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 $120$.

  • $\begingroup$ I guess this is the sort of thing that GAP can answer, given they're finite groups. $\endgroup$ – Dan Rust Jan 28 '14 at 15:29

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.


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.

  • 1
    $\begingroup$ +1! Just to note that non-isomorphic groups may have equal structure descriptions. Of course, this is not the case in the example above since it returns SL(2,5), but in general one should remember that StructureDescription only provides an "informal" overview of the structure of a group. More sophisticated functions to check isomorphism of groups are IdGroup; the StandardPresentation function of the ANUPQ package; and IsomorphismGroups (see gap-system.org/Faq/faq.html#7.12) $\endgroup$ – Alexander Konovalov Jan 29 '14 at 0:36

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