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 . Within this broad framework, two classes are of particular interest:
1. The group of biholomorphic maps (those that respect the structure of as a Riemann surface). It is well-known that such maps are given by Möbius transformations, i.e. rational functions of the form
satisfying . The group of Möbius transformations (also known as the Möbius Group and herein denoted ) is naturally isomorphic to , the projective (special) linear group, via:
2. The group of conformal maps , denoted for brevity. To be clear, here we refer to those maps 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 is either biholomorphic (returning to case (1)), or bijective and anti-holomorphic: a biholomorphic function of the complex conjugate .
In this post, we enact a two-part program to calculate the maximal such that the symmetric group injects into (resp. ). Along the way, we study injections of the alternating group into , and highlight some exceptional cases in which our injections can be attached to group actions on a finite invariant set.
— PART I (Injections ) —
Our first goal will be to verify the existence of injective maps . 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 up to inner automorphism. To be specific,
Theorem 1.1: There exists an injection . Moreover, this injection is unique (up to inner automorphism on ).
Proof: Recall the group presentation
Now, suppose that is given, and let denote the images of , respectively. Because is elliptic, we may assume (up to inner automorphism on ), in which is primitive. Since has order two if and only if , takes the form
Secondly, the relation (with plenty of algebra!) forces
as a polynomial in . Moreover, since has order exactly four, we find (else has order dividing two). Next, because, we can rewrite this as , hence . In particular, is non-zero, by a determinant calculation. We may thus projectivize , such that
in which we have used that . Because the centralizer contains the subgroup of diagonal (projective) matrices, we may freely conjugate by diagonal elements. In particular, we note that
hence is conjugate (via ) to a Möbius transformation mapping. In other words, we may assume in line (1) – again, up to inner automorphism on . In this sense, is uniquely determined by , whereby there exist at most two injections modulo inner automorphism: one for each choice of among the roots of .
As it happens, each choice of induces a valid injection (this is finite computation). As a final claim, we leave to the reader to show that conjugation by the Möbius transformation fixes while interchanging the choices for associated to and . That is, any two injections differ by an inner automorphism of , precisely as claimed.
Next, we claim that the constant taken in Theorem 1.1 is maximal. Here, two avenues of proof seem promising:
- A direct proof (along the lines of Theorem 1.1), built from a manageable group presentation for .
- A proof built upon Theorem 1.1, noting that any injection restricts to a map on 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 .
Proof: We begin with the group presentation
known at the time of Burnside’s article Note on the Symmetric Group (1897). If exists, let denote the images of , respectively. For definiteness, we may assume that (up to inner automorphism). For, we recall that if and only if . Thus takes the form
and some computer-assisted algebra quickly gives us the following relations:
The first of these implies , whereas the second yields
Since , we obtain , which contradicts that .
— PART II (Injections ) —
The questions answered in the previous section admit natural analogues with in place of . Answering them, however, will be a bit harder than before, owing to the greater complexity of the group structure on . For this reason, we try when possible to reduce questions regarding to questions of alone.
For example, the identification induces a semi-direct product structure on ; namely,
in which the -action is given by complex conjugation. It follows that any injective map induces a restricted map , i.e. an injection of the alternating group into the Möbius group. To see how this might benefit us, consider the following:
Proposition 2.1: There exists no injection . It follows that does not inject into .
Proof: See the Exercises.
As it turns out, there does exists an injection of into , which is maximal in the sense of the preceding Proposition. Unfortunately, this fact does not a priori imply that injects into (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 .
Proof: As in Theorem 1.2, we begin with the group presentation
Suppose that exists. Since has cycle type , it follows that, hence lies in . Regarded as an element of , is torsion (hence elliptic), so we may assume that , in which (up to inner automorphism). As by Theorem 1.2, we may write
(in general form). The relation holds if and only if
(2) , , and .
