Determine whether the fundamental group of a Mobius band glued to a torus is abelian

algebraic-topologyfundamental-groups

I'm studying for a comprehensive exam in topology, and this is an old question from a few years ago:

Let $X$ be the topological space obtained by gluing together a Mobius band $B$ and a torus $T= S^1 \times S^1$ by identifying the boundary circle $C \subset B$ with the loop $C' = S^1 \times \{x_0\}$ inside $T$. Let $G = \pi_1(X)$.

  1. Describe $G$ in terms of generators and relations.
  2. Is $G$ abelian?
  3. Compute $H_1(X)$.

I'm unsure of how to see whether $G$ is abelian… Using Van Kampen's theorem (by taking open neighborhoods of $B, T, C$ which deformation retract onto each respective space), we see that $G = \pi_1(X) = \langle b, c : c^2 b = bc^2 \rangle$. My question is: how one can use this presentation of $G$ to deduce whether it is abelian?

I know that $H_1(X) \cong G^{ab}$ and I believe that I can, in theory, deduce what $H_1(X)$ is via an argument using Mayer-Vietoris… so, if $H_1(X)$ ends up equaling $G$ then I guess I could call it a day. However, all of the things I searched on justifying why a fundamental group is abelian involves some abstract nonsense that doesn't really make use of how it is defined by its generators and relations, so I feel like there should be a fairly easy way of deducing it from that. The only algebra I have under my belt is commensurate with my university's undergraduate algebra sequence, which is (allegedly) sufficient for these questions.

Any guidance would be greatly appreciated! Even if it is just a link to a reference for how to reconcile these ugly presentations with what the groups 'actually' are.

Best Answer

To show that a group is not abelian, it suffices to find a nonabelian quotient, eg by adding relations. Now there are many obvious quotients of the group you found that are easily seen to be nonabelian, eg $\langle b, c : b^2 = c^2 = 1 \rangle \cong \mathbb{Z}/2 * \mathbb{Z}/2$ or $S_3$ by mapping $c$ to a transposition and $b$ to a $3$-cycle.

As @Ben mentions in the comments, to find the abelianization, it suffices to add relations making every pair of generators commute, and in this case it should be recognizable as $\mathbb{Z}^2$. Let $\langle S : R \rangle$ be a presentation of a group $G$ and let $R'$ be the union of $R$ with $[S, S]$, ie the set of commutators of pairs in $S$. Then $G' = \langle S : R' \rangle$ is an abelian quotient of $G$ and it is not hard to check that the kernel of the quotient map $G \to G'$ is $[G, G]$.

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