Group $G$ such that $|G| = 385$

Question

I have a question about Dietrich Burde's answer in the following link:

Group of an order 385

In this answer, there is a step where the user concludes that from $|Aut(P_7)| = 6 = 2 \cdot 3$ we have $|G/C_G(P_7)| = 1$. I understand the rest of their proof, but I'm not understanding why this step follows. Since $G/C_G(P_7)$ is a subgroup of $Aut(P_7)$, couldn't $|G/C_G(P_7)|$ also be of order $2$, $3$, or $6$ ? How can we rule out these possibilities?

Thanks!

Answer 1

$G$ is of order $5\cdot 7\cdot 11$ and $P_7$ is a normal subgroup of order $7$. And since $P_7$ is central then $P_7\subseteq C_G(P_7)$ and thus $|G/C_G(P_7)|$ divides $5\cdot 11$. Therefore $2,3$ and $6$ are not possible.