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For background, i've noticed that its easy to show that an equation of form $x^n = a$ in an abelian torsion free group always has either an unique solution for $x$ or none. So i decided to try seeing what happens for non-abelian torsion free groups. The most obvious and simple example to try would be the quotient $G = \frac {F[a,b]}{N}$ where $F[a,b]$ is the free group on two generators, $N$ is the normal subgroup generated by the relation $a^2=b^2$. It is simple to show that $a\neq b$ in $G$ therefore it remains to show whether $G$ is torsion free.

It is easy to show that there is no $x$ such that $x^n = a^2b^{-2}$ in the free group because $x$ would have to have reduced form $ayb^{-1}$ and therefore any non-trivial power would have $b^{-1}a$ in its reduced form. Of course this is not enough as $N$ has elements other than $a^2b^{-2}$. I'm not sure what else to do. I could try to find an invariant that classifies whenever an element is in $N$ and then try showing that if $x^n$ is in $N$ then $x$ is in $N$, but that doesn't seem easy. Is there an easier way? What are some more systematic techniques to solve this kind of problems?

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  • $\begingroup$ In $\mathbb{Z}$ the equation $3x=2$ has no solution at all. $\endgroup$
    – Randall
    Apr 3, 2023 at 14:32
  • $\begingroup$ @Randall Thank you, fixed. $\endgroup$
    – Carla_
    Apr 3, 2023 at 14:35
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    $\begingroup$ What is "$F[a,b]$"? The free group on $a$ and $b$? $\endgroup$ Apr 3, 2023 at 15:57
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    $\begingroup$ That's way too easy to confuse with a polynomial ring over a field $F$; you really should specify that is what you mean. $\endgroup$ Apr 3, 2023 at 17:06
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    $\begingroup$ @ArturoMagidin I've edited to specify that after you commented. Thank you for the feedback. $\endgroup$
    – Carla_
    Apr 3, 2023 at 17:16

2 Answers 2

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In some sense it is provable that there is no systematic technique which solves all problems of this type: it is a theorem that there does not exist an algorithm to solve the word problem in group theory.

There are nonetheless many specific techniques that you can try, although it helps to have a large knowledge base to know what technique will work on which group. You seem to have more-or-less picked this group presentation out of a hat, and it's not such an easy example to analyze, so it might be hard to stumble upon a good technique.

My own knowledge base of geometry and topology leads me to the solution I outline here: your group $G$ is representable as a subgroup of the group $\text{Isom}(\mathbb E^2)$ of isometries of the Euclidean plane.

Draw the square $[-1,+1] \times [-1,+1]$. Your element $a$ will act as the glide reflector isometry of $\mathbb E^2$ that takes the lower side $[-1,1] \times \{-1\}$ to the right side $\{1\} \times [-1,+1]$, given by the formula $$A(x,y) = (y+2,x) \qquad $$ Your element $b$ will act as the glide reflector that takes the left side $\{-1\} \times [-1,+1]$ to the upper side $[-1,+1] \times \{1\}$ given by the formula $$B(x,y) = (x+2,y) $$ You can easily see from this that $A^2=B^2$, namely both are equal to the translation isometry given by $(x,y) \mapsto (x+2,y+2)$. One therefore obtains a homomorphism $$\phi : F[a,b] / N \to \text{Isom}(\mathbb E^2) $$ defined by $\phi(a)=A$ and $\phi(b)=B$. In particular, for any element $w \in F[a,b]$, if $w \in N$ then $\phi(w)$ is the identity isometry of $\mathbb E^2$.

Now here comes the hard work, namely to prove the converse:

If $w \in F[a,b]$ and if $\phi(w)$ is the identity isometry of $\mathbb E^2$ then $w \in N$. In other words, $w$ can be written as a relator $$w = \prod_{i=1}^K v_i r_i v_i^{-1} $$ for some elements $v_1,\ldots,v_K \in F[a,b]$ and for elements $r_i$ each of which is one of the words $a^2b^2$ or $ab^2a$ or $b^2a^2$ or $ba^2b$ or their inverses $b^{-2}a^{-2}$ or $a^{-1}b^{-2}a^{-1}$ or $b^{-2}a^{-2}$ or $b^{-1}a^{-2}b^{-1}$.

