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

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.

• Hint: In terms of generators and relations, abelianization is just adding the relations $g_1 g_2 = g_2 g_1$ for all generators $g_1$ and $g_2$, so in your case we have $G^{\mathrm{ab}} = \langle b, c \mid c^2 b = b c^2, bc = cb \rangle = \langle b, c \mid bc = cb \rangle \cong \mathbb{Z}^2$. Commented Jul 22 at 20:33
• @BenSteffan This is very helpful, thank you!!! So since $G^{\mathrm{ab}} = G / [G,G]$ and (hopefully...) $G^\mathrm{ab} \not\cong G$, this group can't be abelian? Commented Jul 22 at 21:06
• A group is abelian if and only if $G^{\mathrm{ab}} \cong G$, so that's what you're trying to show, yes. Finding non-abelian quotients of $G$ as in ronno's answer is one (quite sleek & effective) way to do that. Commented Jul 22 at 21:27

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 $$[G, G]$$.

• Thank you so much, that makes life much easier! Would you happen to know a reference for finding nonabelian quotients via adding relations as you mention in the first paragraph? I tried Googling, and unfortunately most results were over my head or not directly relevant. Commented Jul 22 at 21:09
• @homieo'morphic added a sketch, please ask for more details if you need Commented Jul 22 at 22:37
• This was incredibly illuminating, thank you! :) Commented Jul 23 at 3:39