# Quotient of group given by presentation is finite

Consider the group $$G = \langle a,b,c ~ \mid ~ a^2 = b^3 = c^5 = abc \rangle$$ Prove that $\mathbf{(a)}$ $abc$ is an element of the center of $G$; and $\mathbf{(b)}$ $G/ \langle abc \rangle$ is a finite group.

To prove $\mathbf{(a)}$, we can show that $abc$ commutes with the generators $a,b,c$: $$a(abc) = a(a^2) = a^3 = a^2(a) = abc(a)$$ $$b(abc) = b(b^3) = b^4 = b^3(b) = abc(b)$$ $$c(abc) = c(c^5) = c^6 = c^5(c) = abc(c)$$

But I'm not sure how to prove part $\mathbf{(b)}$... any hints would be appreciated.

• Interesting fact: A group given by a presentation of the form $\langle a, b, c; a^i, b^j, c^k, abc\rangle$ is called a triangle group, and is finite if and only if $1/i+i/j+1/k>1$. Clearly $G/\langle abc\rangle$ has a presentation of this form (but using this classification to answer your problem is somewhat boring). [Wikipedia called these groups "von Dyke groups", but I have never seen them called that in the literature.] – user1729 Dec 15 '17 at 9:31

In the quotient: $abc=1$, which means that $abc=a^2=b^3=c^5=1$.
Using the fact that $c=(ab)^{-1}$, we get that $(ab)^5=1$, meaning that the group is really $$\langle a,b \mid a^2=b^3=(ab)^5=1 \rangle.$$
the Cayley graph for this group is the truncated dodecahedron where $b$ can be seen in the the triangles, $(ab)$ is identifies with decahedrons and the edges connected them should be understood as two edges, $a^2$. Since the polyhderon is simply connected, this is the Cayley Complex, and hence we deduce that the group is finite. Hence, the group is finite with 60 elements. it is in fact $A_5$.
• What you know is that the group is generated by torsion elements. This doesn't imply that every element is torsion; for example, $\langle a, b \mid a^2 = b^2 = 1 \rangle$ is the infinite dihedral group, which contains elements of infinite order such as $ab$. – Qiaochu Yuan Dec 15 '17 at 1:27
$G/<abc>$ is a finite Triangle group. (See Wikipedia for helpful info on these groups).