Sylow not implies local divisibility-closed

Statement
It is possible to have a finite group $$G$$, a prime number $$p$$, and a $$p$$-Sylow subgroup $$H$$ of $$G$$ such that $$H$$ is not a local divisibility-closed subgroup of $$G$$. In particular, we can set our example so that there are elements of $$H$$ that have $$p^{th}$$ roots in $$G$$ but such that none of the $$p^{th}$$ roots is in $$H$$.

Proof
Consider the following:


 * $$G$$ is symmetric group:S4, acting on the set $$\{1,2,3,4,\}$$.
 * $$p = 2$$.
 * $$H$$ is the subgroup D8 in S4, i.e., the 2-Sylow subgroup. Explicitly, it is the subset:

$$\! H = \{, (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1,2)(3,4), (1,4)(2,3), (1,3), (2,4) \}$$

The element $$(1,4)(2,3) \in H$$ has square roots $$(1,2,4,3)$$ and $$(1,3,4,2)$$ in $$G$$, but neither of these square roots is in $$H$$.