In this post, we discuss a few ways in which the symmetric and alternating groups can be realized as finite collections of self-maps on the Riemann sphere.  Throughout, our group operation will be composition of functions: as such, the maps we choose will necessarily be homeomorphisms of $\mathbb{P}^1$.  Within this broad framework, two classes are of particular interest:

1. The group of biholomorphic maps $f: \mathbb{P}^1 \to \mathbb{P}^1$ (those that respect the structure of $\mathbb{P}^1$ as a Riemann surface). It is well-known that such maps are given by Möbius transformations, i.e. rational functions of the form

$\displaystyle f(z) = \frac{a z +b}{cz +d},$

satisfying $ad-bc \neq 0 \in \mathbb{C}$. The group of Möbius transformations (also known as the Möbius Group and herein denoted $\mathcal{M}$) is naturally isomorphic to $\mathrm{PSL}_2(\mathbb{C})$, the projective (special) linear group, via:

$\displaystyle\frac{az+b}{cz+d} \mapsto \left(\begin{matrix} a & b \\ c & d \end{matrix} \right).$

2. The group of conformal maps $f: \mathbb{P}^1 \to \mathbb{P}^1$, denoted $\mathcal{C}$ for brevity.  To be clear, here we refer to those maps $f$ which preserve unsigned angle measure. (In contrast, some authors require conformal maps to preserve orientation as well.)  We recall the fundamental result that such maps contain the Möbius group as a subgroup of index two.  To be specific, any conformal self-map on $\mathbb{P}^1$ is either biholomorphic (returning to case (1)), or bijective and anti-holomorphic: a biholomorphic function of the complex conjugate $\overline{z}$.

After the fold, we begin a two-part program to calculate the maximal $n$ such that the symmetric group $S_n$ injects into $\mathcal{M}$ (resp. $\mathcal{C}$).  Along the way, we study injections of the alternating group into $\mathcal{M}$, and highlight some exceptional cases in which our injections can be attached to group actions on a finite invariant set.

— PART I (Injections $\varphi: S_n \to \mathcal{M}$)

Our first goal will be to verify the existence of injective maps $\varphi: S_4 \to \mathcal{M}$. While it suffices here to provide a single example, it is infinitely more enlightening to show how such an example may be found.  As a happy consequence, we’ll end up classifying all possible images $\varphi(S_4)$ up to inner automorphism.  To be specific,

Theorem 1.1: There exists an injection $\varphi: S_4 \to \mathcal{M}$. Moreover, this injection is unique (up to inner automorphism on $\mathcal{M}$).

Proof: Recall the group presentation

$S_4 \cong \langle \alpha,\beta \mid \alpha^2 =\beta^3 = (\beta\alpha)^4\rangle.$

Now, suppose that $\varphi: S_4 \to \mathcal{M}$ is given, and let $\sigma,\tau \in \mathcal{M}$ denote the images of $\alpha,\beta$, respectively.  Because $\tau(z) \in \mathrm{PSL}_2(\mathbb{C})$ is elliptic, we may assume $\tau(z)=\lambda z$ (up to inner automorphism on $\mathcal{M}$), in which $\lambda \in \mu_3$ is primitive.  Since $f \in \mathcal{M}$ has order two if and only if $\mathrm{tr} f =0$, $\sigma(z)$ takes the form

$\displaystyle\sigma(z) =\frac{az+b}{cz-a},$

Secondly, the relation $(\tau \sigma)^4(z)=z$ (with plenty of algebra!) forces

$a(\lambda-1)(a^2+2bc+a^2\lambda^2)(cz^2-(a+a\lambda)z-b\lambda)=0,$

as a polynomial in $\mathbb{C}[z]$.  Moreover, since $\tau\sigma$ has order exactly four, we find$a^2+2bc+a^2\lambda^2=0$ (else $\tau \sigma$ has order dividing two). Next, because$\lambda^2+\lambda+1=0$, we can rewrite this as $\lambda(2bc-a^2)=0$, hence $a^2=2bc$. In particular, $a$ is non-zero, by a determinant calculation.  We may thus projectivize $a=1$, such that

(1)                                        $\displaystyle\sigma(z)=\frac{z+1/2c}{cz-1},$

in which we have used that $2bc=1$.  Because the centralizer $C_\mathcal{M}(\tau)$ contains the subgroup of diagonal (projective) matrices, we may freely conjugate $\sigma$ by diagonal elements.  In particular, we note that

