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The Klein bottle $K$ is the connected sum of two real projective planes $K = \mathbb{RP}^2\#\mathbb{RP}^2$ and has fundamental group $\pi_1(K) = \pi_1(\mathbb{RP}^2\#\mathbb{RP}^2) = \langle a, b\mid a^2b^2 = 1\rangle$. Then by the Hurewicz Theorem, we should have $$ H_1(K ;\mathbb{Z}) = \pi_1(K)/(aba^{-1}b^{-1}). $$

I know that $H_1(K;\mathbb{Z}) = \mathbb{Z}\oplus\mathbb{Z}_2$. But why is the following is true? $$ \langle a, b\mid a^2b^2 = 1\rangle/(aba^{-1}b^{-1}) = \mathbb{Z}\oplus\mathbb{Z}_2 $$

Shaun
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Bryan Li
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    Just to be clear, are you asking for a direct proof of the fact that the abelianisation of $\langle a, b \mid a^2b^2 = 1\rangle$ is $\mathbb{Z}\oplus\mathbb{Z}_2$ (i.e. without the Hurewicz Theorem)? – Michael Albanese Apr 25 '22 at 11:04
  • Yes! I need a direct proof of $\langle a, b| a^2b^2 = 1\rangle/(aba^{-1}b^{-1}) = \mathbb{Z}\oplus\mathbb{Z}_2$ without using Hurewicz Theorem. Sorry for my poor algebra. – Bryan Li Apr 25 '22 at 11:08
  • Is $\langle a, b| a^2b^2 = 1\rangle/(aba^{-1}b^{-1}) = \langle a, b| a^2b^2 = 1, aba^{-1}b^{-1} = 1\rangle $? – Bryan Li Apr 25 '22 at 11:12
  • Another representation of the fundamental group $\pi_1(K)$ is $\langle a, b| abab^{-1} = 1\rangle$. Why is that equals to $\langle a, b| a^2b^2 = 1 \rangle$? – Bryan Li Apr 25 '22 at 11:16
  • Your title question (which, it seems, does not match the algebraic question in your first comment) is answered here and here. – Lee Mosher Apr 25 '22 at 14:26

1 Answers1

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First note that $\langle a, b \mid a^2b^2, aba^{-1}b^{-1}\rangle \cong \langle a, b \mid (ab)^2, aba^{-1}b^{-1}\rangle$ and that a presentation for $\mathbb{Z}\oplus\mathbb{Z}_2$ is $\langle c, d \mid d^2, cdc^{-1}d^{-1}\rangle$. From these presentations, it seems that an isomorphism should map $ab$ to $d$. As $a$ and $c$ have infinite order, we can also try mapping $a$ to $c$. In order to define a homomorphism, we see that $b$ must be mapped to $c^{-1}d$.

We first need to check that

\begin{align*} \phi : \langle a, b\mid a^2b^2, aba^{-1}b^{-1}\rangle &\to \langle c, d \mid d^2, cdc^{-1}d^{-1}\rangle\\ a &\mapsto c\\ b &\mapsto c^{-1}d \end{align*}

is a well-defined group homomorphism, and then verify that it is an isomorphism.

Consider the homomorphism

\begin{align*} \Phi : \langle a, b\rangle &\to \langle c, d\rangle\\ a &\mapsto c\\ b &\mapsto c^{-1}d. \end{align*}

Note that $\Phi((ab)^2) = \Phi(ab)^2 = (\Phi(a)\Phi(b))^2 = (cc^{-1}d)^2 = d^2$ and

\begin{align*} \Phi(aba^{-1}b^{-1}) &= \Phi(a)\Phi(b)\Phi(a^{-1})\Phi(b^{-1})\\ &= \Phi(a)\Phi(b)\Phi(a)^{-1}\Phi(b)^{-1}\\ &= cc^{-1}dc^{-1}(c^{-1}d)^{-1}\\ &= dc^{-1}d^{-1}c\\ &= c^{-1}(cdc^{-1}d^{-1})c. \end{align*}

As $\Phi((ab)^2)$ and $\Phi(aba^{-1}b^{-1})$ are in the kernel of the natural projection $\langle c, d\rangle \to \langle c, d \mid d^2, cdc^{-1}d^{-1}\rangle$, the homomorphism descends to a homomorphism, namely $\phi : \langle a, b\mid a^2b^2, aba^{-1}b^{-1}\rangle \to \langle c, d \mid d^2, cdc^{-1}d^{-1}\rangle$ as defined above.

Suppose $w \in \langle a, b \mid (ab)^2, aba^{-1}b^{-1}\rangle$ is in the kernel of $\phi$. Note that $w = a^mb^n$ for some $m, n \in \mathbb{Z}$. As $\phi(a^mb^n) = \phi(a)^m\phi(b)^n = c^m(c^{-1}d)^n = c^{m-n}d^n$ we must have $m - n = 0$ and $n \in 2\mathbb{Z}$ - to see this, note that the isomorphism $\langle c, d \mid d^2, cdc^{-1}d^{-1}\rangle \to \mathbb{Z}\oplus\mathbb{Z}_2$ is given by $c^rd^s \mapsto (r, s)$. Therefore $m = n = 2k$ for some $k \in \mathbb{Z}$. Now note that $w = a^mb^n = a^{2k}b^{2k} = (ab)^{2k} = ((ab)^2)^k = 1$, so $\phi$ is injective. On the other hand, $\phi$ is surjective since $\phi(a) = c$ and $\phi(ab) = d$. Therefore $\phi$ is an isomorphism.

Michael Albanese
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