Element of infinite order for a given group presentation

Question

Let $G=\langle a,b,c,d \mid abcda^{-1}b^{-1}c^{-1}d^{-1}\rangle$ be our presentation. The claim is that the commutator $[a,b]$ has inifinite order in $G$. I think this might be related to small cancellation, but not really sure how promising that approach really is. I am grateful for any hints, thanks!

Answer 1

Yes you can do this by small cancellation. The presentation satisfies $C'(1/6)$ (in fact $C'(1/7)$), so by Greendlinger's lemma any word equal to the identity in the group must have a subword consisting of more than half of a cyclic conjugate of a relator or its inverse. Since $[a,b]^k$ does not contain such a subword for any $0 \ne k \in {\mathbb Z}$, it cannot be equal to the identity.

Answer 2

You can also do an ad-hoc approach (without small cancelation).

We take two (more or less) arbitrary noncommuting matrices $A:=\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$ and $B:=\begin{pmatrix} 1&0\\1&1 \end{pmatrix}$ (living over, say, the rationals).

The map $a\mapsto A,b\mapsto B, c\mapsto B^{-1}, d \mapsto A^{-1}$ extends to a representation of $G$ and the image of the commutator of $a$ and $b$ under this representation is $ABA^{-1}B^{-1}=\begin{pmatrix} 3&-1\\1&0 \end{pmatrix}$ which is easily checked to be of infinite order.

Answer 3

If you don't mind "a hammer"(although not really much more difficult than small cancellation machinery), you can use some one relator theory, and apply the Freiheitssatz, to get that $\langle a,b\rangle$ generates a free subgroup on two generators in the whole group. This result can be found in both of "the" Combinatorial Group Theory books (one by Lyndon and Schupp, and the one by Magnus, Karrass, and Solitar). You can also look at this blog post for an introduction (and the other posts one relator posts are interesting as well).

Answer 4

Here is a topological/geometric proof.

Label the sides of an octagon by this relator. You will see that by gluing opposite sides there is a single vertex in the quotient. The quotient space $S$ is a surface, and your group is isomorphic to $\pi_1 S$. The element $[a,b]$ is represented by a simple closed curve $c$ on $S$ which separates $S$ into two one-holed tori.

The surface $S$ has a complete hyperbolic metric, and the curve $c$ is isotopic to a unique simple closed geodesic. All its powers are simple closed geodesics, and therefore represent nontrivial elements of $\pi_1 S$.