PSL(2,7) is a nonabelian simple group of order 168

Direct expansion of a product shows that the determinant of the product of two such matrices is the product of their determinants. Their inverses are $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$; together with F1 this makes $S$ a group. Its first column can be any of the $7^2-1=48$ nonzero vectors. For first column $(a,c)$ with $a\mathrm{\ne }0$, there are exactly seven second columns: $b$ is arbitrary and $d=(1+bc)\mathrm{/}a$; for $a=0$ one has $c\mathrm{\ne }0$, $b=-1\mathrm{/}c$, and $d$ arbitrary. Thus $\mathrm{\mid }S\mathrm{\mid }=48\cdot 7=336$. The central subgroup $Z=\{I,-I\}$ has two elements; its cosets pair each matrix with its negative, so $G=S\mathrm{/}Z$ has order $168$.

Write $U(t)=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$, $L(t)=\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$, $D(a)=\left(\begin{array}{cc}a& 0\\ 0& a^{-1}\end{array}\right)$, and $W(a)=\left(\begin{array}{cc}0& a\\ -a^{-1}& 0\end{array}\right)$ for $a\mathrm{\ne }0$. Multiplication gives $W(a)=U(a)L(-a^{-1})U(a)$ and $D(a)=W(a)W(1{)}^{-1}$. These upper and lower matrices generate $S$: if the top left entry $a$ of $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S$ is nonzero, then $M=L(c\mathrm{/}a)D(a)U(b\mathrm{/}a)$; if it is zero, left multiplication by $W(1)$ makes it nonzero, since $c\mathrm{\ne }0$. Every displayed factor is generated by the $U$ and $L$, proving the assertion.

In $S$, $[D(2),U(t)]=U(3t)$ since $2^2-1=3$ in ${\mathbb{F}}_7$. As multiplication by $3$ permutes the field, every $U(t)$ is a commutator. Also $W(1)U(t)W(1{)}^{-1}=L(-t)$, so normality of the commutator subgroup puts every $L(t)$ in it. Step 2.1 gives $[S,S]=S$. The same calculations after quotienting, and the images of the same generators, give $[G,G]=G$. Since $\mathrm{\mid }G\mathrm{\mid }=168>1$, $G$ cannot be abelian: all commutators in an abelian group are the identity.

Let $\mathrm{\Omega }$ be the one-dimensional subspaces of ${\mathbb{F}}_7^2$, written $\mathrm{\infty }=[1:0]$ and $x=[x:1]$ for $x\in {\mathbb{F}}_7$. The matrices in $S$ permute these lines, and $Z$ fixes them, giving an action of $G$ on these eight points. It is faithful: a matrix fixing all lines fixes $[1:0]$ and $[0:1]$, hence is diagonal; fixing $[1:1]$ makes its two diagonal entries equal, and determinant one then makes the matrix $I$ or $-I$. The action is transitive because $W(1)$ sends $\mathrm{\infty }$ to $0$ and $U(t)$ translates $x$ to $x+t$. The stabilizer $H$ of $\mathrm{\infty }$ contains all these translations and is transitive on the other seven points. Thus the action is two-transitive: first move one entry of an ordered pair to its target, then use the target stabilizer to move the other entry.

If $N⊴G$, its orbits form a $G$-invariant equivalence relation on $\mathrm{\Omega }$, since $g(Nx)=N(gx)$. Under two-transitivity any equivalent pair of distinct points makes every pair of distinct points equivalent, by transporting that ordered pair. Thus either all $N$-orbits are singletons, or $N$ is transitive. In the singleton case faithfulness forces $N=1$. If $N\mathrm{\ne }1$, then $N$ is transitive and $G=NH$: for any $g\in G$ take $n\in N$ with $n\mathrm{\infty }=g\mathrm{\infty }$ and then $n^{-1}g\in H$.

Let $A$ be the image of $\{U(t):t\in {\mathbb{F}}_7\}$ in $G$. It is abelian, since $U(s)U(t)=U(s+t)$. Every matrix representing a member of $H$ is upper triangular, of the form $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$; it conjugates $U(t)$ to $U(a^{2}t)$, so $A⊴H$. The conjugates of $A$ generate $G$, since they include the images of all $U(t)$ and $L(t)$ by step 3.1 and these generate by step 2.1. For nontrivial normal $N$, let $\pi :G\to G\mathrm{/}N$ be the quotient map. By step 4.1 any $g$ is $nh$ with $n\in N$, $h\in H$, and therefore $\pi (gA{g}^{-1})=\pi (hA{h}^{-1})=\pi (A)$. Hence $G\mathrm{/}N$, generated by the images of these conjugates, equals the abelian subgroup $\pi (A)$. F3 and step 3.1 imply $G=[G,G]\subseteq N$, so $N=G$. With step 1.1 this proves simplicity, order $168$, and nonabelianness. All sets and selections used are finite.