Poincaré dodecahedron space The Poincaré homology sphere $X$ is constructed by identifying opposite faces of a dodecahedron by a (clockwise) twist of 36 degree. 
Many books say its fundamental group $\pi_1(X)$ is the binary icosahedron group, my question is how to prove this. In Ratcliffe's hyperbolic manifold book, there is a theorem of Poincaré about computing $\pi_1$ of such spaces, following this method one can check $\pi_1(X)$ has a presentation of $6$ generators, $a, b, c, d, e, f$ and relations $a=bd=ce=df=eb=fc$ and $f=be$, how to show this is isomorphic to the 120-order icosahedron group? Is there any easy way to see this? Thanks!
 A: The  presentation for $\pi_1(X)$ that you write down simplifies to $\langle b, e \mid beb=eb^2e, ebe=be^2b\rangle$. As is well-known, the binary icosahedral group $I^*$ is isomorphic to the group of unimodular, $2\times 2$ matrices over $\mathbb{Z}_5$. An explicit isomorphism $\pi_1(M) \to {\rm SL}(2,5)$ is given by
$$
b \mapsto \begin{pmatrix} 3 & 1 \\\\ -1 & 0 \end{pmatrix}, \quad 
e \mapsto \begin{pmatrix} -1 & -1 \\\\ 0 & -1 \end{pmatrix}.
$$ 
A: You can read the book by Seifert and Threlfall "A textbook of topology", pages 223-225: They 
start by writing down a presentation of $\pi_1$ of the Poincare homology sphere (by reading off generators and relators from the identification of faces of the spherical dodecahedron). Then they describe how to transform this presentation to the one of the icosahedral group. No computers are required, just some patience with their somewhat archaic style.  
A: A quite direct argument is to solve exercise 4.4.17 in page 252 of Thurston's 
book 3-D Geom & Topo - the published one in 1997; in this exercise Thurston 
considered the action of the icosahedral group on the unit circle bundle $UTS^2=RP^3$; 
this is a free action, and the point is to describe a fundamental domain and show that 
the pattern of gluing of its boundary is exactly the same as in the Poincare dodecahedral 
space. When doing this exercise, 
one may want to notice two things: 1. On $S^2$, each of the $5$ angles of a 
spherical pentagon face of a dodecahedron is $120$ degrees, in contrast to 
$108$ degrees for a pentagon on plane - this is useful when you try to create a local coordinate system on the circle bundle $UTS^2$; 2. The 
fundamental domain Thurston created here does not look like a dodecahedron 
in the metric sense - it look like a cylinder over a pentagon base; however 
other fundamental domains will mark a line, with quite a slope, on each 
of the five side faces and therefore each side face splits into two "pentagons"... 
With these in mind, exercise 4.4.17 should be very pleasant to work out.     
A: Here we take a 3D pentagon instead of a "spherical pentagon" - does it correspond to Thurston's idea of the fundamental domain?
http://quantumcinema.uni-ak.ac.at/site/research/current-topics/
