# $G=\langle a,b \mid abab^{-1}\rangle$ and $H=\langle c,d \mid c^2d^2\rangle$ are isomorphic (Can't use Seifert/van Kampen Theorem)

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.

• 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. – Pedro Tamaroff Apr 3 '16 at 2:48
• @PedroTamaroff Got it, fixed – Mike Apr 3 '16 at 2:51

• 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. – Mike Apr 3 '16 at 2:02
• 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 $\pi_1(K)$. – Alex S Apr 3 '16 at 2:05
• 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. – Mike Apr 3 '16 at 2:09
• 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. – 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: $f(g(c))=f(ab)=cdd^{-1}=c$, $f(g(d))=f(b^{-1})=(d^{-1})^{-1}=d$; $g(f(a))=g(cd)=abb^{-1}=a$, $g(f(b))=g(d^{-1})=(b^{-1})^{-1}=b$.
• 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. – Mike Apr 3 '16 at 2:23
• 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). – Omar Antolín-Camarena Apr 3 '16 at 2:35