$\displaystyle\left(\begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right)\left( \begin{matrix} 1 & 1/2c \\ c & -1 \end{matrix} \right)\left( \begin{matrix} c^{-1} & 0 \\ 0 & 1 \end{matrix} \right)=\left(\begin{matrix} 1 & c^2/2 \\ 1 & -1 \end{matrix} \right),$

hence $\sigma(z)$ is conjugate (via $C_\mathcal{M}(\tau)$) to a Möbius transformation mapping$\infty \mapsto 1$. In other words, we may assume $c=1$ in line (1) – again, up to inner automorphism on $\mathcal{M}$. In this sense, $\sigma$ is uniquely determined by $\tau$, whereby there exist at most two injections $\varphi: S_4 \to \mathcal{M}$ modulo inner automorphism: one for each choice of $\lambda$ among the roots $\lambda_1,\lambda_2$ of $f(t)=t^2+t+1$.

As it happens, each choice of $\lambda$ induces a valid injection $\varphi: S_4 \to \mathcal{M}$ (this is finite computation). As a final claim, we leave to the reader to show that conjugation by the Möbius transformation $f(z)=-1/(2z)$ fixes $\sigma(z)$ while interchanging the choices for $\tau(z)$ associated to $\lambda_1$ and $\lambda_2$. That is, any two injections $\varphi,\varphi': S_4 \to \mathcal{M}$ differ by an inner automorphism of $\mathcal{M}$, precisely as claimed.                                                                                             $\square$

Next, we claim that the constant $n=4$ taken in Theorem 1.1 is maximal. Here, two avenues of proof seem promising:

1. A direct proof (along the lines of Theorem 1.1), built from a manageable group presentation for $S_5$.
2. A proof built upon Theorem 1.1, noting that any injection $\varphi: S_5 \to \mathcal{M}$ restricts to a map on $S_4$ of the type studied in Theorem 1.1.

Both work well.  The first is given below, while the second is sketched in the Exercises:

Theorem 1.2: There exists no injection $\varphi: S_5 \to \mathcal{M}$.

Proof: We begin with the group presentation

$S_5 \cong \langle \alpha,\beta \mid \alpha^2=\beta^5=(\beta \alpha)^4=[\beta,\alpha]^3 \rangle,$

known at the time of Burnside’s article Note on the Symmetric Group (1897). If $\varphi$ exists, let $\sigma,\tau \in \mathcal{M}$ denote the images of $\alpha,\beta$, respectively.  For definiteness, we may assume that $\tau(z)=\lambda z$ (up to inner automorphism). For$f \in \mathcal{M}$, we recall that $\mathrm{tr} f =0$ if and only if $\mathrm{ord} f =2$.  Thus $\sigma(z)$ takes the form

$\displaystyle\sigma(z)=\frac{az+b}{cz-a},$

and some computer-assisted algebra quickly gives us the following relations:

$(\tau\sigma)^4: \quad a^2+2bc\lambda+a^2 \lambda^2=0;$

$[\tau,\sigma]^3: \quad (bc+bc\lambda^2+a^2 \lambda-bc\lambda)(bc+bc\lambda^2+3a\lambda^2+bc\lambda)=0.$

The first of these implies $a^2(1-\lambda)^2=2\lambda \det \sigma$, whereas the second yields

$\lambda^2(\det \sigma)^2 = (2a^2\lambda + bc(1+\lambda^2))^2=((1+\lambda^2)(a^2+\det \sigma)-2a^2\lambda)^2$

$= (a^2(1-\lambda)^2+(1+\lambda^2)\det \sigma)^2=(2\lambda \det \sigma + (1+\lambda^2)\det \sigma)^2$

$= (1+\lambda)^4 (\det \sigma)^2.$

Since $\det \sigma \neq 0$, we obtain $(1+\lambda)^4=\lambda^2$, which contradicts that $\lambda^5=1$.   $\square$

— PART II (Injections $\varphi: S_n \to \mathcal{C}$

The questions answered in the previous section admit natural analogues with $\mathcal{C}$ in place of $\mathcal{M}$.  Answering them, however, will be a bit harder than before, owing to the greater complexity of the group structure on $\mathcal{C}$.  For this reason, we try when possible to reduce questions regarding $\mathcal{C}$ to questions of $\mathcal{M}$alone.

