TITLE: Why $PSL_3(\mathbb F_2)\cong PSL_2(\mathbb F_7)$?
QUESTION [19 upvotes]: 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?

REPLY [5 votes]: (A sketch of a proof from http://www.math.vt.edu/people/brown/doc/PSL(2,7)_GL(3,2).pdf by E.Brown and N.Loehr.)
$PSL_3(\mathbb F_2)$ is the group of automorphisms of $\mathbb F_8$ as a vector space over $\mathbb F_2$. Fix a generator $x$ in $\mathbb F_8^{\times}$. It defines a map $\mathbb P^1(\mathbb F_7)\to\mathbb F_8$: $k\mapsto x^k$ (we define $x^\infty:=0$).
Now, for $f\in PSL_2(\mathbb F_7)$ the map $x^k\mapsto x^{f(k)}$ is, in general, not $\mathbb F_2$-linear, but the map $x^k\mapsto x^{f(k)}+x^{f(\infty)}$ is$^1$. And the map $T:f\mapsto(x^k\mapsto x^{f(k)}+x^{f(\infty)})$ gives a desired isomorphism $PSL_2(\mathbb F_7)\to PSL_3(\mathbb F_2)$.
$^1$ There is no conceptual explanation in the paper, but a check for generators of $PSL_2(\mathbb F_7)$ isn't hard --- and a check that $T$ is a homomorphism finishes the proof.