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Let $K$ the Klein Bottle and $\pi_1(K) = \langle a,b \mid b a b a^{-1} \rangle $ be the fundamental group of the Klein bottle. Observe that $\langle b \rangle $ is a normal subgroup of $\pi_1(K)$, hence, the quotient $G = \frac{\pi_1(K)}{\langle b\rangle} $ is a group. I need to show that $G$ is isomorphic to $\mathbb{Z}$.

In fact, let $\phi : F(a,b) \to \mathbb{Z}$ a homomorphism of groups (By universal property of free groups) s.t. $a\mapsto1, b\mapsto 0 $, where $F(a,b)$ is a free group of two generators. My strategy is use the the first isomorphism theorem for a induced map of $\phi$, $\tilde{\phi}: \pi_1(K) \to \mathbb{Z}$. My problem is show that $\ker \tilde{\phi}$ is $\langle b\rangle$.

Let $g\in \ker \tilde{\phi}$ then, $\tilde{\phi}(g) = \tilde{\phi}(a^{n_1} b^{m_1} \cdots a^{n_k} b^{m_k}) = (n_1+\cdots + n_k)\tilde{\phi}(a) + (m_1+\cdots m_k)\tilde{\phi}(b) = m \tilde{\phi}(a) + n \tilde{\phi}(b)=0$, But $\tilde{\phi}(a) =1$ and $\tilde{\phi}(b) =0$ so $n =0$ and $m\in \mathbb{Z}$. Observe that, if $g$ assume the form $a^n b^m$ then, I can concluded that $\ker \tilde{\phi}$ is exactly $\langle b\rangle$. But does every $g$ in $\pi_1(K)$ have this form?

  • In general, to get the quotient of a finitely presented group by the normal closure of some of the generators, you just replace those generators by the identity in the presentation. In your example, this results in the presentation $\langle a \mid \rangle$ which is indeed isomorphic to $({\mathbb Z},+)$. – Derek Holt May 19 '23 at 13:58
  • I know this is true, but I'm looking to build an "explicit" function without using this fact. – Horned Sphere May 19 '23 at 14:00
  • We have $\pi_1(K)\cong \Bbb Z\rtimes \Bbb Z$, see also here, and hence $G\cong \Bbb Z$. – Dietrich Burde May 19 '23 at 14:14
  • If you know that the fact mentioned by @DerekHolt is true, and if you know the general proof of that fact, then the explicit function you want should be present in that proof. – Lee Mosher May 19 '23 at 14:24
  • @LeeMosher, Could you indicate a reference so I can read it? – Horned Sphere May 19 '23 at 15:02
  • The general property I referred to follows directly from the definition of the group defined by a presentation as a quotient of the free group on the generators by the normal closure of the defining relators. – Derek Holt May 19 '23 at 15:16

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