(Case 1): If , we may scale ; then and . It follows that , in which or . If , we then have, which implies that , a contradiction. Thus , and we may write
On the other hand, that forces , an impossibility. This case is therefore untenable, and we turn to:
(Case 2): If , we may projectivize to assume ; then and . This last expression is both real and (purely) imaginary, hence . Thus , and takes the form
As for the relation , it follows that . Since, this implies . After substitutiting this into our expression for , we find that the relation forces
Coupled with , we find , but this contradicts that has order five. Thus there exists no injection .
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 . 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 if and only if .
— PART III (Group Actions and Invariant Sets) —
We’ve seen that contains no symmetric subgroups beyond those that inject into , despite the fact that contains properly as an index two subgroup. Why, then, might we be interested in subgroups not contained in ?
For starters, let be any injection, and consider the sequence of maps
in which denotes the quotient of by its normal subgroup . As mentioned previously, we have . (In particular, we cannot have, Klein’s four group, despite the fact that is normal in .) Thus if and only if . When these equivalent conditions hold, membership in is detected by the sign character on . In this one sense, the theory of symmetric subgroups in is more robust than the analogue theory in .
There is a second, much more compelling reason to study symmetric subgroups in , which begins with the following construction:
Let be any finite set, , and suppose that is a subgroup of conformal mappings such that each permutes the points of . (In other words, is an invariant set for the action of .) This gives an induced map , known as the permutation representation (associated to the group action ). Our interest in such maps is simple: if bijects, then is exactly the sort of map we’ve been looking for.
If this permutation representation bijects, then our general theory implies. Of these , the extremal case naturally presents the greatest interest. Here, there are two cases to consider:
- All points in lie on a circle in . Up to inner automorphism, we may assume that , with .
- The points in lie on no common circle, but we may assume after conjugation that is given by , with .
If case (1) holds, note that we may assume , by modding out by the trivial action of . As it turns out, however, case (1) cannot actually occur:
Example 3.1: Suppose that surjects, and that case (1) holds. We may assume . If corresponds to the 3-cycle , brief computation shows that
Since fixes , we must have . But this polynomial has no real roots, which contradicts that . It follows that case (1) cannot occur.
Note: If case (2) can be realized, this gives strong motive to consider injections of into , as opposed to injections into the smaller group . 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 merely surjects. By the previous Example, it follows that the invariant set has (after inner automorphism). As the following (general) Lemma shows, the map injects without further hypotheses:
Lemma 3.2: Suppose that contains four points not lying on a circle in . If fixes each point in , then . In other words, the permutation representation injects.
Proof: Suppose that a non-identity element fixes . Conjugating by an element of , we may assume that contains , with. We note that , as acts sharply 3-transitively on . Thus, writing with , it follows that fixes the three points , and . Because acts sharply 3-transitively on , we have. Thus , which fixes (and nothing else). This contradicts that fixes , whence no such exists.
And now, our final Theorem:
Theorem 3.3: There exists a subgroup and an invariant set ,, such that the permutation representation bijects. If and are as and above, there exists an element such that and . It follows that the associated injection is unique up to inner automorphism.
Proof: Suppose that and exist as above. Up to inner automorphism by, we may assume that , with . Moreover, Example 3.1 implies that is a root of . In the group presentation
we may take , wherein is as in Example 3.1. Since the image of (, say) is anti-holomorphic, hence a simple transposition (by order). Moreover, can be chosen to fix while transposing , as is complete. Thus takes the form , for which we note that surjects.
By Lemma 3.2, we have . As such, is uniquely determined by, and is determined by choice of (among the roots of ). These two choices of are related by conjugation by , which gives uniqueness of (in the sense of this Theorem). It follows that is unique up to inner automorphism, as claimed.
— EXERCISES —
Exercise: Herein, we outline a second proof of Theorem 1.2. Assuming that injects, let denote any subgroup isomorphic to . By Theorem 1.1, we may assume that is generated by
in which is some primitive cube root of unity. Show that contains an element that commutes with such that and. Use these relations to find a contradiction.
Exercise: Using the group presentation
show that contains a subgroup isomorphic to (which is unique up to inner automorphism). Then, using the group presentation
show that no injection exists.