subgroup if its order is a power of the prime $$p$$ and its index is not a multiple of $$p$$.

Related facts

 * Hall satisfies intermediate subgroup condition
 * Sylow does not satisfy transfer condition

Facts used

 * 1) uses::Index is multiplicative

Proof
Given: A finite group $$G$$, a $$p$$-Sylow subgroup $$H$$ of $$G$$, a subgroup $$K$$ of $$G$$ containing $$H$$.

To prove: $$H$$ is a Sylow subgroup (in fact, a $$p$$-Sylow subgroup) inside $$K$$.

Proof: Clearly, $$H$$ continues to be a $$p$$-subgroup inside $$K$$, because it is a $$p$$-group. Thus, it suffices to show that the index $$[K:H]$$ is relatively prime to $$p$$. For this, observe that by fact (1):

$$[G:H] = [G:K][K:H]$$

Thus, $$[K:H]$$ is a divisor of $$[G:H]$$. Since $$[G:H]$$ is relatively prime to $$p$$, so is $$[K:H]$$.