We introduce this with the following observation. Let $f,F$ be “fractional linear transformations”, that is, functions of the form $f(x) = (ax+b)/(cx+d)$, $F(x) = (Ax+B)/(Cx+D)$ (with coefficients $a,b,c,d$ and $A,B,C,D$ in some given field $K$). Then the composition $F \circ f$ is of the same form. Explicitly, $$ F(f(x)) = \frac{A f(x) + B}{C f(x) + D} = \frac{A \frac{ax+b}{cx+d} + B}{C \frac{ax+b}{cx+d} + D} = \frac{(Aa + Bc)\,x + (Ab + Bd)}{(Ca + Dc)\,x + (Cb + Dd)}. $$ The pattern of the coefficients $({Aa + Bc \; Ab + Bd \atop Aa + Dc \; Cb + Dd})$ should look familiar: it exactly matches the entries of the product of $2 \times 2$ matrices $({A \; B \atop C \; D}) \, ({a \; b \atop c \; d}).$ 😮 What’s going on here, and why?

We shall first give a more precise account of “what”, using some of the group-theoretic terminology that we have developed during the past two months, and then — after a scenic detour to construct an outer automorphism of $S_6$ and note a Rubik’s Cube curiosity — explain “why”.

We already have some terminology to describe the situation. For starters, we claim that the nonconstant fractional linear transformations form a group, call it $\Gamma.$ (That’s a capital Greek letter Gamma; LaTeX: \Gamma.) As we know, this means not only that $\Gamma$ is closed under composition but also that it has an identity element and inverses. Well, the identity map ${\rm id}(x) = x$ is in $\Gamma$ because $x = (1 \cdot x + 0) \, / \, (0\cdot x + 1).$ As for inverses, if $f(x) = (ax+b) \, / \, (cx+d)$ then $f^{-1}(x)$ is the solution of $f(y)=x;$ solving for $y$ yields $y = (dx-b) / (-ax+c)$ which is again a fractional linear transformation, so indeed $f^{-1} \in \Gamma.$

To go further it will be convenient to introduce the following notation: to any $2 \times 2$ matrix $M = ({a_{11} \; a_{12} \atop a_{21} \; a_{22}})$ associate the fractional linear transformation $\Phi_M$ (we’ve already seen $\Phi$ = Phi = \Phi), defined by $$ \Phi_M(x) = F(x) = \frac{a_{11} x + a_{12}}{a_{21} x + a_{22}}. $$ We have observed that $\Phi_M(\Phi_{M'}(x)) = \Phi_{MM'}(x),$ i.e. $\Phi_M \Phi_{M'} = \Phi_{MM'}.$ This suggests that the map taking $M$ to $\Phi_M$ is a homomorphism to $\Gamma.$ But we need to be careful here: what are the domain and codomain of this homomorphism? Recall that $M_2(K)$ does not quite form a group under matrix multiplication: we must remove the singular matrices (matrices $M = ({a_{11} \; a_{12} \atop a_{21} \; a_{22}})$ with $a_{11} a_{22} = a_{12} a_{21})$ to get the group ${\rm GL}_2(K).$ Thankfully the nonsingularity condition $a_{11} a_{22} \neq a_{12} a_{21}$ is exactly what we need to assure that the transformation $\Phi_M$ does not simplify to a constant (as in $M = ({2 \; 4 \atop 3 \; 6}),$ $\Phi_M(x) = (2x+4)/(3x+6) = 2/3$) or worse (try $M=0$, i.e. $M=({0 \; 0 \atop 0 \; 0})$ ). Removing singular matrices from $M_2(K),$ and removing constant transformations from $\Gamma$, we get a surjective homomorphism ${\rm GL}_2(K) \to \Gamma.$ Moreover, the kernel of this homomorphism consists exactly of the scalar matrices $k I_2 = ({k \; 0 \atop 0 \; k})$ for $k \in K^\times$: if $M$ is such a matrix then $\Phi_M(x) = kx/k = x$ so $\Phi_M$ is the identity transformation, and we readily check that there are no other $M$ in the kernel. So (by our “first isomorphism theorem”) we have identified $\Gamma$ with the projective linear group ${\rm PGL}_2(K)$. Indeed if we multiply a matrix $({a\;b\atop c\;d}) \in {\rm GL}_2(K)$ by $k I_2$ then $\Phi_M$ is replaced by the transformation $\Phi_{kM}$ that takes $x$ to $(kax+kb) \, / \, (kcx + kd),$ which simplifies to $(ax+b) \, / \, (cx + d) = \Phi_M(x)$ as expected.

