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.
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 $A_5$.