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


as a polynomial in \mathbb{C}[z].  Moreover, since \tau\sigma has order exactly four, we finda^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). Forf \in \mathcal{M}, we recall that \mathrm{tr} f =0 if and only if \mathrm{ord} f =2.  Thus \sigma(z) takes the form


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.  Sinceb \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


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.

For free, we obtain

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.)  ThusA_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 impliesn \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


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\}, withz_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 thatH' = \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), asS_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 byK, 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


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.