# Identifying $\langle a,b\mid a^2b^2\rangle$.

## The Question:

What group is $G=\langle a,b\mid a^2b^2\rangle$?

## Thoughts:

I found that the presentation maps onto $\langle a, b\mid a^2, b^2, 1\cdot 1\rangle\cong \mathbb{Z}_2\ast\mathbb{Z}_2$, which is reassuring since GAP says $G$ is infinite.

## Extra Context:

I'm not sure if that's enough context so here's a Q&A:

• What are you studying?

A PhD in combinatorial group theory, first year.

• What text is this drawn from, if any? If not, how did the question arise?

None. The group arose in my research. Without giving too much away, the group is cyclically presented.

• What kind of approaches (to similar problems) are you familiar with?

Here's a list of related questions of mine:

Identifying $\langle \{x_h\mid h\in H\}\mid \{x_{h'}x_{h'a}x_{h'b}^{-1}\mid h'\in H\}\rangle$ if $H=\langle a,b\mid a^2, b^4, ab=ba\rangle$.

Identifying $\langle r,s\mid srs^{-2}r, r^{-1}srsr^{-1}\rangle$.

Identifying $\langle x_e, x_a, x_b, x_{ab}\mid x_ex_a=x_{ab}, x_ax_e=x_b, x_bx_{ab}=x_a, x_{ab}x_b=x_e\rangle.$

Identifying $\langle x_1, \dots, x_6\mid x_1x_2=x_3, x_2x_3=x_1, x_3x_1=x_2, x_4x_5=x_6, x_5x_6=x_4, x_6x_4=x_5\rangle.$

There's more but I think that's enough.

• What kind of answer are you looking for? Basic approach, hint, explanation, something else?

A simple identification would be great. A detailed explanation of why it is what it is would be better, although (strong) hints are preferred.

• Is this question something you think should be able to answer? Why or why not?

Yes. I've been working closely with presentations in a research context for months now. It looks like Google should have the answer.

• The answer to the question in the title of that supposed duplicate is in the negative, @DietrichBurde. – Shaun May 15 '18 at 17:02
• Upon closer inspection, it says in the accepted answer that $G=\Bbb Z*_{\Bbb Z} \Bbb Z$ is a free product with amalgamation of two infinite cyclic groups generated by $a, b$ amalgamated along their subgroups $a^2, b^2$. – Shaun May 15 '18 at 17:11
• @DietrichBurde's supposed duplicate: math.stackexchange.com/q/191317/104041. – Shaun May 15 '18 at 17:43

It has a normal abelian subgroup of index $2$ and isomorphic to ${\mathbb Z}^2$, with generators $ab$ and $a^2$.

It is a torsion-free group, so it cannot be a split extension. We have $a(ab)a^{-1} = a^4(ab)^{-1}$, so another presentation of the same group, which makes this structure clearer, is $$\langle a,c,d \mid cd=dc,\, a^2=d,\, aca^{-1}=d^2c^{-1} \rangle,$$ where $c=ab$ and $d=a^2$.

I will add that I don't think that it is completely clear what you mean by "identify this group". The presentation is a precise definition of the group, so you could argue that it is already an identification. I guess you mean something like "find an alternative description or definition of a group isomorphic to this one, if possible a group that has been studied already". That's OK, but in many cases, there will not exist such an alternative description.

Since this particular group is solvable, and even polycyclic, a polycyclic presentation, such as the one I have give above in generators $a$, $c$ and $d$ provides an alternative description. You could for example use the polycyclic package in GAP for computing with polycyclic groups on this group.

• How come the group is solvable? – Shaun May 15 '18 at 19:50
• Oh, and thank you. – Shaun May 15 '18 at 20:04
• It has an abelian subgroup of index $2$, so it must be solvable. – Derek Holt May 15 '18 at 21:29

For a topological/geometric point of view, this group is isomorphic to the fundamental group of the Klein bottle.

Here's how to see it. Start by rewriting the presentation $$\langle a,b \mid a^2b^2 = \text{Id}\rangle = \langle a,b \mid a^2 = b^{-2} \rangle = \langle a,c \mid a^2 = c^2 \rangle \quad\text{(c=b^{-1})}$$ By Van Kampen's Theorem, we can construct a topological space with this fundamental group. Start with two circles $C_a$, $C_c$ whose fundamental groups are infinite cyclic with generators represented by closed paths $\gamma_a$ going exactly one time around $C_a$ and $\gamma_c$ going one time around $C_c$. Now attach $S^1 \times [0,1]$ to these two circles: $S^1 \times 0$ is attached to the path $\gamma_c^2$ that goes two times around $C_a$; and $S_1 \times 1$ is attached to the path $\gamma_b^2$ that goes two times around $C_c$. Let $X$ be the resulting quotient space. This space $X$ is a compact, connected, non-orientable 2-manifold of Euler characteristic $0$, and so it is homeomorphic to the Klein bottle. Thus, by Van Kampen's Theorem, your group is isomorphic to the fundamental group of the Klein bottle.

And once you know that this is the Klein bottle group, it also follows that it is one of the 17 famous wallpaper groups, also known as planar crystallographic groups, namely those groups which can be represented as discrete, cocompact subgroups of isometries of the Euclidean plane. To a crystallographer, this group is known as "pg", and can be represented as generated by two isometries of the $xy$-plane, as shown in this picture: $g_x$ which is a glide reflection along the $x$ axis (the "g" in the "pg" notation refers to a glide reflector); and $t_y$ which is a translation along the $y$ axis. To connect these generators with your two original generators, you can take $a=g_x$ and $b=t_y g_x$.