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.

  • 4
    $\begingroup$ The word is extEnt, by the way. Extant does exist but most probably is not what you had in mind. $\endgroup$ – Mariano Suárez-Álvarez Aug 19 '16 at 19:17
  • $\begingroup$ Thanks for that, I should point that english is not my native language, so any comments about my grammar or orthography will be well received. $\endgroup$ – Cristian Baeza Aug 19 '16 at 19:23

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 :-)

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.


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.


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.

  • 7
    $\begingroup$ But in principle you can also expand things in order to get opportunuties to reduce. Consider the group generated by $a$ and $b$ subject to $abbbbb=1$ and $bbbbb=1$: the word $a$ can be expanded first and then reduced to $1$. Yes: this is a silly example. $\endgroup$ – Mariano Suárez-Álvarez Aug 19 '16 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.