Here's the way I know to do this, expressed rather roughly (it would take a LOT of work to formalize this, roughly speaking a beginning course in geometric group theory). Extend the original square to a tiling of the Euclidean plane by squares of the form $[2m-1,2m+1] \times [2n-1,2n+1]$. Use the given word $w$ to trace out a path in the grid of sides of these squares. The fact that $\phi(w)$ is the identity isometry will imply that this path closes up to form a loop. Use the grid as a guideline for popping (or homotoping) the loop through one square at a time until it becomes the trivial loop. Use that sequence of "popping" steps to build the required relator expression for $w$.

In essence, what one is happening in the above argument is that one is demonstrating that the grid of sides of the square tiling is the Cayley graph of $G$ with respect to the generating set $\{a,b\}$.

It will follow that your group $G$ is isomorphic to the subgroup $\text{image}(\phi) < \text{Isom}(\mathbb E^2)$ that is generated by $A$ and $B$.

Now comes some more work: you can use the theory of Euclidean isometries to analyze the elements of this subgroup. Your two generators are glide reflectors with parallel glide axes of slope $+1$. It follows that every non-identity element of your group is either a Euclidean translation or a Euclidean glide reflection along a glide axis of slope $+1$. Therefore, every nontrivial group element is nontorsion.

Your group also comes up in topology: it happens to be isomorphic to the fundamental group of the Klein bottle. That, in fact, is how I recognized your group. The topology of the Klein bottle is closely connected to Euclidean geometry, and that connection is what led me to the solution I've outlined.

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We can prove that $G$ is torsion-free by using an action of the group on a tree. The theory of groups acting on trees is called Bass-Serre Theory, and the standard reference in Serre's book Trees. This method generalises to all groups of the form $F[a, b]/(a^m=b^n)$, $m, n>1$.

Consider the infinite tree $T$ with all vertices of degree $4=2+2$. Direct the edges so that every vertex has two "incoming" edges and two "outgoing" edges.

I will now construct an action of $G$ on $T$. I will leave it to you to prove that this is indeed an action of $G=\langle a, b\rangle$ on $T$. (c.f. Theorem 7 of "Trees".)

Fix two adjacent vertices $v_a$ and $v_b$. Let the element $a\in G$ act on $T$ by:

  1. fixing the vertex $v_a$,
  2. swapping the two incoming edges, and
  3. swapping the two outgoing edges. Similarly, let $b\in G$ act on $T$ by fixing the vertex $v_b$ and swapping edges as in the action of $a$. This extends to an action of $G$ on $T$, as claimed.

To see that $G$ is torsion-free, we require two standard facts from Bass-Serre Theory (both could be exercises, if you are feeling confident!):

  1. Elements of finite order must fix a vertex of $T$.
  2. Vertex stabilisers $G_v = \{g\in G: gv=v\}$ are precisely the conjugates of $\langle a\rangle$ and $\langle b\rangle$. As $\langle a\rangle$ and $\langle b\rangle$ are torsion-free, their conjugates are torsion-free, and hence $G$ is torsion-free.

The group you are considering is called "free product with amalgamation" of two torsion-free groups. The above is one way of thinking about free products with amalgamation, but a more classical, combinatorial theory also exists (see for example Lyndon and Schupp's book "Combinatorial Group Theory").

Another viewpoint of your group, but one which really requires free product with amalgamation and their sibling HNN-extensions in order to prove things, is to think of it as a one-relator group. One-relator groups have presentations of the form $\langle \mathbf{x}\mid R=1\rangle$, and more details can also be found in Lyndon and Schupp's book. Your group has presentation $\langle a, b\mid a^2b^{-2}=1\rangle$, and so has the required form. Now, a one relator group $\langle \mathbf{x}\mid R=1\rangle$ has torsion if and only if there exists some $R_0\in F(\mathbf{x})$ and some $n>1$ such that $R=R_0^n$. You remarked that $a^2b^{-2}$ cannot be of this form, and hence your group is torsion-free.

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