No simple groups of order 9555: proof While reading through crazyproject, I came across the following proof that there were no simple groups of order $9555$. However, I do not understand the step that says:
"Moreover, since 7 does not divide 12 and 13 does not divide 48, $P_7P_{13}$ is abelian."
I don't quite follow this reasoning. If someone could explain this to me I'd be very grateful.

The idea is to show that elements of $P_7$ and $P_{13}$ commute. The proof seems to be using that $|\mathrm{Aut}(P_7)| = 48$ but this is wrong. Because $|P_7| = 49$ we know that $P_7$ isomorphic to one of $\mathbb{Z}_{49}$ or $\mathbb{Z}_{7} \times \mathbb{Z}_{7}$ and so $|\mathrm{Aut}(P_7)|$ is either $42$ or $48 \cdot 42$.
The idea of the proof can still be used. We know that $P_{13} \leq N_G(P_7)$ and $13$ does not divide $|\mathrm{Aut}(P_7)|$ so we must have $P_{13} \leq C_G(P_7)$. Therefore $P_7P_{13}$ is abelian, since both $P_7$ and $P_{13}$ are.
We could change the proof to avoid thinking about $P_7P_{13}$ as follows. Since $P_{13} \leq C_G(P_7)$ we also have that $P_7 \leq C_G(P_{13}) \leq N_G(P_{13})$ and then pick up the proof in the last sentence.
