Trying to prove that $H=\langle a,b:a^{3}=b^{3}=(ab)^{3}=1\rangle$ is a group of infinite order.

Question

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.

Answer 1

Consider an equilateral triangle in the plane and let $r$, $s$ and $t$ be the motions of the plane given by reflection with respect to each of the sides of the triangle. Then $a=rs$ is a rotation of angle $2\pi/3$ around the vertex of the triangle which is the intersection of the sides with respect to which $r$ and $s$ reflect. Similarly, $b=st$ is a rotation of that same angle around another of the vertices, and so is $c=rt$. Notice that $a^3=b^3=c^3$ and that $c=ab$.

It follows that there is a surjective group homomorphism from your group to the subgroup of the group $\Gamma$ of motions of the plane generated by the three rotations $a$, $b$ and $c$. To show your group is infinite it is enough to show that $\Gamma$ has an infinite orbit in the plane, and you can do this by making pictures :-)

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Later. It is important to note that this is not a random fact. The group generated by the three reflections on the sides of my triangle, which has presentation $\langle r, s, t: r^2=s^2=t^2=(rs)^3=(st)^3=(tr)^3\rangle$ is what we call a Coxeter group, and the subgroup generated by the three rotations $a$, $b$ and $c$ is its positive part. This type of group is very well-known, and a Google search will show.

Answer 2

A string of $a$'s and $b$'s represents a nonzero element of $H$ as long as it does not have a substring of the form $sss$, where $s$ is a smaller string. This is because all of the relations defining $H$ are of the form $s^3=1$ for some string $s$, so if no $s^3$ appears, the string cannot be reduced further.

Now, consider the following sequence of strings:

$$ a, ab, abba, abbabaab,abbabaabbaababba, \dots $$

The first string is $a$. To obtain the successor of $s$, concatenate $s$ with $s'$, where $s'$ is $s$ but with $a$ and $b$ interchanged. This is the the Thue-Morse sequence. Importantly, the Thue-Morse sequence is cube free, i.e. it does not have any substrings of the form $sss$. This fact is a bit tricky to prove, but is well-known.

Letting $s_i$ be the $i^{th}$ string in the above list, for all $i<j$, $s_i^{-1}s_j$ is a substring of $s_j$, which is therefore cube free so by the first paragraph nonzero in $H$. Thus, these strings represent pairwise distinct elements of $H$, so $H$ is infinite.

Answer 3

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.