Assume that $G$ is a finite simple group and that $n_7(G)=8$ (where $n_7(G)$ is the number of Sylow $7$-subgroups). Prove that $G$ does not contain elements of order 5.
I state that since $G$ is finite, then: $|G|=7\cdot p_1^{n_1}\cdot\dots\cdot p_k^{n_k}$, where $p_1,\dots ,p_k$ are primes. For the sylow theorems I know that $7|n_7(G)-1$ and $n_7(G)|p_1^{n_1}\cdot\dots\cdot p_k^{n_k}$. Now suppose by contraddiction that $p_i=5$ for a certain $i\in\{1,\dots, k\}$. So since $G$ is simple, then $n_5(G)\neq 1$. At this point I don't know how to conclude the proof... Any ideas or hints?
Let $P \in {\rm Syl}_7(G)$, then $|G:N_G(P)|=8$ and $G$ embeds isomorphically ($G$ is simple!) in $A_8$. Also, since $7$ is the highest power of $7$ dividing $|A_8|=20160$, we have $P=\langle ( 1 2 3 4 5 6 7) \rangle$ or a conjugate.
Now it is not hard to see that the centralizer in $S_8$ of $P$ is $P$, and so $N_G(P)/P$ must be isomorphic to a subgroup of ${\rm Aut}(P)$, which is cyclic of order $6$. So $|N_G(P)|$ is not divisible by $5$, and hence neither is $G$.
In fact there is a unique (up to isomorphism) simple group that satisfies these conditions, and that is ${\rm PSL}(2,7)$, which is defined naturally as a subgroup of $A_8$. In that example $|N_G(P)|=21$ and $|G|=168$.