Why $PSL_3(\mathbb F_2)\cong PSL_2(\mathbb F_7)$? Why are groups $PSL_3(\mathbb{F}_2)$ and $PSL_2(\mathbb{F}_7)$ isomorphic?
Update. There is a group-theoretic proof (see answer). But is there any geometric proof? Or some proof using octonions, maybe?

Both are simple groups of order 168, and each simple group of order $168$
is isomorphic to $PSL_2(7)$. An extended exercise, with hints. Prove the following:
Let $G$ be a simple group of order $168$.
It has $8$ Sylow $7$-subgroups.
It can be identified with a subgroup of $A_8$.
Labelling the objects it acts on as $\infty,0,1,\ldots,6$
one Sylow $7$-subgroup is generated by $g=(0\ 1\ 2\ 3\ 4\ 5\ 6)$.
The group $G$ is $2$-transitive.
The normalizer of $\langle g\rangle$ is generated by $g$ and
$h=(1\ 2\ 4)(3\ 6\ 5)$.
The setwise stabilizer $H$ of $\{\infty,0\}$ is generated by $h$
and another element $k$ which is the product of $(\infty\ 0)$ and
three other disjoint transpositions.
If $H$ is cyclic, then the Sylow $2$-subgroup of $G$ would be unique,
leading to a contradiction.
So $H$ is nonabelian and we can take $k=(\infty\ 0)(1\ 6)(2\ 3)(4\ 5)$.
Finally $G$ is $PSL_2(7)$.