For example, the identification $\mathcal{M} \cong \mathrm{PSL}_2(\mathbb{C})$ induces a semi-direct product structure on $\mathcal{C}$; namely,

$\mathcal{C} \cong \mathrm{PSL}_2(\mathbb{C}) \rtimes \mathbb{Z}/2\mathbb{Z}$,

in which the $\mathbb{Z}/2\mathbb{Z}$-action is given by complex conjugation. It follows that any injective map $\varphi: S_n \to \mathcal{C}$ induces a restricted map $\varphi: A_n \to \mathcal{M}$, i.e. an injection of the alternating group $A_n$ into the Möbius group. To see how this might benefit us, consider the following:

Proposition 2.1: There exists no injection $\varphi: A_6 \to \mathcal{M}$.  It follows that $S_6$ does not inject into $\mathcal{C}$.

Proof: See the Exercises.                                                                         $\square$

As it turns out, there does exists an injection of $A_5$ into $\mathcal{M}$, which is maximal in the sense of the preceding Proposition.  Unfortunately, this fact does not a priori imply that $S_5$ injects into $\mathcal{C}$ (so Proposition 2.1 won’t serve a major role in our classification of symmetric subgroups).  In fact, the opposite is true:

Theorem 2.2: There exists no injection $\varphi: S_5 \to \mathcal{C}$.

Proof: As in Theorem 1.2, we begin with the group presentation

$S_5 \cong \langle \alpha,\beta \mid \alpha^2=\beta^5=(\beta \alpha)^4=[\beta,\alpha]^3\rangle$.

Suppose that $\varphi : S_5 \to \mathcal{C}$ exists.  Since $\beta$ has cycle type $(5)$, it follows that$\beta \in A_5$, hence $\tau:=\varphi(\beta)$ lies in $\mathcal{M}$. Regarded as an element of $\mathrm{PSL}_2(\mathbb{C})$,$\tau(z)$ is torsion (hence elliptic), so we may assume that $\tau(z) =\lambda z$, in which$\lambda^5=1$ (up to inner automorphism).  As $\sigma(z) \notin \mathcal{M}$ by Theorem 1.2, we may write

$\displaystyle\sigma(z) = \overline{z} \circ \frac{az+b}{cz+d}$

(in general form).  The relation $\sigma^2(z)=z$ holds if and only if

(2)           $\overline{a}b+\overline{b}d=0$,    $\vert a \vert^2-\vert d \vert^2=b \overline{c}-\overline{b}c$,    and   $a \overline{c}+c\overline{d}=0$.

(Case 1): If $c=0$, we may scale $a=1$; then $b+\overline{b}d=0$ and $\vert d \vert =1$.  It follows that $b^2=-d \vert b \vert^2$, in which $\vert b \vert =0$ or $\vert b \vert =1$. If $b=0$, we then have$\sigma(z)=\overline{z/d}$, which implies that $(\tau \sigma)^2(z)=z$, a contradiction.  Thus $\vert b \vert=1$, and we may write

$\displaystyle\sigma(z) = \overline{z} \circ \frac{z+b}{-b^2}.$

On the other hand, that $\mathrm{ord}(\tau \sigma) =4$ forces $\vert b \vert^4+\vert \lambda \vert^2 =0$, an impossibility. This case is therefore untenable, and we turn to:

(Case 2): If $c\neq 0$, we may projectivize to assume $c=1$; then $a+\overline{d}=0$ and $\vert a \vert^2-\vert d \vert^2=b-\overline{b}$. This last expression is both real and (purely) imaginary, hence $0$. Thus $b \in \mathbb{R}$, and $\sigma(z)$ takes the form

$\displaystyle\sigma(z)= \overline{z} \circ \frac{a z+b}{z-\overline{a}}.$

As for the relation $(\tau \sigma)^4(z)=z$, it follows that $2\vert a \vert^2+\lambda \overline{b}+b \overline{\lambda}=0$.  Since$b \in \mathbb{R}$, this implies $b=-\vert a \vert^2/\mathrm{Re}(\lambda)$.  After substitutiting this into our expression for $\sigma$, we find that the relation $[\tau,\sigma]^3(z)=z$ forces

$(\lambda-1)^4(1+\lambda+\lambda^2)(1+3\lambda+\lambda^2)=0.$

Coupled with $\lambda^5=1$, we find $\lambda=1$, but this contradicts that $\tau(z)$ has order five.  Thus there exists no injection $\varphi: S_5 \to \mathcal{C}$.                                        $\square$

We note that Theorem 2.2 recovers Theorem 1.2. Regardless, Theorem 2.2 can’t be used to prove Theorem 1.2, because we have invoked that first theorem in assuming $\sigma \notin \mathcal{M}$. Avoiding such circularity would require — in essence — reproving Theorem 1.2 as an early claim.

