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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.

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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.

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