---- > [!theorem] Theorem. ([[there is a unique simple group of order 168, up to isomorphism]]) > The [[automorphism|automorphism group]] of the **Klein quartic** $X^{3}Y + Y^{3}Z + Z^{3}X$ > is the unique [[simple group]] of order $168$. > \ > The equation above defines an [[variety|algebraic curve]] of genus 3 in the projective space $P^2$. Over $\mathbb{C}$, it may be thought of as a Riemann surface, and topologically it is a sphere with three handles attached. Now there is a theorem in algebraic geometry that if $C$ is a curve of genus $g \geq 2$ over the complex numbers, then its (algebraic) automorphism group $\text{Aut}(C)$ is a finite group of order $\leq 84(g - 1)$. If $g = 3$, this gives an upper bound of 168, and this upper bound is exactly realized for the curve $C$ above, its automorphism group being isomorphic to $G$ (or $G'$). While it is somewhat tricky to see that $\text{Aut}(C) \simeq G$, we can amuse ourselves with the following exercises. ^2045cc > [!proposition] Exploring the Klein Quartic. > >1. Construct a nontrivial element $\sigma$ in $\text{Aut}(C)$ of order 3. You want to define by a formula of the sort $(X,Y,Z) \mapsto (F_1(X,Y,Z), F_2(X,Y,Z), F_3(X,Y,Z))$ where $F_1, F_2, F_3$ are polynomials in the variables $X,Y,Z$. > One such element is choosing $\sigma$ to be given by the map $(X,Y,Z) \mapsto (Y,Z,X)$. This is essentially a 3-cycle so it has order 3. > >2. Construct a nontrivial element $\tau$ in $\text{Aut}(C)$ of order 7. (Consider using a a 7th root of unity.) > >Let $r=e^{2\pi i / 7}$ denote the $7^{th}$ root of unity. > One such element is choosing $\tau$ to be given by the map $(X,Y,Z) \mapsto (r^{2}X, rY, r^{4}Z)$. > >3. (4 pts) Consider the subgroup of $\text{Aut}(C)$ generated by $\tau$ and $\sigma$. What can you say about the order of this group and its structure? > > This is the group $H=\langle \tau \rangle \langle \sigma \rangle$ ; since these subgroups overlap trivially (prime orders) by [[cardinality of frobenius product of subgroups]] and since $\tau$ is [[normal subgroup|normal]] (so $H \leq G$). Thus $H$ has order 21 and is the [[internal semi-direct product]] $\langle \tau \rangle \rtimes \langle \sigma \rangle$. ^bb7fc5 > [!proof]- Proof. ([[there is a unique simple group of order 168, up to isomorphism]]) > # $G=PSL_{2}(\mathbb{F_{}}_{7})= SL_{2}(\mathbb{F}_{7} )/ \{ \pm I \}$ > **Part a. Show that $|G|=168=2^{3} \cdot 3 \cdot 7$.** Let $\text{GL}_{n}$ denote the [[general linear group]]. Recall that $|\text{GL}_{2}(\mathbb{F}_{7})|=(7^{2}-1)(7^{2}-7)$, by [[general linear group over a prime field]] Now, [[determinant|det]]$: \text{GL}_{2}(\mathbb{F}_{7}) \to \mathbb{F}_{7}^{\times}$ is a [[surjection|surjective]] [[group homomorphism|homomorphism]] whose [[kernel]] is $\text{SL}_{2}(\mathbb{F}_{7})$. By the [[first isomorphism theorem]], then, we have $\text{GL}_{2}(\mathbb{F}_{5}) / \text{SL}_{2}(\mathbb{F}_{7}) \cong \mathbb{F}_{7}^{\times}$. So $|\text{SL}_{2}(\mathbb{F}_{7})|=\frac{|\text{GL}_{2}(\mathbb{F}_{7})|}{|\mathbb{F}_{7}^{\times}|}=\frac{(7^{2}-1)(7^{2}-7)}{6}=336.$ [[order of quotient group is quotient of orders|Conclude that]] $|\text{PSL}_{2}(\mathbb{F}_{7})|=\frac{336}{2}=168$. > **Part b.