Corollary 2.3: There exists an injection $\varphi: S_n \to \mathcal{C}$ if and only if $n \leq 4$.

— PART III (Group Actions and Invariant Sets)

We’ve seen that $\mathcal{C}$ contains no symmetric subgroups $S_n$ beyond those that inject into $\mathcal{M}$ , despite the fact that $\mathcal{C}$ contains $\mathcal{M}$ properly as an index two subgroup. Why, then, might we be interested in subgroups $\varphi(S_n) \subset \mathcal{C}$ not contained in $\mathcal{M}$?

For starters, let $\varphi : S_n \to \mathcal{C}$ be any injection, and consider the sequence of maps

$f : S_n \stackrel{\varphi}{\longrightarrow} \mathcal{C} \stackrel{\pi}{\longrightarrow} \mathcal{C}/\mathcal{M} \cong \mathbb{Z}/2\mathbb{Z},$

in which $\pi$ denotes the quotient of $\mathcal{C}$ by its normal subgroup $\mathcal{M}$.  As mentioned previously, we have $A_n \subset \ker f$. (In particular, we cannot have$\ker f =V_4$, Klein’s four group, despite the fact that $V_4$ is normal in $S_4$.)  Thus$A_n = \ker f$ if and only if $\varphi(S_n) \not\subset \mathcal{M}$.  When these equivalent conditions hold, membership in $\mathcal{M}$ is detected by the sign character on $S_n$. In this one sense, the theory of symmetric subgroups in $\mathcal{C}$ is more robust than the analogue theory in $\mathcal{M}$.

There is a second, much more compelling reason to study symmetric subgroups in $\mathcal{C}$, which begins with the following construction:

Let $K \subset \mathbb{P}^1$ be any finite set, $\# K =n$, and suppose that $H \leq \mathcal{C}$ is a subgroup of conformal mappings such that each $\sigma \in H$ permutes the points of $K$.  (In other words, $K$ is an invariant set for the action of $H$.)  This gives an induced map $\rho : H \to S_n$, known as the permutation representation (associated to the group action $H \circlearrowright K$).  Our interest in such maps is simple: if $\rho: H \to S_n$ bijects, then $\rho^{-1}: S_n \to \mathcal{C}$ is exactly the sort of map we’ve been looking for.

If this permutation representation bijects, then our general theory implies$n \leq 4$. Of these $n$, the extremal case $n=4$ naturally presents the greatest interest.  Here, there are two cases to consider:

1. All points in $K$ lie on a circle in $\mathbb{P}^1$.  Up to inner automorphism, we may assume that $K=\{0,1,\infty,z_0\}$, with $z_0 \in \mathbb{R}$.
2. The points in $K$ lie on no common circle, but we may assume after conjugation that $K$ is given by $\{0,1,\infty,z_0\}$, with $z_0 \notin \mathbb{R}$.

If case (1) holds, note that we may assume $H \subset \mathcal{M}$, by modding out by the trivial action of $\overline{z}$.  As it turns out, however, case (1) cannot actually occur:

Example 3.1: Suppose that $\rho: H \to S_4$ surjects, and that case (1) holds. We may assume $H \subset \mathcal{M}$. If $\tau(z)$ corresponds to the 3-cycle $(0\,1\,\infty)$, brief computation shows that

$\displaystyle\tau(z)=\frac{1}{1-z}.$

Since $\tau(z)$ fixes $z_0$, we must have $z_0^2-z_0+1=0$. But this polynomial has no real roots, which contradicts that $z_0 \in \mathbb{R}$. It follows that case (1) cannot occur.

Note: If case (2) can be realized, this gives strong motive to consider injections of $S_n$ into $\mathcal{C}$, as opposed to injections into the smaller group $\mathcal{M}$.  As it happens, case (2) is realized, in an essentially unique way.

To prove this result, it will be advantageous to present first a general Lemma. For the moment, let’s relax our assumptions and suppose that the permutation representation $\rho: H \to S_4$ merely surjects.  By the previous Example, it follows that the invariant set $K= \{0,1,\infty,z_0\}$ has $z_0 \notin \mathbb{R}$ (after inner automorphism).  As the following (general) Lemma shows, the map $\rho$ injects without further hypotheses:

Lemma 3.2: Suppose that $K$ contains four points not lying on a circle in $\mathbb{P}^1$. If $\psi(z) \in \mathcal{C}$ fixes each point in $K$, then $\psi(z)=z$.  In other words, the permutation representation $\rho : H \to S_n$ injects.

