# Simple Groups of Order 168

Two large families of finite simple groups are the alternating groups $A_n$, and the projective special linear groups $\operatorname{PSL} (n, q) = \operatorname{PSL} (n, \mathbb{F}_q)$. All but a few of these groups are nonabelian simple groups: $A_3$ is abelian and $A_4$ has a normal subgroup of order 4, but $A_n$ is simple if $n \ge 5$. The group $\operatorname{PSL} (n, \mathbb{F}_q)$ is simple unless $n = 2$ and $q = 2$ or $3$. We will not prove this here, but proofs may be found in many places, such as

The groups $\operatorname{GL} (n, \mathbb{F}_q)$, $\operatorname{PGL} (n, \mathbb{F}_q)$, $\operatorname{SL} (n, \mathbb{F}_q)$ and $\operatorname{PSL} (n, \mathbb{F}_q)$ are generally all distinct, though the second two have the same order. The group $\operatorname{PGL} (n, \mathbb{F}_q)$ is by definition $\operatorname{GL} (n, \mathbb{F}_q) / Z$, where $Z \cong \mathbb{F}_q^{\times}$ is the center of $\operatorname{GL} (n, \mathbb{F}_q)$, the set of scalar matrices, that is, matrices of the form
 $\label{zzzenter} \left(\begin{array}{ccc} z & & \\ & \ddots & \\ & & z \end{array}\right) .$ (1.10.1)

So we have a short exact sequence $1 \longrightarrow Z \longrightarrow \operatorname{GL} (n, \mathbb{F}_q) \longrightarrow \operatorname{PGL} (n, \mathbb{F}_q) \longrightarrow 1.$ The intersection $Z_1$ of $Z$ with $\operatorname{SL} (n, \mathbb{F}_q)$ consists of those matrices (1.10.1) with $z^n = 1$; the order of this group is $\gcd (q - 1, n)$. The group $\operatorname{PSL} (n, \mathbb{F}_q)$ is defined to be the quotient $\operatorname{SL} (n, \mathbb{F}_q) / Z_1$, so we have a short exact sequence $1 \longrightarrow Z_1 \longrightarrow \operatorname{SL} (n, \mathbb{F}_q) \longrightarrow \operatorname{PSL} (n, \mathbb{F}_q) \longrightarrow 1.$

Proposition 1.10.1: The group $\operatorname{PSL} (n, \mathbb{F}_q)$ is isomorphic to a subgroup of $\operatorname{PGL} (n, \mathbb{F}_q)$.

Proof. (Click to Expand/Collapse)

Consider the homomorphism $\operatorname{SL} (n, \mathbb{F}_q) \longrightarrow \operatorname{GL} (n, \mathbb{F}_q) \longrightarrow \operatorname{PGL} (n, \mathbb{F}_q),$ where the first map is the inclusion and the second projection. The kernel of the composition is clearly $\operatorname{SL} (n, \mathbb{F}_q) \cap Z = Z_1$, so $\operatorname{PSL} (n, \mathbb{F}_q) = \operatorname{SL} (n, \mathbb{F}_q) / Z_1$ is isomorphic to the image of this homomorphism, a subgroup of $\operatorname{PGL} (n, \mathbb{F}_q)$.

Since $|\mathbb{F}_2^{\times} | = 1$ in the special case where $q = 2$, so in this special case the groups $\operatorname{GL} (n, \mathbb{F}_2)$, $\operatorname{SL} (n, \mathbb{F}_2)$, $\operatorname{PGL} (n, \mathbb{F}_2)$ and $\operatorname{PSL} (n, \mathbb{F}_2)$ are all isomorphic.

Let us consider the group $\operatorname{PGL} (3, F)$, where $F$ is an arbitrary field. Let $V$ be a fixed three-dimensional vector space over $F$. The projective plane is the set $\mathbb{P}(V)$ of one-dimensional subspaces of $V$. Let elements of $\mathbb{P}(V)$ be called points. These are organized into sets called lines. If $v^{\ast} \in V^{\ast}$ is a linear functional on $V$, then the set of points of the form $F v$ where $v \in V$ with $v^{\ast} (v) = 0$ is called a line. Two linear functionals determine the same line if they are proportional, so the lines are in bijection with the one-dimensional vector spaces in $V^{\ast}$; that is, the set of lines may be identified with $\mathbb{P}(V^{\ast})$.

