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I know that the Klein bottle $K$ is obtained from the unit square by making identifications on the boundary with the appropriate directional arrows. Usually, what is done is that we identify the point $(0,t) \sim (1,1-t)$ and the point $(t,0) \sim (t,1)$ $\forall t \in [0,1]$.

Now suppose we let $a \in \pi_{1}(K)$ (fundamental group of $K$) be represented by the path $\gamma(t)= [(0,t)]$, where $[(x,y)]$ is the element of $K$ represented by $(x,y) \in I^2$. Similarly, we let $b \in \pi_{1}(K)$ be represented by the path $\delta(t)= [(t,0)]$. Now I know that $\pi_{1}(K)$ has the presentation $\langle a, b $ $\vert$ $bab^{-1}a= 1 \rangle$. If we let $N= \langle a \rangle= \{a^k: k \in \mathbb{Z}\}$, since $bab^{-1}= a^{-1} \in N$, it follows that $N$ is a normal subgroup of $\pi_{1}(K)$. Does this seem about right so far?

My question is that what then is the quotient group $\pi_{1}(K)/N$? I'm not sure if it is infinite cyclic but I think it has got to be generated by $b$. I would like to know how I can prove this claim. I encountered this question on an old topology prelim from about five years ago. Looks like I need to brush up on some group theory!

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From the relation $bab^{-1}a = 1$, you obtain $ba = a^{-1}b$ and $ab = ba^{-1}$.

Thus you can write each element of $\pi_1(K)$ uniquely as $b^ma^n$, multiplying two such elements produces

$$b^ma^nb^pa^r = b^{m+1}a^{-n}b^{p-1}a^r = \dotsc = b^{m+p}a^{(-1)^pn + r},$$

thus indeed you have

$$\pi_1(K)/\langle a\rangle \cong \langle [b]\rangle$$

infinite cyclic.

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  • $\begingroup$ I take it that $p$ is an integer, and not a prime number. $\endgroup$ – Libertron Aug 15 '13 at 21:18
  • $\begingroup$ Yes. Well, it might be prime, that wouldn't hurt ;) $\endgroup$ – Daniel Fischer Aug 15 '13 at 21:21
  • $\begingroup$ Hello, I have a question. Are $\langle b\rangle$ and $\langle a\rangle$ infinite and cyclic?. Why?. $\endgroup$ – Andrés Felipe Sep 11 '18 at 9:02
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Let me give a slightly "snappier" solution than Daniel Fischer's. It may seem slightly longer, but once you get used to the method you can do it off the top of your head. It will take you literally ten seconds! The point of this answer is to say: You have been given a presentation so use it!

Recall the presentation of the Klein bottle group is $G=\langle a, b; b^{-1}ab=a^{-1}\rangle$. Then the presentation $$H=\langle a, b; b^{-1}ab=a^{-1}, a=1\rangle$$ is, by definition, the group $$G/\langle\langle a\rangle\rangle$$ where $\langle\langle a\rangle\rangle$ is the normal closure of $a$, that is, the smallest normal subgroup of $G$ containing $a$. Therefore, $\langle\langle a\rangle\rangle=N$, and so $G/N\cong H$ has the above presentation, which is clearly seen to be infinite cyclic after applying Tietze transformations (that is, remove $a$ from the list of generators and replace it with the identity wherever it appears in the relations, so you get $$\langle b; b^{-1}b=1\rangle\cong\langle b; -\rangle\cong\mathbb{Z}$$ as required).

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The fundamental group of the Klein bottle is a split group extension of $\mathbb{Z}$ by $\mathbb{Z}$ $$0 \to \mathbb{Z} \to K \to \mathbb{Z} \to 0,$$ where the generator $b$ of the quotient $\mathbb{Z}$ acts by negation on the kernel $\mathbb{Z}$ of the exact sequence of groups.

This is easy to see from the relation $b^{-1}ab=a^{-1}.$ There are no relations on the powers of $a$ and $b$ separately, so they each generate infinite cyclic groups, and $b$ negates the powers of $a$ under conjugation.

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