Why $PSL_3(\mathbb F_2)\cong PSL_2(\mathbb F_7)$?

Solution 1:

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 $PSL_2(7)$.

Solution 2:

The group $G=\operatorname{PSL}_2(7)$ acts on $X=P^1(\mathbb{F}_7)$. Fix $p\in X$, and consider the action of the stabilizer subgroup $G_p=\{g\in G:g\cdot p=p\}$ on the set $\binom{X\setminus\{p\}}{3}$ of $3$-element subsets of $X\setminus\{p\}$. It has three orbits, of sizes $7$, $7$ and $21$; this can be checked by considering cross-ratios. Pick one of the small ones: one can check that it is a triple Steiner system $S(2,3,7)$, so it is isomorphic as a design, to $P^2(\mathbb{F}_2)$.

Playing a bit with this construction can be used to realize the isomorphism explicitely.

Later. An observation, which helps explainwhy this works, is that if $o\in\binom{X}{4}$ is one of the $G$-orbits of size $14$, then the automorphism group of $o$ (that is, the set of permutations of $X$ which map elements of $o$ to elements of $o$) is $\operatorname{GL}_3(\mathbb{F}_2)\rtimes\mathbb{F}_2^3$, the group of affine maps of $\mathbb{F}_2^3$ (here \rtimes is supposed to mean crossed product). Fixing an element $p\in X$, as I did above, amounts to picking a 'zero' in $\mathbb{F}_2^3$ viewed as an affine space, that is, looking at it as a vector space. (This puts a structure of affine $3$-space over $\mathbb{F}_2$ on $X$, and if we start with the other $14$-element orbit $o'\in\binom{X}{4}$ we get another structure of affine $3$-space over $\mathbb{F}_2$; $\operatorname{PGL}_2(7)$ is precisely the set of permutations of $X$ which preserves both affine structures)