Show that the order of $G$ cannot be divisible by $7$. Show that the order of a proper primitive group of degree $19$ cannot be divisible by $7$.
It means as following-
This means that a group $G$ which acts on a set $\Omega$ of cardinality $19$ (this is what degree means) and action is transitive and primitive  , and proper means that when we see its permutation representation, $\phi :G \to S_{19}$ then $\phi(G) \neq A_{19}$ or $S_{19}$.
For the solution of this, If i let $7 ||G|$, then there will be a sylow $7$ subgroup , say, $P$, now if I can show, as per hint in the book, that $C_G(P)$ contains a $5$- cycle, then obviously, $P=1$ and I am done, but how to show $C_G(P)$ contains a $5$- cycle?
 A: I can't make any sense at all of your comment "if $C_G(P)$ contains a $5$-cycle then obviously $P=1$". You are trying to prove that $A_n \le G$.
There is an old result of Jordan that says that, if a primitive permutation group $G \le S_n$ contains a $p$-cycle with $p<n-2$, then $A_n \le G$, and I am guessing that you are expected to use that.
So, if $G \le S_{19}$ is primitive and $G$ contains a $7$-cycle, then $A_{19} \le G$ and we are done, so assume not. Then, assuming that $7$ divides $|G|$, all elements of order $7$ in $G$ consist of $2$ disjoint $7$-cycles, and a Sylow $7$-subgroup $P$ of $G$ has order $7$ and fixes exactly $5$ points.
There is a general result that the normalizer of a Sylow $p$-subgroup $P$ in a transitive permutation group acts transitively on the fixed points of $P$. It's proof is an easy application of Sylow's Theorem.
Using that, $N_G(P)$ acts transitively on the $5$ fixed points of $P$, so $5$ divides $|N_G(P)|$ and there exists $g \in N_G(P)$ of order $5$. Now since $5$ does not divide $|{\rm Aut}(P)|=6$, $g \in C_G(P)$. Clearly $g$ must fix both of the $2$ orbits of $P$ of size $7$, and hence it must fix all points of both orbits. So $g$ is a $5$-cycle and $A_n \le G$ by Jordan's Theorem.
