Prove there is no element of order 6 in a simple group of order 168 Let $G$ be a simple group of order 168. Let $n_p$ be the number of Sylow $p$ subgroups in $G$.
I have already shown: $n_7 = 8$, $n_3 = 28$, $n_2 \in \left\{7, 21 \right\}$
Need to show: $n_2 = 21$ (showing there is no element of order 6 In $G$ will suffice)
Attempt so far: If $P$ is a Sylow-2 subgroup of $G$, $|P| = 8$.
 Assume for contradiction that $n_2 = 7$. Then the normalizer $N(P)$ has order 24. Let $k_p$ be the number of Sylow-$p$ subgroups in $N(P)$. Then $k_3 \in \left\{1,4 \right\}$ and $k_2 \in \left\{1,3 \right\}$. Then I showed $k_3 = 4, k_2 = 1$. Counting argument shows there is an element of order 6 in $N(P)$, and thus in $G$ too.
I don't know how to proceed from here.
I am told that there cannot be an element of order 6 in $G$, but I don't know how to prove it, if someone could help me prove this I would very much appreciate it.
Can someone help me?

If there is an element of order 6, then that centralizes the Sylow $3$-subgroup $P_3$ generated by its square. You have already shown that $|N(P_3)|=168/n_3=6$. Therefore the normalizer of any Sylow $3$-subgroup would have to be cyclic of order 6, and an element of order 6 belongs to exactly one such normalizer. Thus your group would have $56=2\cdot n_3$ elements of order $3$, $56=2\cdot n_3$ elements of order $6$, $48=6\cdot n_7$ elements of order $7$, and therefore only eight other elements. Those eight would have to form a Sylow $2$-subgroup, and that would be unique, so...