Now since $\Gamma$ is a group of functions under composition, the map $(g,x) \mapsto g(x)$ must be a group action. Again some care is needed: $g$ is in $\Gamma$, and $x$ should be in the set $\Omega$ that $\Gamma$ acts on; but what exactly is this $\Omega$? It cannot be simply $K$ because the denominator in $cx+d$ might be zero.

For once the answer suggested by elementary calculus is correct: if $g\in\Gamma$ acts by $g(x) = (ax+b)/(cx+d)$ and $cx+d=0$ then we say $g(x) = \infty.$ That is, $\Omega$ consists of all $x \in K$ together with an extra element called $\infty$. For that to work, we need to also define $g(\infty)$ and check that the group-action axiom $g(h(x)) = (gh)(x)$ still holds if one or more of $x,h(x),g(h(x))$ is $\infty$. Again you won't be surprised that we define $g(\infty) = a/c$ which (at least for $K=\bf R$) is the limit of $g(x)$ as $x \to \infty.$ If $c=0$ then naturally we interpret $a/c$ as $\infty$ so $g(\infty) = \infty.$ You might worry about what happens if also $a=0$ — but we cannot have both $a=0$ and $c=0$ because then $g(x)$ is the constant $b/d$ (and indeed the determinant $ad-bc$ is zero). It is then routine, albeit tedious, to check that $g(h(x)) = (gh)(x)$ still works in all cases; we’ll be able to give a better (or at least more structural) explanation once we have reached the “why” section below. For now, we remark on the transitivity of the action of $\Gamma$ on $\Omega$. Certainly all points of $\Omega$ are in the same orbit: it is easy to get from any $x \in K$ to any other, or to $\infty$. The stabilizer $\Gamma_\infty$ of $\infty$ is just the group of linear substitutions $f(x) = ax+b,$ corresponding to scalar multiples of the matrix $({a\;b\atop 0\;1})$ (any $a \in K^\times$ and $b \in K$), which we have already encountered as a subgroup of ${\rm GL}_2(K)$. [What is $\Gamma_0?$ Can you check that it is the conjugate of $\Gamma_\infty$ by a group element such as $g(x) = 1/x$ which switches $0$ and $\infty$?] The action of that group $\Gamma_\infty$ on $K = \Omega - \{\infty\}$ is in turn transitive, and indeed “2-transitive” (a.k.a. “doubly transitive”): for any $x,y,x',y' \in K$ with $x\neq y$ and $x'\neq y'$ we can find $a\in K^\times$ and $b\in K$ such that $x'=ax+b$ and $y'=ay+b.$ In fact the choice is unique (“two points determine a unique line”), so this action of $\Gamma_\infty$ on $K$ is “sharply 2-transitive”. This makes the action of $\Gamma$ on $\Omega$ sharply 3-transitive: for all pairwise distinct $x,y,z$ and pairwise distinct $x',y',z'$ in $\Omega$, there exists a unique $f \in \Gamma$ such that $f(x)=x',$ $f(y)=y',$ and $f(z)=z'.$

In particular if $K$ is a finite field of $q$ elements then $|\Gamma| = (q+1) q (q-1) = q^3-q.$ Happily this agrees with our formulas $|{\rm PGL}_2(K)| = (q^2-1)(q^2-q)$ and $|{\rm GL}_2(K) : {\rm PGL}_2(K)| = |K^\times| = q-1$. For example, if $q=2$ or $q=3$ then $|\Gamma| = (q+1)!$ so the homomorphism $\Omega \to S_\Omega$ is an isomorphism (it’s easy to see that even for finite $K$ the kernel is trivial — you might remember that this is equivalent to the action being “faithful”). For $q = 4$ we have $|\Gamma| = 60 = \frac12 5!$ so the image is an index-2 subgroup of $S_5;$ as we know, such a subgroup must be normal and thus equal to $A_5.$ We next consider the most interesting case, $q=5.$ (There is no field of $6$ elements because $6$ is not a prime power; for $q=7$ and larger, $|\Gamma|$ soon becomes very small compared with $|S_{q+1}|.$)

