# Sylow not implies local divisibility-closed

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Sylow subgroup) need not satisfy the second subgroup property (i.e., local divisibility-closed subgroup)
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## 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

Further information: symmetric group:S4, D8 in S4

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 $H$.