** Let $x$ and $y$ be the classes of [[matrix|matrices]] $x=A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \text{ and } y=B=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ in $G$. Show that $x$ and $y$ [[generating set of a group|generate]] distinct $7$-sylow subgroups $H$ and $K$. Deduce that $n_{7}(G)=8$. > For $k \in [7]$, $A^{k}=\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \text{ and } B^{k}=\begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix},$ so that $H:=\langle A \rangle = \{ \begin{bmatrix}1 & k \\0 & 1 \end{bmatrix} : k =0,1,\dots,6 \} \leq \text{SL}_{2}(\mathbb{F}_{7})$ and $K:=\langle B \rangle = \{ \begin{bmatrix}1 & 0 \\k & 1 \end{bmatrix} : k =0,1,\dots,6 \} \leq \text{SL}_{2}(\mathbb{F}_{7})$ clearly these are distinct. By [[the Sylow theorems]], $n_{7} \in \{ 1, 8 \}$. But we just showed $n_{7} \geq 2$. Thus $n_{7}=8$. > **Part c.** Let $N \trianglelefteq G$ be a [[normal subgroup]]. Show that $[G : N] \neq 2$. > Suppose that $[G : N]=2$, so that $|N|=84=2^{2} \cdot 3 \cdot 7$. Then by [[the Sylow theorems]] $N$ has exactly one $7$-[[p-Sylow subgroup|Sylow subgroup]] $K$. $K$ is also a $p$-Sylow of $G$; by the [[the Sylow theorems|the second Sylow theorem]] we may [[conjugate]] it to obtain $A$, (or if it equals $A$ already, $B$). Then since $N$ is invariant under [[conjugate|conjugation]], we see that $N$ contains both $7$-Sylows, a contradiction. > **Part d.** Let $N \trianglelefteq G$ be a [[normal subgroup]]. Show that $[G : N] \neq 3$. > Suppose that $[G : N]=3$, so $|N|=56$. By [[the Sylow theorems]] $N$ has either $1$ or $8$ 7-Sylows. It can't be $1$ else we get a contradiction by the reasoning from **part c**, so it is $8$— i.e., $N$ contains all $7$-Sylows of $G$. In particular, $H \leq N$ and $K \leq N$, thus $N \ni u:=xy^{2}x \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix}.$ $|u|=3$, since $\begin{bmatrix}3&4\\2&3\end{bmatrix}^3=I_{2}$ when $\mathbb{F}_{7}$ is the ground [[field]]. But $3 \not {|} 56$, [[Lagrange's Theorem|so this is a contradiction]]. > **Part e.** Let $N \trianglelefteq G$ be a [[normal subgroup]]. Show that $[G:N] \neq 6$. > Suppo[](field.md)md)=6$. Let $G$ [[group action|act on]] the set $G / N$ of [[coset|cosets]] of $N$ by left multiplication, $G \times G / N \to G / N, \ \ (g, aN ) \mapsto gaN.$ This [[homomorphism induced by group action|induces a homomorphism]] $\phi:G \to S_{6}$ that is an [[group embedding|embedding]] since that [[group action|action is faithful]]. [[homomorphisms preserve structure|Thus]] $|\phi(G)|=168$. But $168 \not{|}|S_{6}|$, so we have a contradiction. > *Suppose now that $G$ is not simple. We will derive a contradiction.* > **Part f.** Let $N \trianglelefteq G$ be a maximal proper [[normal subgroup]]. Show that the only possib[](group%20homomorphisms%20preserve%20structure.md)]] is $[G:N]=7$, so that $|N|=24$. > Since $N$ is maximal, [[the correspondence theorem]] tells us that $G / N$ is [[simple group|simple]]. In light of previous parts, we have $|G / N| \in \{7,8,12,14,21,24,28,42,56,84 \}$. From our work in [[there is a single nonabelian simple group of order less than 168]] we know the only option is $|G / N|=7$ since that is the only case where $G / N$ is [[simple group|simple]] (it's [[group isomorphism|isomorphic]] to $C_{7}$). Thus $[G : N]=7$. > **Part g.** Since $|N|=24$, we must have $n_{2}(N) \in \{ 1,3 \}$. Show that $n_{2}(N) \neq 1$. > Suppose $n_{2}(N)=1$. Then $N$ has a unique 2-Sylow of order $8$, $H$, that is [[normal subgroup|normal in]] $N$. By [[p-Sylow normality nests]], $H$ is also [[normal subgroup|normal in]] $G$. Now, $|G / H|=21$ and [[the Sylow theorems|hence]] has a [[normal subgroup|normal]] $7$-Sylow $\overline{K}$. $\overline{K}$ [[the correspondence theorem|corresponds to]] a [[normal subgroup]] $K$ of $G$ containing $H$. ![[CleanShot 2023-10-26 at [email protected]]] > From the definition of the [[bijection]] in [[the correspondence theorem]] we see that $[G:K]=[G / H: \overline{K}]=3$. Then **part d** gives a contradiction. > **Part h.