TITLE: Trying to prove that $H=\langle a,b:a^{3}=b^{3}=(ab)^{3}=1\rangle$ is a group of infinite order.
QUESTION [12 upvotes]: I'm trying to prove that the following group has infinite order: $$H=\langle a,b\mid a^{3}=b^{3}=(ab)^{3}=1\rangle.$$
Currently I'm checking on some cases using the relations, but my problem is the reducibility for large products.
Naively I started to check that $ab$ is different than $1,a,b$ and then $ba$ than $1,a,b,ab$, just to understand $H$ in some extent. 
Now I'm wondering about some more effective method to prove that $|H|=\infty$, I'm tempted to look for an injection from some group of infinite order into $H$, but I'm still stuck.
More than asking for a solution I'd rather appreciate some hints or thoughts about it. Thanks a lot.

REPLY [5 votes]: Let $p$ be a prime congruent to $1\ mod\ 3$. Find a homomorphism from your group $H$ to the non-abelian group of order $3p$. Then use the fact that there are infinitely many primes congruent to $1\ mod\ 3$ to show that $H$ is infinite. 
This is a problem from Dummit and Foote's Abstract Algebra, 3rd ed, 6.3.14, pg 221.