# Infinite and non-abelian fundamental group

$$\require{AMScd}$$ I ran into trouble while trying to answer this question. I am trying to prove the following:

Suppose $$U_1,U_2$$ and $$U_3 := U_1 \cap U_2$$ are open, path-connected subsets of $$X = U_1 \cup U_2$$. Suppose that $$\pi_1(U_1) \cong \mathbb{Z}/3$$, $$\pi_1(U_2) \cong \mathbb{Z}/4$$ and $$\pi_1(U_3) \cong \mathbb{Z}/2$$. Show that $$\pi_1(X)$$ is infinite and non-abelian.

Using van Kampen, we get the following commutative diagram:

$$\begin{CD} \mathbb{Z}/2 @>\varphi_1>> \mathbb{Z}/3\\ @V\varphi_2VV @VV\psi_1V \\ \mathbb{Z}/4 @>\psi_2>> \pi_1(X) \end{CD}$$

Here, $$\varphi_1,\varphi_2,\psi_1,\psi_2$$ are the homomorphisms induced by the canonical inclusions $$U_3 \hookrightarrow U_1$$, $$U_3 \hookrightarrow U_2$$, $$U_1 \hookrightarrow X$$ and $$U_2 \hookrightarrow X$$ respectively. Now, since $$\# \text{Hom}_\mathbb{Z}(\mathbb{Z}/2,\mathbb{Z}/3) = \gcd(2,3) = 1,$$ we must have that $$\varphi_1 = 0$$ is the zero map. Since the diagram commutes, we also get $$\psi_2 \circ \varphi_2 = 0$$. Using group presentation, we can write $$\pi_1(U_1) \cong \langle \alpha \mid \alpha^3\rangle$$, $$\pi_1(U_2) \cong \langle \beta \mid \beta^4\rangle$$ and $$\pi_1(U_3) \cong \langle \gamma \mid \gamma^2\rangle$$. Again by van Kampen, we get \begin{align} \pi_1(X) \cong (\mathbb{Z}/3) *_{\mathbb{Z}/2} (\mathbb{Z}/4) \cong \left\langle \alpha, \beta \mid \alpha^3,\; \beta^4,\; 0=\varphi_2(\gamma)\right\rangle. \end{align} My problem is that I'm not sure how to continue. Can we say anything about the map $$\varphi_2$$? Since $$\# \text{Hom}_\mathbb{Z}(\mathbb{Z}/2,\mathbb{Z}/4) = \gcd(2,4)=2$$ there are two possibilities for $$\varphi_2$$. One is the zero map again, and the other is the map $$0 \mapsto 0$$ and $$1\mapsto 2$$. In the case that $$\varphi_2$$ is indeed the zero map, the claim follows immediately. Is there something that we can infer from $$\psi_2 \circ \varphi_2=0$$?

• math.stackexchange.com/q/4276473 was asked 9 hours ago. Just a coincidence? Oct 14, 2021 at 22:51
• @PaulFrost No, I linked to that post in my question for a reason. Very first line even. Oct 14, 2021 at 22:58

I will write multiplicatively the operation in all involved groups. So the diagram: $$\begin{CD} \mathbb{Z}/2 @>\varphi_1>> \mathbb{Z}/3\\ @V\varphi_2VV @VV\psi_1V \\ \mathbb{Z}/4 @>\psi_2>> \pi_1(X) \end{CD}$$ is $$\begin{CD} \langle \ \gamma\ |\ \gamma^2 =1\ \rangle @>\varphi_1>> \langle \ \alpha\ |\ \alpha^3 =1\ \rangle \\ @V\varphi_2VV @VVV \\ \langle \ \beta\ |\ \beta^4 =1\ \rangle @>>> \pi_1(X) \end{CD}$$ and by van Kampen $$\pi_1(X)$$ is the group with the generators and relations copied from (the multiplicative versions of) $$\Bbb Z/4$$ and $$\Bbb Z/3$$, amalgamated w.r.t. $$\varphi_1(\gamma)=\varphi_2(\gamma)$$. Of course, $$\varphi_1(\gamma)=1$$. For $$\varphi_2(\gamma)$$ we have two chances, either $$1$$ or $$\beta^2$$. So the chances for $$\pi_1(X)$$ are either \begin{aligned} \pi_1(X) &=\langle \ \alpha,\beta\ |\ \alpha^3=1\ ,\ \beta^4=1\ ;\ 1=1\ \rangle \\ &=\langle \ \alpha,\beta\ |\ \alpha^3=1\ ,\ \beta^4=1\ \rangle \ , \\[3mm] &\qquad\text{ or } \\[3mm] \pi_1(X) &=\langle \ \alpha,\beta\ |\ \alpha^3=1\ ,\ \beta^4=1\ ;\ \beta^2=1\ \rangle \\ &=\langle \ \alpha,\beta\ |\ \alpha^3=1\ ,\ \beta^2=1\ \rangle \ . \end{aligned} Two infinite groups with above presentations.
The last relation $$\phi_2 (\gamma) = 0$$ is either $$\beta^2 = 0$$ or $$\beta^0 = 0$$ (the last of which is trivial so doesn't add any information)
This implies that $$\pi_1(X) = \langle\alpha, \beta \mid \alpha^3, \beta^n \rangle$$ where $$n = 2$$ or $$4$$. This is the case since $$\langle\alpha, \beta \mid \alpha^3, \beta^2 \rangle =\langle\alpha, \beta \mid \alpha^3, \beta^4, \beta^2 \rangle$$.
This means that in either case $$\pi_1(X)$$ is a free product of two nonzero groups ($$\mathbb Z_3 * \mathbb Z_4$$ or $$\mathbb Z_3 * \mathbb Z_2$$) which is always infinite and non-abelian.
• Thanks, your answer was helpful. I was confused, because getting something like $\pi_1(X) \cong \mathbb{Z}_3 * \mathbb{Z}_4$ felt wrong at first. I see why it makes sense now. Oct 14, 2021 at 23:07