Permutably complemented satisfies intermediate subgroup condition

Statement
Suppose $$H$$ is a permutably complemented subgroup of $$G$$: in other words, there exists a subgroup $$K \le G$$ such that $$H \cap K$$ is trivial and $$HK = G$$. Then, if $$H \le L \le G$$, $$H$$ is a permutably complemented subgroup of $$L$$. In fact, the subgroup $$K \cap L$$ is a permutable complement to $$H$$ in $$L$$.

Facts used

 * 1) Modular property of groups: This states that if $$H, K, L \le G$$, and $$H \le L$$, we have:

$$H(K \cap L) = HK \cap L$$.

Proof
Given: $$H, K \le G$$, $$HK = G$$ and $$H \cap K$$ is trivial. $$H \le L \le G$$.

To prove: $$H \cap (K \cap L)$$ is trivial and $$H(K \cap L) = L$$.

Proof: Since $$H \cap K$$ is trivial, $$H \cap (K \cap L)$$ is also trivial. As for the second part, observe that by fact (1):

$$H(K \cap L) = HK \cap L = G \cap L = L$$

as required.