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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.

In differential calculus, the product rule is both simple in form and high in utility.  As such, it is typically presented early on in calculus courses — soon after the linearity of the derivative, in fact.  Moreover, the product rule is easy to derive from first principles:

Theorem (Product Rule): Let $f$ and $g$ be differentiable on the open set $U$. Then $fg$ is differentiable on $U$, and we have

$(fg)'(x)=f(x)g'(x)+g(x)f'(x)$  for all  $x \in U$.

Proof: For $x \in U$, we have (by definition of the derivative)

$\displaystyle(fg)'(x) = \lim_{\epsilon \to 0} \frac{f(x+\epsilon)g(x+\epsilon)-f(x)g(x)}{\epsilon}$

$\displaystyle = \lim_{\epsilon \to 0} \frac{\left(f(x+\epsilon)-f(x)\right)g(x+\epsilon)+f(x)\left(g(x+\epsilon)-g(x)\right)}{\epsilon}$

$\displaystyle = \lim_{\epsilon \to 0} \frac{g(x\!+\!\epsilon)\left(f(x\!+\!\epsilon)-f(x)\right)}{\epsilon}\!+\!\lim_{\epsilon \to 0} \frac{f(x)(g(x\!+\!\epsilon)-g(x))}{\epsilon},$

under the assumption that each of these last two limits exists.  This of course holds, as these limits are $g(x)f'(x)$ and $f(x)g'(x)$, respectively.   $\square$

All in all, then, the product rule is easy to prove and easy to use.  But — and this is of utmost pedagogical importance — is the product rule intuitive? By this proof alone, I would argue not; the manipulation of the numerator is weakly-motivated and our result falls out without reference to more general phenomena.

In this post, we’ll explore the merits of a second proof of the product rule, one that I hope presents a motivated and compelling argument as to why the product rule should look the way it does.