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

Why are teams $PSL_3(\mathbb{F}_2)$ and also $PSL_2(\mathbb{F}_7)$ isomorphic?

Update. There is a group-theoretic evidence (see solution). Yet exists any kind of geometric evidence? Or some evidence making use of octonions, possibly?

2019-05-05 13:42:47
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Answers: 6

There is a sensibly geometric evidence in Section 1.4 of Noam Elkies' The Klein Quartic in number theory. A really harsh recap is as adheres to. The one-of-a-kind straightforward team $G$ of order $168$ has a $3$ - dimensional irreducible depiction $V$ which is specified over a number area $k$. By lowering this depiction modulo a prime over $2$ we get a $3$ - dimensional depiction of $G$ over $\mathbb{F}_2$ which recognizes $G$ with $\text{GL}_3(\mathbb{F}_2)$ by simpleness (and also a counting argument). By lowering this depiction modulo a prime over $7$ we get a $3$ - dimensional depiction of $G$ over $\mathbb{F}_7$ which is, as it ends up, the symmetrical square of the specifying depiction of $\text{PSL}_2(\mathbb{F}_7)$.

2019-05-08 09:52:40

At the last joint conferences in SF there was a speak about specifically this.


I mosted likely to the talk yet I do not bear in mind the specific isomorphism ; yet at least a really concrete solution exists.

2019-05-08 09:25:23

Can't leave comments yet, yet the information of there being just one straightforward team of order 168, and also why PSL (2,7) and also PSL (3,2) are order 168 and also straightforward, are defined on web pages 141 - 147 in Smith and also Tabachnikova's "Topics in Group Theory".


2019-05-08 07:44:26

The team $G=\operatorname{PSL}_2(7)$ acts upon $X=P^1(\mathbb{F}_7)$. Deal with $p\in X$, and also take into consideration the activity of the stabilizer subgroup $G_p=\{g\in G:g\cdot p=p\}$ on the set $\binom{X\setminus\{p\}}{3}$ of $3$ - component parts of $X\setminus\{p\}$. It has 3 orbits, of dimensions $7$, $7$ and also $21$ ; this can be examined by taking into consideration cross - proportions. Select among the tiny ones : one can examine that it is a three-way Steiner system $S(2,3,7)$, so it is isomorphic as a layout, to $P^2(\mathbb{F}_2)$.

Playing a little bit with this building and construction can be made use of to understand the isomorphism explicitely.

Later on. A monitoring, which aids explainwhy this functions, is that if $o\in\binom{X}{4}$ is just one of the $G$ - orbits of dimension $14$, after that the automorphism team of $o$ (that is, the set of permutations of $X$ which map components of $o$ to components of $o$) is $\operatorname{GL}_3(\mathbb{F}_2)\rtimes\mathbb{F}_2^3$, the team of affine maps of $\mathbb{F}_2^3$ (below \rtimes is intended to suggest gone across item ). Dealing with a component $p\in X$, as I did above, totals up to selecting a 'absolutely no' in $\mathbb{F}_2^3$ considered as an affine room, that is, considering it as a vector room. (This places a framework of affine $3$ - room over $\mathbb{F}_2$ on $X$, and also if we start with the various other $14$ - component orbit $o'\in\binom{X}{4}$ we get an additional framework of affine $3$ - room over $\mathbb{F}_2$ ; $\operatorname{PGL}_2(7)$ is specifically the set of permutations of $X$ which maintains both affine frameworks)

2019-05-08 07:25:52

Both are simple groups of order 168, and also each straightforward team of order $168$ is isomorphic to $PSL_2(7)$. An extensive workout, with tips. Confirm the adhering to :

Let $G$ be a straightforward team of order $168$.

It has $8$ Sylow $7$ - subgroups.

It can be understood a subgroup of $A_8$.

Classifying the things it acts upon as $\infty,0,1,\ldots,6$ one Sylow $7$ - subgroup is created by $g=(0\ 1\ 2\ 3\ 4\ 5\ 6)$.

The team $G$ is $2$ - transitive.

The normalizer of $\langle g\rangle$ is created by $g$ and also $h=(1\ 2\ 4)(3\ 6\ 5)$.

The setwise stabilizer $H$ of $\{\infty,0\}$ is created by $h$ and also an additional component $k$ which is the item of $(\infty\ 0)$ and also 3 various other disjoint transpositions.

If $H$ is cyclic, after that the Sylow $2$ - subgroup of $G$ would certainly be one-of-a-kind, bring about an opposition.

So $H$ is nonabelian and also we can take $k=(\infty\ 0)(1\ 6)(2\ 3)(4\ 5)$.

Ultimately $G$ is $PSL_2(7)$.

2019-05-08 06:41:51

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

Consider 28 - component set of 4 - tuples of factors on $\mathbb{P}^1(\mathbb{F}_7)$ with cross - proportion equivalent to 3. Recognizing 4 - tuple with its enhance (recall that there are specifically 8 factors on $\mathbb{P}^1(\mathbb{F}_7)$) one obtains 14 - component set $X$-- specifically the variety of factors + the variety of lines on $\mathbb{P}^2(\mathbb{F}_2)$. The set $X$ contains 2 orbits of $PSL_2(\mathbb F_7)$, $P$ and also $L$. Specify a tuple from $P$ and also a tuple from $L$ to be case if they converge by 2 components. Case : 1) result is without a doubt $\mathbb{P}^2(\mathbb{F}_2)$ ; 2) generated homomorphism $PSL_2(\mathbb{F}_7)\to PSL_3(\mathbb F_2)$ is an isomorphism.

Update (2 years later on). This is simply Klein document!

For any kind of 8 - component set $X$ its powerset $2^X$ is a vector room over $\mathbb F_2$. Allow $V$ be the ratio of the subspace created by part with also variety of components by $\langle X\rangle$. Currently the set $Q$ of 4 - tuples of factors of X approximately enhance can be considered as a Klein quadric (also known as $Gr(2,4)$, also known as $PGr(1,3)$) inside $\mathbb P(V)$. Permutations of $X$ act upon aircrafts on $Q$, and also these aircrafts represent factors and also aircrafts of $\mathbb P^3$. Currently also permutations map indicate factors and also aircrafts to levels and also this offers isomorphism $A_8\to PSL_4(\mathbb F_2)$.

Currently for $X=\mathbb P^1(\mathbb F_7)$ its constraint on $PSL_2(\mathbb F_7)\subset A_8$ offers isomorphism $PSL_2(\mathbb F_7)\to PSL_3(\mathbb F_2)\subset PSL_4(\mathbb F_2)$.

2019-05-08 06:35:02