Showing $\left<a,b,c : a^2=b^3=c^5=abc \right>/\left<abc\right>$ is finite

Question

I have had hard time to show that

$$G=\left<a,b,c : a^2=b^3=c^5=abc \right>/\left< abc \right>$$ is a finite group.

I guess that $G=\mathbb{Z}_2*\mathbb{Z}_3*\mathbb{Z}_5$. Even in this situation, it is not clear if it is finite or not. Could anybody help me for this problem? Thanks in advance!

Edit :

Apparently, this question has been answered in Proving Finiteness of Group from Presentation. However, the answer given in the link used the property of triangle group which is not given in Dummit and Foote algebra. Since the prerequisite of the given problem is the Dummit and Foote's Algebra book, I think there should be another way to solve it. So.. does anybody know how to solve this in different way with the answer of given link?

Answer 1

Thanks to ParclyTaxel and Ycor, I was able to figure out the answer hoping that there is no error.

Firstly, we know $G=\left<a,b,c : a^2=b^3=c^5=abc=1 \right>$. Now, note that we can reduce the number of generators and relations by following way. $$b=a^2bc^5=a(abc)c^4=ac^4=ac^{-1}$$ Now, renaming $a=p$ and $c^{-1}=q$, we have $$G=\left< p,q : p^2=q^5=(pq)^3=1 \right>$$

Then by the answer from this link, Group presentation of $A_5$ with two generators, we can prove that $G\cong A_5$ so $G$ is a finite group.