When $K$ is the field ${\bf Z} / 5 {\bf Z}$ of $5$ elements, the action $\Gamma \to S_\Omega$ makes ${\rm PGL}_2(K)$ a subgroup of $S_6$ of index $6! / 120 = 6.$ So $\Gamma$ has 6 cosets, which $S_6$ permutes as usual, giving a homomorphism $h: S_6 \to S_{|S_6:\Gamma|} = S_6.$ We readily see that $h$ is injective (since we already now that $\ker h$ is a normal subgroup, and that the only normal subgroups of $S_6$ are $\{1\},$ $A_6,$ and $S_6$ itself). This makes $h$ an automorphism of $S_6$. But it cannot be an inner automorphism: the stabilizer of $\Gamma$ (considered as a coset of itself) is $\Gamma,$ which we’ve seen is transitive (indeed triply transitive); while if $h$ were an inner automorphism the stabilizer would be the point-stabilizing $S_5 \lt S_6.$ So we have constructed an outer automorphism of $S_6$ — and also found an isomorphism between $\Gamma = {\rm PGL}_2 ({\bf Z} / 5{\bf Z})$ and $S_5$.

This 3-transitive copy of $S_5$ in $S_6$ also happens to arise in Rubik’s Cube, as the corner permutations accessible by rotations of two adjacent faces. Here is the relevant part of the diagram from Problem 5 in the Eighth problem set, with the six corners now labeled by $\Omega = ({\bf Z} / 5{\bf Z}) \cup \{\infty\}$: $$ \begin{array}{|cc||cc|} \hline {\bf \bar 3} & {\bf \bar 1} & \bar 1 & \bar 0 \cr {\bf \bar 4} & {\bf \bar 2} & \bar 2 & \!\!\infty\!\! \cr \hline \end{array} $$ With this labeling, the clockwise rotation $s$ of the left face is readily seen to act by $s(x) = 2x$ (NB this also takes $\bar 0$ to $\bar 0$ and $\infty$ to $\infty$). The clockwise rotation $t$ of the right face turns out to be $t(x) = (2x-2)/x$ (again also fixing the two points off that face, here $\bar 3$ and $\bar 4$). Thus the subgroup $\langle s,t \rangle$ of $S_6$ accessible by just rotating those two faces is contained in ${\rm PGL}_2({\bf Z} / 5{\bf Z}).$ With a bit more work we can show that in fact $\langle s,t \rangle$ is all of ${\rm PGL}_2({\bf Z} / 5{\bf Z}).$ For example, find elements of order $3$ and $5$ to deduce (as on that midterm problem …) that $\left| \langle s,t \rangle \right|$ is a multiple of $3 \cdot 4 \cdot 5 = 60$ so is either all of ${\rm PGL}_2({\bf Z} / 5{\bf Z})$ or a normal subgroup of index 2; but since ${\rm PGL}_2({\bf Z} / 5{\bf Z}) \cong S_5$ an index-2 subgroup would have to be $\cong A_5$ which is impossible because $A_5$ contains no elements of order 4. (Using the fact that $S_5$ is sharply 3-transitive we then deduce that we cannot use just $s$ and $t$ to switch two corner cubes, with the others ending up where they began.)