If $V = F^3$ then we will use the notation $\mathbb{P}^2 (F)$ for $\mathbb{P}(V)$. Points might be represented in homogenous notation: if $a, b, c \in F$, not all zero, then $a : b : c$ denotes the point that is the one-dimensional vector space $(a, b, c)$. Thus if $\lambda$ is a nonzero scalar, then $\lambda a : \lambda b : \lambda c = a : b : c.$ The projective plane $\mathbb{P}^2 (\mathbb{F}_2)$ has seven elements:

We have drawn six of the seven lines through the seven points as ordinary lines; the seventh "line at infinity'' we have drawn as a dashed curve connecting the three points
 $l_{\infty} =\{0 : 0 : 1, \hspace{1em} 0 : 1 : 1, \hspace{1em} 0 : 1 : 0\}$ (1.10.2)

The group of automorphisms of this configuration of seven points and seven lines has order 168, and may be identified with $\operatorname{PGL} (3, \mathbb{F}_2)$.

Let us give a combinatorial definition of a projective plane. A projective plane is an ordered pair $(\Pi, \Lambda)$ consisting of two sets $\Pi$ and $\Lambda$, whose elements are called points and lines respectively. It is assumed that there is given an incidence relation between the elements of $\Pi$ and the elements of $\Lambda$; if $p \in \Pi$ and $L \in \Lambda$ we say that $p$ is on $L$ or that $L$ is on $p$ if this is satisfied. It is assumed that:

Axiom 1.10.1: Given any two distinct points $p, Q$ there is a unique line $L$ on both;

Axiom 1.10.2: Given any two distinct lines $L, M$ there is a unique point $p$ on both;

Axiom 1.10.3: There exist four points $p_1, p_2, p_3, p_4$ such that no three are on any line.

Exercise 1.10.1: Show that these axioms imply that there exist four lines $L_1, L_2, L_3, L_4$ such that no three are on any line. Deduce that if $(\Pi, \Lambda)$ is a projective plane, then so is $(\Lambda, \Pi)$, with the roles of points and lines interchanged.

The finite projective plane $\mathbb{P}^2 (\mathbb{F}_2)$ with seven elements, which we have already constructed, is called the Fano plane. It is useful to know that it is the unique finite projective plane with seven elements. If $(\Pi, \Lambda)$ and $(\Pi', \Lambda')$ are projective planes, an isomorphism between them consists of a pair of bijections $\Pi \longrightarrow \Pi'$ and $\Lambda \longrightarrow \Lambda'$ that preserve the incidence relation; that is, we require that if $(p, l) \in \Pi \times \Lambda$ are mapped to $(p', l') \in \Pi' \times \Lambda'$ under the given bijections, then $p$ is on $l$ if and only if $p'$ is on $l'$.

Theorem 1.10.1: Every projective plane with exactly seven points is isomorphic to the Fano plane.

Proof. (Click to Expand/Collapse)

By the third axiom, there exist four points $p_1, p_2, p_3, p_4$ in $\Pi$ such that no three are on any line. If $i, j \in \{1, 2, 3, 4\}$ let $l_{i j}$ denote the unique line through $p_i$ and $p_j$. The six lines $l_{i j}$ are all distinct, because if two were the same, that line would be on three points. Let $p_5$ be the intersection of $l_{12}$ and $l_{34}$, let $p_6$ be the intersection of $l_{13}$ and $l_{24}$, and let $p_7$ be the intersection of $l_{14}$ and $l_{23}$.

We will argue that the seven points $p_1, \cdots, p_7$ are all distinct. We already know that $\{p_1, p_2, p_3, p_4 \}$ are distinct.

Let us show that $p_5 \notin \{p_1, p_2, p_3, p_4 \}$. Suppose for example that $p_5 = p_1$. Then $p_1$ would lie on $l_{34}$, so $l_{34}$ would be on three points $\{p_1, p_3, p_4 \}$ which is a contradiction. The same argument rules out $p_5$ being $p_2, p_3$ or $p_4$.

