Prove that some group is infinite based on its presentation. Suppose $G$ is a group with presentation $$G = \langle a, b, c\mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ca)^3=1\rangle.$$
I want to show that this group is of infinite order.
I think since the element $abc$ is of infinite order, we can conclude that $G$ is infinite. I feel that for any $n \in\mathbb{N},$ we have that $(abc)^n = (abc)(abc) \cdots (abc)$ (with $n$ factors) cannot be reduced further, as we do not have any relations involving $a, b,$ and $c$ consecutively. But I also think that this reasoning is not that convincing, and hence, I am asking for help.

You want to give a homomorphism from $G$ to a suitable group, such that the image of $abc$ has infinite order. For $G$ I cannot think of one without having to go into nasty matrix coefficients, but the following works:
Let $U=\langle b,abc\rangle$. This subgroup has index $[G:U]=3$. Then (by Reidemeister rewriting) $U$ has a presentation
$$
\langle b,d=abc\mid b^2,d^2bd^{-2}b\rangle
$$
in these two generators, which means that for the infinite cyclic group $C_\infty=\langle z\rangle$ the map $U\to C_\infty$, $b\mapsto 1$, $d\mapsto z$ is a homomorphism that proves that the order of $abc$ must be infinite.
Addendum: Actually, by inducing this representation up to $G$, we get a homomorphism into a matrix group, in which the image of $abc$ has infinite order:
Take the homomorphism $G\to\mbox{GL}_3(\mathbb{Q})$ given by
$$
a\mapsto \left(\begin{array}{rrr}%
0&1&0\\%
1&0&0\\%
0&0&1\\%
\end{array}\right)%
,\qquad
b\mapsto \left(\begin{array}{rrr}%
1&0&0\\%
0&0&2\\%
0&1/2&0\\%
\end{array}\right)%
,\qquad
c\mapsto \left(\begin{array}{rrr}%
0&0&1\\%
0&1&0\\%
1&0&0\\%
\end{array}\right).
$$
It is easy to verify that the images satisfy the relations, so it is a homomorphism. But it maps $abc$ to
$$
\left(\begin{array}{rrr}%
2&0&0\\%
0&0&1\\%
0&1/2&0\\%
\end{array}\right)%
$$
which clearly has infinite order.
