Sylow and 168

The third Sylow theorem says that the number $n_7$ of Sylow $7$ -subgroups is $\equiv 1 \pmod 7$ and divides $168/7 = 24.$ So $n_7=1$ or $8.$ If $n_7=1$ then the unique Sylow $7$ -subgroup is a normal subgroup , so $G$ is not simple.

So $n_7=8.$ There are $8$ distinct subgroups of order $7.$ The order of every non-identity element in each subgroup is $7$ (it divides $7$ by Lagrange's theorem ), so there are six non-identity elements in each subgroup with order $7.$ Also, any two Sylow $7$ -subgroups have trivial intersection (their intersection has order dividing $7$ by Lagrange's theorem, and if it equals $7$ then they are the same subgroup). So these six elements are distinct. There are thus a total of $6 \cdot 8 = \fbox{48}$ elements of order $7$ in these subgroups.

Finally, any element $g$ of order $7$ generates a cyclic subgroup of order $7,$ namely $\{1,g,g^2,\ldots,g^6\}.$ This is a Sylow $7$ -subgroup. So $g$ is contained in one of these Sylow $7$ -subgroups, so these $48$ elements are all the elements of order $7.$

( Aside: It can be shown that $G \cong GL_3({\mathbb F}_2),$ the group of invertible $3\times 3$ matrices with entries in the field with two elements. For extra extra credit, try to write down the $48$ elements of order $7$ in $G$ explicitly!)