Showing that the Binary Icosahedral Group (given by a presentation) has order $120$. Say we have a group generated by $a, b$, with the relations $(ab)^2=a^3=b^5$ (note that these are not necessarily equal to the identity). How do I show that the group has $120$ elements? Without having orders given to any of the generators, why can't there be infinitely many elements?

This problem is from Stillwell's "Mathematics and its History," after Section 22.8. I'm first asked to show that adding the relation that all $3$ equal the identity gives the icosahedral group, isomorphic to $A_5$, and then to show that this implies the group without that relation has at least $60$ elements. Then, a parenthetical after that problem says "Harder: Show that it in fact has $120$ elements".

Consider the group $G=\langle a, b, c; a^2=b^3=c^5=abc\rangle^{\ast}$. Clearly, $\langle abc\rangle$, which is the same subgroup as $\langle a^2\rangle$, as $\langle b^3\rangle$, and as $\langle c^5\rangle$, is a central subgroup of $G$. As we know from the earlier part of the problem, quotienting out this subgroup gives $A_5$, which has order $60$. Hence, in order to prove the result it is sufficient to prove that $(abc)^{2}=1$ and that $G\not\cong A_5$.
A proof of these facts can be found on the Groupprops website. Their proof shows that $G$ contains a subgroup which is a homomorphic image of the dicyclic group $\langle x, y, z; x^2=y^2=z^5=xyz\rangle$, and the subgroup is embedded such that $x_0y_0z_0=abc$. In a dicyclic group, $(x_0y_0z_0)^2=1$, hence $(abc)^2=1$ as required. The subgroup is given by the following embedding.
$$
\begin{align*}
x_0&\mapsto a^{-1}bab^{-1}a\\
y_0&\mapsto a\\
z_0&\mapsto cac^{-3}
\end{align*}
$$
I will leave you to prove that $G\not\cong A_5$. There are a number of ways of doing this. For example, you could prove that $abc\neq 1$ in $G$, or you could "realise" $G$ as a group of matrices (hint: $\operatorname{SL}_2(5)$) or permutations$^{\dagger}$. Read the wikipedia page to get yourself started.
$^{\ast}$The presentation given in this answer differs from the one you gave. However, proving that $(abc)^2=1$ (which is the purpose of this answer) proves that these presentations define isomorphic groups. Can you see why this is?
$^{\dagger}$Another possible, elegant, way would have been to have found a non-trivial homomorphic image of $G$ which was not $A_5$. However, in reality $G$ has only one proper, non-trivial subgroup so this won't work!
