# Characteristic subgroup of Sylow subgroup is weakly closed iff it is normal in every Sylow subgroup containing it

## Statement

Suppose $G$ is a finite group, $p$ is a prime number, $P$ is a $p$-Sylow subgroup (?), and $K$ is a Characteristic subgroup (?) of $P$. Then, $K$ is a Weakly closed subgroup (?) in $P$ (relative to $G$) if and only if $K$ is a normal subgroup in every $p$-Sylow subgroup containing it.

## Facts used

1. Weakly closed implies conjugation-invariantly relatively normal in finite group
2. Sylow implies order-conjugate: For a prime $p$, any two $p$-Sylow subgroups are conjugate.
3. Sylow implies WNSCDIN
4. WNSCDIN implies every normalizer-relatively normal conjugation-invariantly relatively normal subgroup is weakly closed

## Proof

Given: A finite group $G$, a prime $p$, a $p$-Sylow subgroup $P$ of $G$. A characteristic subgroup $K$ of $P$.

### Weakly closed implies normal in every Sylow subgroup containing it

To prove: Given that $K$ is weakly closed in $P$, and $Q$ is a $p$-Sylow subgroup of $G$ containing $K$, $K$ is normal in $Q$.

Proof (quick version): This follows from facts (1) and (2). By fact (2), $P$ and $Q$ are conjugate, and by fact (1), $K$ is normal in $Q$. (Note that this part does not use the assumption that $K$ is a characteristic subgroup of $P$).

To prove: Given that $K$ is normal in every $p$-Sylow subgroup of $G$ containing it, $K$ is weakly closed in $P$. In other words, if $g \in G$ is such that $gKg^{-1} \le P$, then $gKg^{-1} = K$.
Proof (quick version): The proof follows from facts (3) and (4), and the observation (again stemming from fact (2)) that the conjugates of a particular $p$-Sylow subgroup are precisely all the $p$-Sylow subgroups.