Similarly $p_6$ and $p_7$ are not in $\{p_1, p_2, p_3, p_4 \}$. To show that the seven points $p_1, \cdots, p_7$ are all distinct, we have only to show that $p_5, p_6$ and $p_7$ are distinct. If for example $p_5 = p_6$, then since $p_5$ is on $l_{12}$ and $p_6$ is on $l_{13}$, the point $p_5$ would be the unique common point of $l_{12}$ and $l_{13}$, which is $p_1$. But we have already shown $p_5 \neq p_1$. Similarly $p_5 \neq p_7$ and $p_6 \neq p_7$.

Since $\Pi$ has seven elements we see that $\Pi =\{p_1, p_2, p_3, p_4, p_5, p_6, p_7 \}$. We have identified six lines $l_{i j}$ with $1 \le i, j \le 4$, and each of these lines is on three of the seven points. Let us argue that none of the lines contains more than three points. For example, we know that $l_{12}$ is on $p_1, p_2, p_5$. It is not on $p_3$ or $p_4$ because no three of $p_1, p_2, p_3, p_4$ are on the same line. Let us show that $l_{12}$ is not on $p_6$. Indeed, $p_6$ is on $l_{13}$ and since $l_{12}, l_{13}$ are on a unique point $p_1$ we would have $p_6 = p_1$, which we have ruled out. The same reasoning shows that $p_7$ is not on $l_{12}$, and so $p_1, p_2, p_5$ are the only points on $l_{12}$. The same reasoning shows that each of the six lines $l_{i j}$ is on exactly three points.

Let $l_{\infty}$ denote the unique line containing $p_5$ and $p_6$. We will show that $l_{\infty}$ also contains $p_7$. Indeed, $l_{\infty}$ must intersect $l_{14}$ in some point $X$. We will show $X = p_7$. Since $l_{14}$ is on exactly three points, if $X \neq p_7$ then $X$ is one of the other two points $p_1$ or $p_4$ on $l_{14}$. If $p_1$ is on $l_{\infty}$ then both $p_1$ and $p_5$ are on $l_{\infty}$, so $l_{\infty}$ is the unique line $l_{12}$ on both $p_1$ and $p_5$. Since $p_6$ is on $l_{\infty}$ it follows that $p_6$ is on $l_{12}$, but we have already shown that this is not true. Similarly $X \neq p_4$ and therefore $X = p_7$, proving that $p_5, p_6$ and $p_7$ are on the line $l_{\infty}$.

We have now found seven lines $l_{i j}$ and $l_{\infty}$, each on 3 points. There can be no incidence relations besides the ones we have found without contradicting Axioms 1.10.1 and 1.10.2. We may now map $(\Pi, \Lambda)$ isomorphically to the Fano plane thus: $\begin{array}{llll} p_1 \longmapsto 1 : 1 : 1 & \hspace{2em} & p_4 \longmapsto 1 : 0 : 0 & \hspace{2em} p_7 \longmapsto 0 : 1 : 1\\ p_2 \longmapsto 1 : 0 : 1 & & p_5 \longmapsto 0 : 1 : 0 & \\ p_3 \longmapsto 1 : 1 : 0 & & p_6 \longmapsto 0 : 0 : 1 & \end{array}$

Let us see what the geometry of the Fano plane tells us about parabolic subgroups of $\operatorname{GL} (3, \mathbb{F}_2)$. In general, the standard Borel subgroup $B$ of $\operatorname{GL} (n, F)$. A subgroup conjugate to $B$ is called a Borel subgroup. A subgroup containing $B$ is called a standard parabolic subgroup. A subgroup containing a conjugate of $B$ is called parabolic.

Exercise 1.10.2: Let $q = p^k$ where $p$ is prime. Show that the Borel subgroups of $\operatorname{GL} (n, \mathbb{F}_q)$ are exactly the $p$-Sylow subgroups.

In the case of $\operatorname{GL} (3, \mathbb{F}_2)$, the standard Borel subgroup consists of matrices of the form $B = \left\{ \left(\begin{array}{ccc} 1 & \ast & \ast\\ & 1 & \ast\\ & & 1 \end{array}\right) \right\} .$ It has order 8, and is a 2-Sylow subgroup. There are two conjugacy classes of parabolic subgroups, namely the conjugates of $P = \left\{\left(\begin{array}{ccc} \ast & \ast & \ast\\ \ast & \ast & \ast\\ & & 1 \end{array}\right)\right\}, \qquad Q = \left\{\left(\begin{array}{ccc} 1 & \ast & \ast\\ & \ast & \ast\\ & \ast & \ast \end{array}\right)\right\} .$ These have interpretations in terms of the Fano plane: thus $P$ is the stabilizer of $0 : 0 : 1$ and $Q$ is the stabilizer of the line $l_{\infty}$ at infinity defined in (1.10.2). We have $P \cap Q = B$.