Finally, to explain why ${\rm PGL}_2(K)$ has the same structure as fractional linear transformations. In general, we know that ${\rm GL}_n(K)$ acts on the vector space $K^n,$ and the action is almost transitive: the orbits are $\{\vec 0\}$ and everything else. The quotient group ${\rm PGL}_n(K) = {\rm GL}_n(K) / (K^\times I_n)$ does not act on $K^n,$ but it does act on the orbits of $K^\times I_n$. We remove the uninformative $\vec 0;$ the remaining orbits are the nonzero vectors on lines through the origin (a.k.a. one-dimensional subspaces of $K^n$). The set of such lines is called projective space of dimension $n-1$ over $K$, and denoted by ${\bf P}^{n-1}(K).$ NB the dimension is $n-1$ rather than $n$ because we lose one dimension by identifying proportional vectors. In fact if a vector $\vec x = (x_1,\ldots,x_n)^* \in K^n$ [using the star to remind that we’re working with column vectors] has $x_n \neq 0$ then the line through $\vec 0$ and $\vec x$ contains a unique vector whose last coordinate is $1$ (namely $x_n^{-1} \vec x$); this identifies “most” of ${\bf P}^{n-1}(K)$ with $K^{n-1},$ leaving only a “hyperplane at infinity” where $x_n = 0,$ which is a projective space of dimension one less. Multiplication of $n \times n$ matrices by column vectors of length $n$ yields a well-defined action of ${\rm PGL}_n(K)$ on ${\bf P}^{n-1}(K).$

In particular, if $n=2$ then we get a projective line that consists of $K$ together with a “point at infinity” $\infty,$ with the line through the nonzero vector $(x_1,x_2)^*$ identified with $x = x_1/x_2$ if $x_2 \neq 0,$ and with $\infty$ if $x_2 = 0.$ Any $({a\;b\atop c\;d}) \in {\rm GL}_2(K)$ takes $(x_1,x_2)^*$ to $(ax_1+bx_2, cx_1+dx_2)^*.$ The ratio $(ax_1+bx_2) / (cx_1+dx_2)$ must depend only on $x=x_1/x_2,$ and indeed we compute that if $x_2 \neq 0$ then $$ \frac{ax_1+bx_2}{cx_1+dx_2} = \frac{(ax_1+bx_2) / x_2}{(cx_1+dx_2) / x_2} = \frac{a \frac{x_1}{x_2} + b}{c \frac{x_1}{x_2} + d} = \frac{ax + b}{cx + d} $$ which explains the isomorphism beteween $\Gamma$ and ${\rm PGL}_2(K)$: the action of $\Gamma$ on $\Omega$ is just the quotient by $K^\times$ of the action of ${\rm GL}_2(K)$ on $K^2 - \{0\}.$ (We still need to check that our identification also works in the special case $x_2 = 0$ and $x = \infty,$ but that is routine.)

Once $n \gt 2$ the action of ${\rm PGL}_n(K)$ on ${\bf P}^{n-1}(K)$ is still doubly transitive, but not triply transitive. That is because the action must preserve collinearity: three points in ${\bf P}^{n-1}(K)$ are collinear if they come from vectors in $K^n$ that are linearly dependent (so span a vector subspace of dimension 2 rather than 3). Collinearity in ${\bf P}^{n-1}(K)$ (and coplanarity for $n \gt 3$ etc.) gives projective space a geometric structure that was not so easy to see for $n=2.$ For example, the projective plane (i.e. ${\bf P}^{n-1}(K)$ with $n-1=2$, so $n=3)$ has the nice property that any two lines meet in a unique point. Indeed the projective plane is a copy of the ordinary plane $K^2$ together with a projective “line at infinity” $\ell_\infty;$ lines in ${\bf P}^2(K)$ other than $\ell_\infty$ are ordinary lines in $K^2$ together with a “point at infinity“ on $\ell_\infty,$ and if they are parallel in $K^2$ then they have the same point at infinity so they meet there.

Historically ${\bf P}^2(K)$ was first discovered (in the case $K = {\bf R}$) in the Renaissance visual arts, where it provides the mathematical setting for the geometry of perspective. A painting or drawing can reproduce your view of a three-dimensional scene by projecting it from your eye to the plane $\Pi$ of the canvas or paper — hence the term “projective“ for ${\rm PGL}_n$ and ${\bf P}^{n-1}.$ Changing $\Pi$ to some other $\Pi'$ requires a projection from $\Pi$ to $\Pi'$ which (unless $\Pi,\Pi'$ are parallel) misbehaves by sending a line on $\Pi$ to infinity and missing a line of $\Pi'$. Completing $\Pi$ and $\Pi'$ to projective planes cures this misbehavior. You are familiar with this line at infinity as the “horizon” where parallel train tracks seem to meet.