# 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?

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There have been, I think, multiple proofs in the AMM, at least one of which is geometric. – Qiaochu Yuan Aug 2 '10 at 17:02
The American Mathematical Monthly: maa.org/pubs/monthly.html . But it is not easy to search and your institution might not have a subscription. – Qiaochu Yuan Aug 2 '10 at 17:50
@Qiaochu: indeed. One of the greatest pities of the Age of the Internet as far as math is involved is that searchig the AMM is more or less useless :( – Mariano Suárez-Alvarez Apr 20 '11 at 17:43

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)

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It's more or less the same thing as math.stackexchange.com/questions/1401/why-psl-3f-2psl-2f-7/… isn't it? – Grigory M Aug 4 '10 at 6:48
Yup. It all comes down to the fact that the action of $G$ on $\binom{X}{k}$ is transitive for all $k\in\{0,\dots,8\}$ except for $k=4$, in which we get three orbits of length $14$, $14$ and $42$. I like my construction better, for some reason :) – Mariano Suárez-Alvarez Aug 4 '10 at 6:56
aha, it's becoming less mysterious now, thank you – Grigory M Aug 4 '10 at 7:12
Some details seem to be wrong (in your last comment, at least). There are only two orbits of PSL on $\binom X4$: one with cross-ratio 3 aka 5 (of size 28), and one with cross-ratio 2 aka 4 aka 6 (of size 42). – Grigory M Nov 11 '12 at 21:12
@GrigoryM, GAP tells me that: OrbitLengths(PSL(2,7), Combinations([1..8], 4), OnSets) is [ 14, 42, 14 ]. – Mariano Suárez-Alvarez Nov 12 '12 at 5:29

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 $PSL_2(7)$.

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Thanks! One hand, it proves even more than I asked for, but one the other hand, I hope for some more... geometric, maybe, proof. – Grigory M Aug 2 '10 at 14:42
For reference: as Steeve D points out, details are spelled out on pages 141-147 in Smith and Tabachnikova's "Topics in Group Theory". – Grigory M Aug 4 '10 at 7:16

There is a reasonably geometric proof in Section 1.4 of Noam Elkies' The Klein Quartic in number theory. A very rough summary is as follows. The unique simple group $G$ of order $168$ has a $3$-dimensional irreducible representation $V$ which is defined over a number field $k$. By reducing this representation modulo a prime over $2$ we obtain a $3$-dimensional representation of $G$ over $\mathbb{F}_2$ which identifies $G$ with $\text{GL}_3(\mathbb{F}_2)$ by simplicity (and a counting argument). By reducing this representation modulo a prime over $7$ we obtain a $3$-dimensional representation of $G$ over $\mathbb{F}_7$ which is, as it turns out, the symmetric square of the defining representation of $\text{PSL}_2(\mathbb{F}_7)$.

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the paper looks interesting, thanks – Grigory M Aug 4 '10 at 6:49

(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.

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At the last joint meetings in SF there was a talk about exactly this.

http://www.ams.org/amsmtgs/2124_abstracts/1056-z1-379.pdf

I went to the talk but I don't remember the explicit isomorphism; but at the very least a very concrete answer exists.

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At present I can only post comments on my own questions or answers so that's why I didn't just post this as a comment. The link includes an email of one of the speakers so the actual isomorphism could probably be tracked down if anyone is interested. – solbap Aug 3 '10 at 0:28

Can't leave comments yet, but the details of there being only one simple group of order 168, and why PSL(2,7) and PSL(3,2) are order 168 and simple, are spelled out on pages 141-147 in Smith and Tabachnikova's "Topics in Group Theory".

Steve

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You might also be interested in the article by Brown and Loehr, published in the AMM 116(8), 2009, pgs. 727--732. – user641 Aug 4 '10 at 5:09

(Idea of a proof from V.Dotsenko's paper in http://www.mccme.ru/free-books/matprosd.html (in Russian))

Consider 28-element set of 4-tuples of points on $\mathbb{P}^1(\mathbb{F}_7)$ with cross-ratio equal to 3. Identifying 4-tuple with its complement (recall that there are exactly 8 points on $\mathbb{P}^1(\mathbb{F}_7)$) one gets 14-element set $X$ — exactly the number of points + the number of lines on $\mathbb{P}^2(\mathbb{F}_2)$. The set $X$ consists of 2 orbits of $PSL_2(\mathbb F_7)$, $P$ and $L$. Define a tuple from $P$ and a tuple from $L$ to be incident if they intersect by 2 elements. Claim: 1) result is indeed $\mathbb{P}^2(\mathbb{F}_2)$; 2) induced homomorphism $PSL_2(\mathbb{F}_7)\to PSL_3(\mathbb F_2)$ is an isomorphism.

Update (2 years later). This is just Klein correspondence!

For any 8-element set $X$ its powerset $2^X$ is a vector space over $\mathbb F_2$. Let $V$ be the quotient of the subspace generated by subset with even number of elements by $\langle X\rangle$. Now the set $Q$ of 4-tuples of points of X up to complement can be viewed as a Klein quadric (aka $Gr(2,4)$, aka $PGr(1,3)$) inside $\mathbb P(V)$. Permutations of $X$ act on planes on $Q$, and these planes correspond to points and planes of $\mathbb P^3$. Now even permutations map points to points and planes to plains and this gives isomorphism $A_8\to PSL_4(\mathbb F_2)$.

Now for $X=\mathbb P^1(\mathbb F_7)$ its restriction on $PSL_2(\mathbb F_7)\subset A_8$ gives isomorphism $PSL_2(\mathbb F_7)\to PSL_3(\mathbb F_2)\subset PSL_4(\mathbb F_2)$.

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