Exercise 1.10.3: Show that $|P| = 24$ and that $P$ has 7 conjugates; show that $|Q| = 24$ and that $Q$ has 7 conjugates; but that $P$ and $Q$ are not conjugate. Show that if $P'$ and $Q'$ are conjugates of $P$ and $Q$ respectively, then $Q'$ and $Q'$ are the isotropy subgroups of a point $p$ and a line $l$ in the Fano plane, and that $|P' \cap Q' | = \left\{ \begin{array}{ll} 8 & \text{if p is on l},\\ 6 & \text{otherwise.} \end{array} \right.$

The groups $P$ and $L$ are semidirect products. We may write $P = MU, \hspace{2em} M = \left\{ \left( \begin{array}{ccc} \ast & \ast & \\ \ast & \ast & \\ & & 1 \end{array} \right) \right\} \hspace{2em} U = \left\{ \left( \begin{array}{ccc} 1 & & \ast\\ & 1 & \ast\\ & & 1 \end{array} \right) \right\}$ and $Q = NV, \hspace{2em} N = \left\{ \left( \begin{array}{ccc} 1 & & \\ & \ast & \ast\\ & \ast & \ast \end{array} \right) \right\} \hspace{2em} V = \left\{ \left( \begin{array}{ccc} 1 & \ast & \ast\\ & 1 & \\ & & 1 \end{array} \right) \right\} .$ We can use these subgroups to "transport'' the structure of the Fano plane into a structure that is intrinsic to the group.

Exercise 1.10.4: Show that $U \vartriangleleft P$ but that $M$ is not normal in $U$.

Exercise 1.10.5: Show that $U$ and $V$ are both isomorphic to $Z_2 \times Z_2$, and that any subgroup of $\operatorname{GL} (3, \mathbb{F}_2)$ that is isomorphic to $Z_2 \times Z_2$ is a conjugate of either $U$ or $V$. Show that $U$ and $V$ are not conjugate in $\operatorname{GL} (3, \mathbb{F}_2)$. Show that $U$ and $V$ have seven conjugates each. Show that if $U'$ and $V'$ are conjugates of $U$ and $V$, and if $U' \cap V'$ is nonempty, then it is cyclic of order $2$. If $z$ is its generator, then the centralizer $C (z)$ is a 2-Sylow subgroup of $\operatorname{GL} (3, \mathbb{F}_2)$ that contains both $U'$ and $V'$.

By an involution in a group we mean an element of order two.

Exercise 1.10.6: Show that $G = \operatorname{GL} (3, \mathbb{F}_2)$ has one conjugacy class of involutions, and there are 21 elements of this conjugacy class. If $z$ is an involution, then its centralizer $C (z)$ is a 2-Sylow subgroup, and $z \longmapsto C (z)$ is a bijection between the involutions of $G$ and its 2-Sylow subgroups.

Exercise 1.10.7: Let $\{U_1, \cdots, U_7 \}$ be the conjugates of $U$ and $\{V_1, \cdots, V_7 \}$ be the conjugates of $V$. Call the $U_i$ points and the $V_j$ lines. Define an incidence relation between the $U_i$ and the $V_j$ by declaring that $U_i$ is on $V_j$ if $U_i \cap V_j$ is nontrivial. Show that the configuration $U_i$ and $V_j$ with this incidence relation is a copy of the Fano plane.

The significance of this is that we have found a structure inside the group $\operatorname{GL} (3, \mathbb{F}_2)$ that is identical to the Fano projective plane. Therefore, suppose we are given another finite group $G$ of order 168, and we want to prove that $G$ isomorphic to $\operatorname{GL} (3, \mathbb{F}_2)$. If what we are trying to prove is true, then $G$ will have two conjugacy classes of $Z_2 \times Z_2$ subgroups, and if we call elements of one class points and the other class lines, with the incidence relation that a point is on a line if the two subgroups have nontrivial intersection, and thus we will have a copy of the Fano plane realized inside $G$. Since $G$ acts on this configuration by conjugating the subgroups, we will have a homomorphism from $G$ to the automorphism group of the Fano plane, that is, to $\operatorname{GL} (3, \mathbb{F}_2)$.

