I'm reading up on Algebraic Topology in preparation for a summer course, and learning about the classification of surfaces I ran across this problem:

Show that the groups $G=\langle a,b \mid abab^{-1}\rangle$ and $H=\langle c,d \mid c^2d^2\rangle$ are isomorphic.

It's at a point in the text where Van Kampen's theorem hasn't yet been covered, so I'd like to resolve this without using any more advanced tools than I already have available.

ATTEMPT: The Euler Characteristic of both is $0$ and they are both non-orientable. So they are homeomorphic to the Klein bottle by the Classification theorem, which means there are bijective maps from either $G$ and $H$ to the fundamental group of the Klein bottle $\pi_1(K)=\langle e,f\mid e^2f^{-2}=1\rangle$. If I can show such maps are also homomorphisms, then they are isomorphisms and I'm done. Would a map that sends, say the a to c and b to d, such that it is a homomorphism, work? Or would I not be guaranteed that that map is also the one that is bijective? (I suppose I could show that manually but I was wondering whether the Classification Theorem provides a homeomorphism canonically)

Another approach is to show that $G$ and $H$ are the same as $\pi_1(K)=\langle a,b\mid a^2b^{-2}\rangle$ by playing around with the words, for example:

Given the relations of $G$, I have $$a^2b^{-2}=ababb^{-1}a^{-1}b^{-1}a^{-1}=1\Rightarrow G\langle a,b \mid abab^{-1}\rangle = G\langle a,b \mid a^2b^{-2}\rangle=\pi_1(K).$$ This shows that the relation for $\pi_1(K)$ is true if I assume the relation of $G$, does that show isomorphism or would I need an explicit map from one set of relations to the other?

If this is the correct approach I suppose it wouldn't be difficult to come up with a similar demonstration for $H$.

Any help or insight is appreciated, introductory texts to algebraic topology dive in without much motivation and I haven't really grasped the connection between free groups and the surfaces they represent.

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    $\begingroup$ Please use $\langle,\rangle$ \langle,\rangle. Compare $<a>$ with $\langle a\rangle$. Even better, compare $\left\langle \begin{matrix}x\\ y\end{matrix}\right\rangle$ with $<\begin{matrix}x\\y\end{matrix}>$. Also, use $\mid$, \mid for a vertical bar. This has an automatic spacing. Compare $\langle x\mid y\rangle$ and $\langle x|y\rangle$. I have edited the title of your post. Please edit the body accordingly. $\endgroup$ – Pedro Tamaroff Apr 3 '16 at 2:48
  • $\begingroup$ @PedroTamaroff Got it, fixed $\endgroup$ – Mike Apr 3 '16 at 2:51

Your first approach is good. Every continuous map induces a homomorphism of fundamental groups. Homeomorphisms, however, induce isomorphisms of fundamental groups (see proof here). Therefore, your groups are isomorphic.

  • $\begingroup$ Thank you for your answer. Reading that link, it says that the induced homomorphism (which the theorem says is an isomorphism) is between fundamental groups. So if I know there is a homeomorphism from $G$ to $\pi_1(K)$, then the induced homomorphism would be between the fundamental group of $G$ and $\pi_1(K)$, but what is the fundamental group of $G$? Does it even make sense to ask what is the fundamental group of a free group like G? I'm pretty sure your link answers my question, but I'm now getting familiar with these concepts, any elaboration would be helpful. $\endgroup$ – Mike Apr 3 '16 at 2:02
  • $\begingroup$ In your problem, $G$ is the fundamental group of some topological space (call it $X$). You showed via the Euler characteristic and the classification of surfaces theorem that $X$ is homeomorphic to $K$. Thus, $\pi_1(X)=G$ is isomorphic to $\pi_1(K)$. $\endgroup$ – Alex S Apr 3 '16 at 2:05
  • $\begingroup$ Oh I see, so when I "represent" a topological surface via a free group, that free group IS the (or a?) fundamental group of that surface. For example, I can draw the rectangle of a torus with edges $aba^{-1}b^{-1}$, then the group $$T=<a,b | aba^{-1}b^{-1}>$$ is the fundamental group of the Torus. In other words if I consider $a$ and $b$ to be paths, any closed loop on the Torus will be of the form $a(t)^mb(t)^n$ for some m,n integers. Is this correct? Thanks. $\endgroup$ – Mike Apr 3 '16 at 2:09
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    $\begingroup$ Essentially, but a few notes. First, the groups you gave are not free. Free groups do not have any relations, as yours do. Second, be careful with the notion that we can go back and forth between spaces and groups. Each compact surface has a different fundamental group, but this is not true for non-surfaces. $\endgroup$ – Alex S Apr 3 '16 at 2:30
  • $\begingroup$ I see. I think I'll have to read my way to the Seifert/van Kampen theorem to get a better understanding. What seems odd to me is that ANY closed loop on the Torus, even anything that does not go around the hole, can be expressed by powers of $aba^{-1}b^{-1}$. Now that I think of it loops that don't go through the hole are contractible to a point and therefore homotopic to the constant path. $\endgroup$ – Mike Apr 3 '16 at 2:35

This is easy to do without any topology at all, just explicitly write down an isomorphism: consider the homomorphism $g : H \to G$ given by $c \mapsto ab$, $d \mapsto b^{-1}$. This is well-defined because it respects the defining relation of $H$: we have $c^2d^2\mapsto(abab)b^{-2}=abab^{-1}=1_G$. To show it's an isomorphism, write down the inverse $f : G \to H$: $a \mapsto cd$, $b\mapsto d^{-1}$. This one is also well-defined (because $abab^{-1} \mapsto cdd^{-1}cdd=c^2d^2=1_H$) and it's easy to check it really is a two-sided inverse for the first homomorphism: $f(g(c))=f(ab)=cdd^{-1}=c$, $f(g(d))=f(b^{-1})=(d^{-1})^{-1}=d$; $g(f(a))=g(cd)=abb^{-1}=a$, $g(f(b))=g(d^{-1})=(b^{-1})^{-1}=b$.

  • $\begingroup$ Thanks, so you are saying that an isomorphism between presentations of groups, $G$ and $H$, is a homomoprhic bijection where the relations of the domain are respected in either direction? I've spent some time here: math.stackexchange.com/questions/1650439/… and I figured isomorphism between group presentations is not as easy as it seems. $\endgroup$ – Mike Apr 3 '16 at 2:23
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    $\begingroup$ Right, @Craig: to define a homomorphism $f : G \to H$ you just need to assign an element of $H$ to each generator of $G$ and check that the assignment would send every relation in the presentation of $G$ to the identity element of $H$. As you said, if $f$ is bijective then it is an isomorphism, but sometimes that might be hard to show directly. One way is to give another homomorphism the other way, $g:H\to G$ and to check that $f\circ g = id_H$ and $g\circ f=id_G$ (for example, by checking that $f(g(x))=x$ for each generator in the presentation of $H$ and the same for the other composite). $\endgroup$ – Omar Antolín-Camarena Apr 3 '16 at 2:35

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