4
$\begingroup$

$\require{AMScd}$ 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 $\psi_2 \circ \varphi_2=0$?

$\endgroup$
2
  • $\begingroup$ math.stackexchange.com/q/4276473 was asked 9 hours ago. Just a coincidence? $\endgroup$
    – Paul Frost
    Oct 14, 2021 at 22:51
  • $\begingroup$ @PaulFrost No, I linked to that post in my question for a reason. Very first line even. $\endgroup$
    – jasnee
    Oct 14, 2021 at 22:58

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you, it makes sense now! $\endgroup$
    – jasnee
    Oct 14, 2021 at 23:05
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – jasnee
    Oct 14, 2021 at 23:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .