Fundamental group of Poincaré sphere

Question

Do the two presentations below, $$G=\langle d,v \mid dv^2d=vdv, dv^3d=v^2 \rangle$$ and $$\langle r,s,t \mid r^2=s^3=t^5=rst \rangle = \langle s,t \mid (st)^2=s^3=t^5 \rangle,$$ define the same group?

Motivation: I am working on Poincaré homology sphere $X$, constructed by identifying the opposite faces of a dodecahedron using the minimal clockwise twist to line up the faces. I was able to verify that its homology groups are the same as the 3-sphere, and now I would like to compute its fundamental group. Using van Kampen theorem, I found the first presentation for $\pi_1(X)$; however, I did not succeed in identifying it with the binary icosahedral group (hoping I computed correctly the fundamental group), given by the second presentation.

Nota Bene: Using a mathematical software, I checked that $G$ has order $120$.

Answer 1

The first presentation I found, before simplifying it, was $$ G= \langle d,v \mid vd^2vd^{-1}v^{-1}d^{-1}=1, dvd^{-1}vdv^{-1}=1 \rangle. \hspace{1cm} (\ast)$$

Writing the second relation as $dvd^{-1}=vd^{-1}v^{-1}$ $(1)$, the first one becomes $$1=vd(dvd^{-1})v^{-1}d^{-1} = vdvd^{-1}v^{-2}d^{-1}$$ hence $vdv=dv^{2}d$. Multiplying $(1)$ by $v$ on the left, $$v^2d^{-1}v^{-1}=(vdv)d^{-1}=(dv^2d)d^{-1}=dv^2$$ hence $dv^3d=v^2$, so one find the presentation of $G$ I gave in my question $$\langle v,d \mid dv^2d=vdv, dv^3d=v^2 \rangle.$$

Reordering the relations of $(\ast)$, we get $$G= \langle d,v \mid d^{-1}v^{-1}d^{-1}v \cdot d^2v=1, v^{-1}dvd^{-1}vd=1 \rangle.$$

If we introduce the second relation into the first one where the dot is, we get $$1= d^{-1}v^{-1}d^{-1}v (v^{-1}dvd^{-1}vd ) d^2v= d^{-2}vd^3v=1,$$ so $$G= \langle d,v \mid d^{-2} vd^3v=1, v^{-1}dvd^{-1}vd=1 \rangle.$$

Setting $v=da$, the relations become $$d^4=a^{-1}da^{-1} \ \text{and} \ a^2=d^{-1}ad^{-1};$$ setting $b=d^{-1}$, the relations become $$b^4=aba \ \text{and} \ a^2=bab.$$ Multiplying the fist relation by $b$ on the right, and the second relation by $a$ on the right, we finally get a presentation of the binary icosahedral group: $$G= \langle a,b \mid a^3=b^5=(ab)^2 \rangle.$$

The main argument comes from A Textbook of Toplogy, Seifert & Threlfall, pp. 224-225.

Answer 2

The answer is yes. I don't have a written proof, but you can check the isomorphism with the following GAP code.

F2 := FreeGroup("d", "v");;
G := F2 / [F2.d*F2.v^2*F2.d*(F2.v*F2.d*F2.v)^-1, F2.d*F2.v^3*F2.d*F2.v^-2];;

F3 := FreeGroup("r", "s", "t");;
H := F3 / [F3.r^2*F3.s^-3, F3.r^2 * F3.t^-5, F3.r^-2 * F3.r * F3.s * F3.t];;

Using the GAP command IdGroup or StructureDescription, you can check that both $G$ and $H$ here are isomorphic to $\operatorname{SmallGroup}(120,5)$ which is the binary icosahedral group.