How did Dehn prove that the trefoil is chiral? I understand that, in 1914, Dehn proved that the trefoil knot is chiral (it doesn't equal its mirror image). However, the paper in which he does so is in German, and I can't find a description of his proof anywhere else. The only proof I know of its chirality is through the Jones polynomial/Kauffman bracket or something stronger, but those were discovered decades later. So how did Dehn do it, with the tools available to him at the time?
 A: Here's a translation: Google books or Springer-Verlag.  See page 200 for the translators notes or 203 for the beginning of the paper.
A very loose summary: with $K$ a trefoil knot, Dehn embeds a (quotient of a) Cayley graph of $G=\pi_1(S^3-K)$ into $\mathbb{H}^2\times\mathbb{R}$ to determine the group of outer automorphisms of $G$, and the group ends up being isomorphic to $\mathbb{Z}/2\mathbb{Z}$.  Using a peripheral system (i.e., oriented meridian and longitude loops), he shows the non-trivial outer automorphism reverses the orientation of the ambient space.

The appendix of the translation contains a more streamlined proof for general torus knots, yet still with the same result about automorphism groups; here I'll give a specialization to the trefoil.
One presentation of the trefoil knot group is $\langle x,y\mid x^2=y^3\rangle$ (Hatcher ch.1 has a proof; it follows from the van Kampen theorem by thinking of the trefoil as being on the surface of a torus).  The element $x^2$ is certainly in the center, and $G/\langle x^2\rangle$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/3\mathbb{Z}$, presented by $\langle\overline{x},\overline{y}\mid\overline{x}^2,\overline{y}^3\rangle$.  Since this quotient has no center, the center of $G$ is $Z(G)=\langle x^2\rangle$.
Let $f:G\to G$ be an automorphism.  The center is mapped onto itself, so we get an induced map $\overline{f}:G/Z(G)\to G/Z(G)$. We see $\overline{f}(\overline{x})^2=1$ and $\overline{f}(\overline{y})^3=1$.  By considering reduced words, there are $\overline{s},\overline{t}\in G/Z(G)$ such that $\overline{f}(\overline{x})=\overline{s}\overline{x}\overline{s}^{-1}$ and $\overline{f}(\overline{y})=\overline{t}\overline{y}^{\pm 1}\overline{t}^{-1}$. (I think reduced words can be substituted for thinking of $G/Z(G)$ acting on its Cayley graph.)  Any choice of $\overline{s},\overline{t}$ and exponent for $\overline{y}$ gives an endomorphism for $G/Z(G)$.  Suppose $\overline{f}$ is an automorphism.  By composing with an inner automorphism, we may assume $\overline{f}(\overline{x})=\overline{u}\overline{x}\overline{u}^{-1}$ and $\overline{f}(\overline{y})=\overline{y}^{\pm 1}$ for some $\overline{u}$.  Since it is an automorphism, $\overline{x}$ is generated by these two elements, and by considering words it follows that $\overline{u}=1$.  Hence the automorphisms are given by $\overline{f}(\overline{x})=\overline{s}\overline{x}\overline{s}^{-1}$ and $\overline{f}(\overline{y})=\overline{s}\overline{y}^{\pm 1}\overline{s}^{-1}$ for $\overline{s}\in G/Z(G)$.
The center is isomorphic to $\mathbb{Z}$, so the automorphism group of $Z(G)$ is generated by $x^2\mapsto x^{-2}$.  Again, the center is mapped onto itself, so $f(x)^2=x^{2\varepsilon}$, where $\varepsilon=\pm 1$.  By the group relation, $x^{2\varepsilon}=f(x)^2=f(y)^3=y^{3\varepsilon}$.  By the result about automorphisms of $G/Z(G)$, there are integers $h,k$ and an element $z\in G$ such that $f(x)=zx^{1+2h}z^{-1}$ and $f(y)=zy^{s+3k}z^{-1}$ where $s=\pm 1$.  Since $x$ and $y$ have infinite order, it follows that $\varepsilon=1+2h$ and $\varepsilon=s+3k$, so $f(x)=zx^{\varepsilon}z^{-1}$ and $f(y)=zy^{\varepsilon}z^{-1}$.
Therefore, the group of outer automorphisms is generated by an $f:G\to G$ defined by $f(x)=x^{-1}$ and $f(y)=y^{-1}$.
Consider an orientation-preserving homeomorphism $F:S^3\to S^3$ that carries $K$ to $\overline{K}$ (the reflection of $K$), and let $F':S^3\to S^3$ be a reflection, so $F'\circ F:S^3\to S^3$ carries $K$ to itself.  The composition carries a longitude to a longitude, but reverses the orientation of a meridian.  It also induces an automorphism $f:G\to G$.  Compose $f$ with an inner automorphism so that $z=1$ in the above notation.
A peripheral system for $K$ is given by $\lambda=x^2$ for a longitude and $\mu=y^{-1}x$ for a meridian.  Then, $f(\lambda)=x^{\pm 2}$, and since we asserted $f$ sends a longitude to a longitude of the same orientation, $f(\lambda)=\lambda$.  Thus, $f(y)=y$ so $f(\mu)=\mu$, contradicting the fact that $f$ needs to reverse the orientation of $\mu$.

In Lickorish's book, exercise 11.10 is to reproduce Dehn's argument, however Lickorish does not seem to cite Dehn when he describes the idea of having the fundamental group act on hyperbolic space, and he instead cites Trotter for a result about certain pretzel knots not being equivalent to their reverses.