Proof: Suppose that a non-identity element $\psi \in \mathcal{C}$ fixes $K$. Conjugating $\psi$ by an element of $\mathcal{M}$, we may assume that $K$ contains $\{0,1,\infty,z_0\}$, with$z_0 \in \mathbb{C} \smallsetminus \mathbb{R}$.  We note that $\psi \notin \mathcal{M}$, as $\mathcal{M}$ acts sharply 3-transitively on $\mathbb{P}^1$. Thus, writing $\psi(z) = \overline{z} \circ \sigma(z)$ with $\sigma \in \mathcal{M}$, it follows that $\sigma$ fixes the three points $0,1$, and $\infty$.  Because $\mathcal{M}$ acts sharply 3-transitively on $\mathbb{P}^1$, we have$\sigma(z)=z$.  Thus $\psi(z)=\overline{z}$, which fixes $\mathbb{R} \cup \{\infty\}$ (and nothing else).  This contradicts that $\psi$ fixes $z_0$, whence no such $\psi$ exists.                                  $\square$

And now, our final Theorem:

Theorem 3.3: There exists a subgroup $H \leq \mathcal{C}$ and an invariant set $K$,$\# K=4$, such that the permutation representation $\rho : H \to S_4$ bijects.  If $H'$ and $K'$ are as $H$ and $K$ above, there exists an element $\psi \in \mathcal{C}$ such that$H' = \psi H \psi^{-1}$ and $K'=\psi K$.  It follows that the associated injection$\rho^{-1} : S_4 \to \mathcal{C}$ is unique up to inner automorphism.

Proof: Suppose that $H$ and $K$ exist as above.  Up to inner automorphism by$\mathcal{M}$, we may assume that $K=\{0,1,\infty,z_0\}$, with $z_0 \in \mathbb{C} \smallsetminus \mathbb{R}$.  Moreover, Example 3.1 implies that $z_0$ is a root of $t^2-t+1$.  In the group presentation

$S_4 \cong \langle \alpha,\beta \mid \alpha^2=\beta^3=(\beta \alpha)^4 \rangle,$

we may take $\beta \leftrightarrow \tau(z)$, wherein $\tau(z)$ is as in Example 3.1.  Since $H \not\subset \mathcal{M}$ the image of $\alpha$ ($\sigma$, say) is anti-holomorphic, hence a simple transposition (by order).  Moreover, $\sigma$ can be chosen to fix $(0,\infty)$ while transposing $(1,z_0)$, as$S_4$ is complete.  Thus $\sigma(z)$ takes the form $\sigma(z)=z_0 \overline{z}$, for which we note that$\rho: \langle \sigma,\tau \rangle \to S_4$ surjects.

By Lemma 3.2, we have $H= \langle \sigma,\tau \rangle$.  As such, $H$ is uniquely determined by$K$, and $K$ is determined by choice of $z_0$ (among the roots of $t^2-t+1$). These two choices of $K$ are related by conjugation by $\overline{z} \in \mathcal{C}$, which gives uniqueness of $H,K$ (in the sense of this Theorem).  It follows that$\rho^{-1}:S_4 \to \mathcal{C}$ is unique up to inner automorphism, as claimed.                    $\square$

— EXERCISES —

Exercise: Herein, we outline a second proof of Theorem 1.2.  Assuming that$\varphi : S_5 \to \mathcal{M}$ injects, let $H \leq S_5$ denote any subgroup isomorphic to $S_4$.  By Theorem 1.1, we may assume that $H$ is generated by

$\displaystyle\sigma(z)= \frac{z+1/2}{z-1}$   and    $\tau(z) = \lambda z$,

in which $\lambda$ is some primitive cube root of unity.  Show that $\varphi(S_5)$ contains an element $\nu \notin H$ that commutes with $\sigma(z)$ such that $\mathrm{ord}(\nu)=2$ and$\mathrm{ord}(\tau \nu)=4$.  Use these relations to find a contradiction.

Exercise: Using the group presentation

$A_5 \cong \langle \alpha, \beta \mid \alpha^2=\beta^3=(\alpha\beta)^5 \rangle$,

show that $\mathcal{M}$ contains a subgroup isomorphic to $A_5$ (which is unique up to inner automorphism).  Then, using the group presentation

$A_6 \cong \langle \alpha,\beta \mid \alpha^2=\beta^4 =(\alpha\beta)^5=(\alpha\beta^2)^5 \rangle$,

show that no injection $\varphi : A_6 \to \mathcal{M}$ exists.