I am looking at Abstract Algebra, 3rd ed., by Dummit and Foote, page 208. In classfying groups of order $168$, we first assume that there is a simple group of order $168$ and prove that
(1) $G $ has no proper subgroup of index less than $7,$
(2) $n_7=8$ so that the normalizer of a Sylow $7$-subgroup has order $21.$ In particular no element of order $2$ normalizes a Sylow $7$-subgroup.
(3) $G $ has nonelements of order $21$.
Here $n_p$ denotes the number of Sylow $p$-subgroups.
I get why everything above is true. But I am having difficulty understanding the bolded lines in the following paragraph:
By Sylow's Theorem $n_3=7$ or $28$; we next rule out the former possibility. Assume $n_3=7$, let $P \in Syl_3 (G) $ and let $T $ be a Sylow $2$-subgroup of the group $N_G (P)$ of order $24$. Each Sylow $3$-subgroup normalizes some Sylow $7$-subgroup of $G $ so $P $ normalizes a Sylow $7$-subgroup $R $ of $G $. For every $t \in T $ we also have that $P=tPt^{-1} $ normalizes $tRt^{-1} $. The subgroup $T $ acts by conjugation on the set of eight Sylow $7$-subgroups of $G $ and since no element of order $2$ in $G $ normalizes a Sylow $7$-subgroup by (2), it follows that $T $ acts transitively, i.e. every Sylow $7$-subgroup of $G $ is one of the $tRt^{-1} $. Hence $P $ normalizes every Sylow $7$-subgroups of $G $.
Again I get everything up until the bolded words. I don't get how the fact (2) is related to the transitivity of the group action, and how $P $ normalizes every Sylow $7$-subgroups of $G $.
You have a group $T$ of order $8$ acting on a set of size $8$ such that no element of order $2$ in $T$ fixes any point in the set.
So the stabilizer in $T$ of a point in the set must be trivial, and hence the orbit of $T$ on the set has size $|T| = 8$, which means that the action is transitive.
We have seen that $P$ normalizes every Sylow $7$-subgroup in this orbit, and the orbit consists of all of the Sylow $7$-subgroups, so it normalizes all of them.