The group $\operatorname{PSL} (2, \mathbb{F}_7)$ has order 168. Let us carry out this program for this special case. We begin by describing the 2-Sylow subgroups of this group. We have an embedding $\mathbb{F}_{49}^{\times} \longrightarrow \operatorname{GL} (2, \mathbb{F}_7)$ as was explained in Section 1.9. Indeed, if $\alpha \in \mathbb{F}_{49}^{\times}$ then $\lambda_{\alpha} : \mathbb{F}_{49} \longrightarrow \mathbb{F}_{49}$ defined by $\lambda_{\alpha} (x) = \alpha x$ is an invertible linear transformation of $\mathbb{F}_{49}$, which is a two-dimensional vector space over $\mathbb{F}_7$, and on chosing a basis of this vector space, $\lambda_{\alpha}$ may be interpreted as an element of $\operatorname{GL} (2, \mathbb{F}_7)$. Moreover the Frobenius endomorphism $\phi : \mathbb{F}_{49} \longrightarrow \mathbb{F}_{49}$ defined by $\phi (x) = x^7$ is also a linear transformation normalizing the image of $\mathbb{F}_{49}^{\times}$, and so we obtain an embedding $\mathbb{F}_{49}^{\times} \rtimes \left\langle \phi \right\rangle \longrightarrow \operatorname{GL} (2, \mathbb{F}_7)$. The image is a subgroup of order 96. The kernel of the determinant $\operatorname{GL} (2, \mathbb{F}_7) \longrightarrow \mathbb{F}^{\times}$ is a subgroup of order $16$, which is a $2$-Sylow subgroup of $\operatorname{SL} (2, \mathbb{F}_7)$. The image of this subgroup in $\operatorname{PSL} (2, \mathbb{F}_7)$ is a dihedral group of order 8.

Theorem 1.10.2: The group $G = \operatorname{PSL} (2, \mathbb{F}_7)$ has 2 conjugacy classes of subgroups isomorphic to $Z_2 \times Z_2$. Each conjugacy class has seven elements. Let us call the elements of the first conjugacy class points and those of the second lines, with the incidence relation that a point is on a line if they have a nontrivial intersection, or equivalently, if they are contained in a 2-Sylow subgroup. Then the points and lines of this configuration form a projective plane isomorphic to the Fano plane. We have $G \cong \operatorname{GL} (3, \mathbb{F}_2)$.

Proof. (Click to Expand/Collapse)