** Show that $n_{2}(N) \neq 3$ either, and thus $G$ is [[simple group|simple]]. > Suppose $n_{2}(N)=3$. Let $H_{1},H_{2},H_{3}$ denote the $2$-Sylows in $N$, which are the same as the $2$-Sylows in $G$. Now let $G$ act on $\{ H_{1},H_{2},H_{3} \}$ by [[conjugate|conjugation]]. The [[homomorphism induced by group action|induced homomorphism]] $\phi: G \to S_{3}$, $g \mapsto c_{g}$, is nontrivial (because the [[group action]] is [[transitive group action|transitive]]). The [[kernel]] $\ker \phi$ is a [[kernel iff normal subgroup|normal subgroup of]] $G$ and by the [[first isomorphism theorem]] $G / \ker \phi \cong \phi(G)$ thus $[G : \ker \phi] \in \{ 2,3,6 \}$. This is a contradiction by parts **c, d, e**. > ># $G=GL_{3}(\mathbb{F_{}}_{2})$ **Part a.** Show that $|G|=168.$ This is immediate from [[general linear group over a prime field]], which says that $|\text{GL}_{3}(\mathbb{F}_{2})|=(2^{3}-1)(2^{3}-2)(2^{3}-4)=168=2^{3} \cdot 3 \cdot 7.$ > **Part b.** Show that $G$ is [[simple group|simple]]. > Define $x:=\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \text{, } y:=x^{\top}=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}, \text{ and } u:= yx=x^{\top}x=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}.$ Straightforward computation shows $|u|=3$. > Now, our argument proceeds similarly to the above. $x$ and $y$ generate distinct 7-Sylow subgroups. So we see that $n_{7}(G)=8$. > Next let $N$ be a [[normal subgroup]] of $G$. Then $[G: N] \notin \{ 2,3,6 \}$ by the exact same reasoning as parts b, c, d, e above using the new $x,y,u$. > Suppose $G$ is not simple. Then we can repeat parts f,g,h above to obtain a contradiction. So $G$ is simple. > **Part c.** Identify the [[group isomorphism|isomorphism class]] of a $2$-Sylow subgroup in $G$. > A $2$-Sylow has order $8$; thus it is isomorphic to one of $C_{8},C_{4} \times C_{2}, (C_{2})^{3}, D_{4}, Q_{}$. > By [[general linear group over a prime field]], one $2$-Sylow subgroup $H$ consists of all elements in $\text{GL}_{3}(\mathbb{F}_{2})$ of the form $\begin{bmatrix} 1 & * & * \\ & 1 & * \\ & & 1 \end{bmatrix}.$ Because $G$ is simple, $H$ is not a [[normal subgroup]] of $G$. $C_{8}, C_{4} \times C_{2}, \text{ and } (C_{2})^{3}$ are all [[abelian group|abelian]], so we must either have $H \cong D_{4}$ or $H \cong Q$, where the former is the [[dihedral group]] and the latter is the [[quaternion group]]. > The set of matrices of the form $\begin{bmatrix} 1 & a & 0 \\ & 1 & 0 \\ & & 1 \end{bmatrix}$ forms a [[subgroup]] $K$ of $H$. But $K$ is not invariant under [[conjugate|conjugation]] since for example $\begin{pmatrix}1&1&1\\ \:\:0&1&1\\ \:\:0&0&1\end{pmatrix}\begin{pmatrix}1&a&0\\ \:\:0&1&0\\ \:\:0&0&1\end{pmatrix}\begin{pmatrix}1&1&1\\ \:\:\:0&1&1\\ \:\:\:0&0&1\end{pmatrix}^{-1}=\begin{pmatrix} 1 & a & a \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \notin K.$ Since [[quaternion group is nonabelian, but all its subgroups are normal|all subgroups of Q]] are [[normal subgroup]]s, we conclude that $K \cong D_{4}$. ^25a6c4 ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```