Show the two fundamental groups representation of the Klein bottle are the same

Question

Show the two different representation of the fundamental groups of the Klein bottle $G=<a,b:aba^{-1} b>$ and $H=<c,d:c^2 d^{-2}>$. As we calculate in different and gett one result as $\pi_1(K)=G$, and other as $\pi_1(K)=H$, and I need to show $G \cong H$.

I try to do this by constructing a function $f:G\to H$ and try to define $f(a)$ is in terms of $c$ and $d$, and $f(b)$ is also defined in terms of $c$ and $d$, such that $f(e_G)=f(aba^{-1} b)=f(a)f(b)f(a^{-1})f(b)=c^2d^{-2}=e_H$.

I am not sure how to define such $f$ for $a$ and $b$, please help me. Thanks a lot!

Answer 1

It seems to me you're on the right track. As a hint, you can rewrite the relation $aba^{-1}b$ as $a^2a^{-1}ba^{-1}b = a^2(a^{-1}b)^2$. This can let you define a map $H \to G$ by sending $c$ to $a$ and $d$ to $(a^{-1}b)^{-1} = b^{-1}a$. I leave it to you to define the inverse function and check the details.

Answer 2

You can solve this problem in a slightly different way by showing that there are sets of generators of the fundamental group of the Klein bottle that satisfy the given relations.

For instance, if you represent the fundamental group of the Klein bottle as the group generated by the translation b:(x,y) -> (x,y+1) of the plane together with the affine transformation a:(x,y) -> ( x+ 1/2, -y) then

$$aba^{-1} = b^{-1}$$

Now find two other generators of the group that satisfy $$c^2d^{-2} = id$$