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

Question

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)

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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.

Answer 1

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.

Answer 2

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: $f(g(c))=f(ab)=cdd^{-1}=c$, $f(g(d))=f(b^{-1})=(d^{-1})^{-1}=d$; $g(f(a))=g(cd)=abb^{-1}=a$, $g(f(b))=g(d^{-1})=(b^{-1})^{-1}=b$.