Permutation modules and the simple group of order 168

Wildon's Weblog 2018-08-02

In an earlier post I decomposed the permutation module for $\mathrm{GL}_3(\mathbb{F}_2)$ acting on the $7$ non-zero lines in $\mathbb{F}_2^3$ , with coefficients in a field of characteristic $2$ . Some related ideas give the shortest proof I know of the exceptional isomorphism between $\mathrm{PSL}_2(\mathbb{F}_7)$ and $\mathrm{GL}_3(\mathbb{F}_2)$ . The motivation section below explains why what we do ‘must work’, but is not logically necessary.

Motivation

Let $h \in \mathrm{GL}_3(\mathbb{F}_2)$ have order $7$ and let

$H = \mathrm{N}_{\mathrm{GL}_3(\mathbb{F}_2)}(\langle h\rangle) \cong C_7 \rtimes C_3.$

The simple modules for $\mathrm{GL}_3(\mathbb{F}_2)$ in characteristic $2$ are the trivial module, the natural module $E$ , its dual $E^\star$ , and the reduction modulo $2$ of a module affording the $8$ -dimensional character induced from either of the two non-trivial linear characters of $H$ . (A short calculation with Mackey’s Formula and Frobenius Reciprocity shows the module is irreducible in characteristic zero; then since $168/8$ is odd, it reduces modulo $2$ to a simple projective. Alternatively one can work over $\mathbb{F}_4$ and perform the analogue of the characteristic zero construction directly.) The permutation module for $\mathrm{GL}_3(\mathbb{F}_2)$ acting on the cosets of $H$ , defined over $\mathbb{F}_2$ , is projective. Since it has the trivial module in its socle and top, the only possible Loewy structure is

$\begin{matrix} \mathbb{F}_2 \\ E \oplus E^\star \\ \mathbb{F}_2 \end{matrix}.$

In particular, it contains a $4$ -dimensional $\mathbb{F}_2\mathrm{GL}_3(\mathbb{F}_2)$ -submodule $U$ having $E$ as a top composition factor. Below we construct $U$ as a module for $\mathrm{PSL}_2(\mathbb{F}_7)$ and hence obtain the exceptional isomorphism in the form $\mathrm{PSL}_2(\mathbb{F}_7) \cong \mathrm{GL}(E)$ .

Construction

Let $G = \mathrm{PSL}_2(\mathbb{F}_7)$ regarded as a permutation group acting on the $8$ non-zero lines in $\mathbb{F}_7^2$ . For $j \in \{0,1,\ldots, 6\}$ , let $j$ denote the line through $(j,1)$ , and let $\infty$ denote the line through $(1,0)$ . Let $M = \langle e_j : j \in \{0,1,\ldots, 6\} \cup \{\infty\} \rangle_{\mathbb{F}_2}$ be the corresponding $\mathbb{F}_2G$ -permutation module.

Let $h = (0, 1, \dots, 6) \in G$ , corresponding to the Möbius map $x \mapsto x + 1$ and to the matrix $\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$ in $\mathrm{SL}_2(\mathbb{F}_7)$ . The simple modules for $\mathbb{F}_2\langle h \rangle$ correspond to the factors of

$x^7 + 1 = (x+1)(x^3+x+1)(x^3+x^2+1).$

Since $\{1,2,4\} \le \mathbb{F}_7^\times$ is permuted by squaring, an idempotent killing the simple modules with minimal polynomial $x^3 + x + 1$ is $h^4 + h^2 + h$ . Therefore the module $U$ we require has codimension $1$ in the $8-3 = 5$ -dimensional module

$M(h + h^2 + h^4) = \langle e_\infty \rangle_{\mathbb{F}_2} \oplus \left\langle \begin{matrix} e_0 + e_1 + e_3 \\ e_1 + e_2 + e_4 \\ e_2 + e_3 + e_5 \\ e_3 + e_4 + e_6 \end{matrix} \right\rangle_{\mathbb{F}_2}$

(For example, applying $h$ to the final basis vector gives $e_4 + e_5 + e_0$ which is the sum of the first three basis vectors.) Since $e_\infty$ must appear, it can only be that $U$ is generated by $u_0$ where

$u_0 = e_\infty + e_0 + e_1 + e_3$

and so $U$ has basis $u_0, u_1, u_2, u_3$ where $u_i = u_0 h^i$ for each $i$ .

Let $t = (1,2,4)(3,5,6) \in \mathrm{N}_G(\langle h \rangle)$ and let

$j = (0,\infty)(1,3)(2,5)(4,6) \in G,$

chosen so that $h^t = h^2$ and $t^j = t^2$ . The corresponding Möbius maps are $x \mapsto 2x$ and $x \mapsto 3/x$ and one choice of corresponding matrices in $\mathrm{SL}_2(\mathbb{F}_7)$ is $\left( \begin{matrix} 4 & 0 \\ 0 & 2 \end{matrix} \right)$ and $\left(\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}\right)$ . Computation shows that in their (right) action on $U$ ,

$h \mapsto \left( \begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}\right), t \mapsto \left( \begin{matrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 \end{matrix}\right), j \mapsto \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{matrix}\right).$

The unique trivial submodule of $U$ is spanned by

$u_0 + u_2 + u_3 = e_0 + e_1 + \cdots + e_6 + e_\infty.$

Quotienting out, we obtain the $3$ -dimensional representation of $G$ where

$h \mapsto \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{matrix} \right), \; t \mapsto \left( \begin{matrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}\right), \; j \mapsto \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right)$

Since $\mathrm{PSL}_2(\mathbb{F}_7)$ has cyclic Sylow $3$ -subgroups and cyclic Sylow $7$ -subgroups, and a unique conjugacy class of elements of order $2$ (which contains $j$ ), one can see just from the matrices above that the representation is faithful. Comparing orders, we get $\mathrm{PSL}_2(\mathbb{F}_7) \cong \mathrm{GL}_3(\mathbb{F}_2)$ .