Let $\pi : \operatorname{SL} (2, \mathbb{F}_7) \longrightarrow \operatorname{PSL} (2, \mathbb{F}_7)$ be the canonical map. The involutions are given by the following table. $\begin{array}{|l|l|l|} \hline z_1 = \pi \left( \begin{array}{cc} 4 & 3\\ 6 & 3 \end{array} \right) & z_2 =\pi \left( \begin{array}{cc} 1 & 1\\ 5 & 6 \end{array} \right) & z_3 =\pi \left( \begin{array}{cc} 1 & 6\\ 2 & 6 \end{array} \right)\\ \hline z_4 = \pi \left( \begin{array}{cc} 1 & 4\\ 3 & 6 \end{array} \right) & z_5 =\pi \left( \begin{array}{cc} 1 & 5\\ 1 & 6 \end{array} \right) & z_6 =\pi \left( \begin{array}{cc} 4 & 2\\ 2 & 3 \end{array} \right)\\ \hline z_7 = \pi \left( \begin{array}{cc} 1 & 3\\ 4 & 6 \end{array} \right) & z_8 =\pi \left( \begin{array}{cc} 0 & 2\\ 3 & 0 \end{array} \right) & z_9 =\pi \left( \begin{array}{cc} 0 & 4\\ 5 & 0 \end{array} \right)\\ \hline z_{10} = \pi \left( \begin{array}{cc} 2 & 1\\ 2 & 5 \end{array} \right) & z_{11} =\pi \left( \begin{array}{cc} 4 & 4\\ 1 & 3 \end{array} \right) & z_{12} =\pi \left( \begin{array}{cc} 4 & 5\\ 5 & 3 \end{array} \right)\\ \hline z_{13} = \pi \left( \begin{array}{cc} 2 & 2\\ 1 & 5 \end{array} \right) & z_{14} =\pi \left( \begin{array}{cc} 4 & 6\\ 3 & 3 \end{array} \right) & z_{15} =\pi \left( \begin{array}{cc} 2 & 4\\ 4 & 5 \end{array} \right)\\ \hline z_{16} = \pi \left( \begin{array}{cc} 1 & 2\\ 6 & 6 \end{array} \right) & z_{17} =\pi \left( \begin{array}{cc} 2 & 5\\ 6 & 5 \end{array} \right) & z_{18} =\pi \left( \begin{array}{cc} 0 & 1\\ 6 & 0 \end{array} \right)\\ \hline z_{19} = \pi \left( \begin{array}{cc} 2 & 6\\ 5 & 5 \end{array} \right) & z_{20} =\pi \left( \begin{array}{cc} 4 & 1\\ 4 & 3 \end{array} \right) & z_{21} =\pi \left( \begin{array}{cc} 2 & 3\\ 3 & 5 \end{array} \right)\\ \hline \end{array}$ We enumerate the $Z_2 \times Z_2$ subgroups of $\operatorname{PSL} (2, \mathbb{F}_7)$. This is straightforward because each such subgroup is contained in a 2-Sylow subgroup, which is the centralizer of one of its elements. We find two conjugacy classes of $Z_2 \times Z_2$ elements, each having seven elements. The two conjugacy classes are: $\text{Points} : \hspace{1em} \begin{array}{|l|} \hline \{1, z_1, z_8, z_{10} \}\\ \hline \{1, z_{2}, z_{7}, z_{17}\}\\ \hline \{1, z_{3}, z_{16}, z_{21}\}\\ \hline \{1, z_{4}, z_{5}, z_{19}\}\\ \hline \{1, z_{6}, z_{11}, z_{20}\}\\ \hline \{1, z_9, z_{13}, z_{14}\}\\ \hline \{1, z_{12}, z_{15}, z_{18} \}\\ \hline \end{array} \hspace{1em} \text{Lines} : \hspace{1em} \begin{array}{|l|} \hline \{1, z_1, z_{12}, z_{14} \}\\ \hline \{1, z_{2}, z_{5}, z_{15}\}\\ \hline \{1, z_{3}, z_{4}, z_{13}\}\\ \hline \{1, z_{6}, z_{18}, z_{21}\}\\ \hline \{1, z_{7}, z_{10}, z_{16}\}\\ \hline \{1, z_{8}, z_{11}, z_{19}\}\\ \hline \{1, z_{9}, z_{17}, z_{20}\}\\ \hline \end{array}$ With the incidence relation that a point is on a line if and only if they have an involution in common, the table shows that every point is on exactly three lines, and every line is on exactly three points. To check that every pair of distinct points is on a unique line, we may conjugate so that one of the points is $\{1,z_1,z_8,z_{10}\}$, and there are only six possibilities for the second point. For example, $\{1, z_{7}, z_{10}, z_{16}\}$ is the unique line on the first two points.

Exercise 1.10.8: Give a second proof that $PSL(2,\mathbb{F}_7)\cong GL(3,\mathbb{F}_2)$ as follows. The group $GL(3,\mathbb{F}_2)$ has eight 7-Sylow subgroups. Label these by the points of $\mathbb{P}^1(\mathbb{F}_7)$ as follows: $\begin{array}{|l|l||l|l|} \hline \infty & \left<\left(\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)\right> & 3 & \left<\left(\begin{array}{rrr} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)\right>\\ \hline 0 & \left<\left(\begin{array}{rrr} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{array}\right)\right> & 4 & \left<\left(\begin{array}{rrr} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right)\right>\\ \hline 1 & \left<\left(\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 0 \end{array}\right)\right> & 5 & \left<\left(\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right)\right>\\ \hline 2 & \left<\left(\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{array}\right)\right> & 6 & \left<\left(\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right)\right>\\ \hline \end{array}$ Find a set of generators of $GL(3,\mathbb{F}_2)$, and show for every $\gamma$ in this set of generators that conjugation by $\gamma$ has the same effect on these eight 7-Sylows as some linear fractional transformation. Explain why this implies the isomorphism. (Compare Exercise 1.9.3.)