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Kosniowski's A First Course in Algebraic Topology sure claims that $$\langle a, b \mid baba^{-1}\rangle$$ and $$\langle a, c \mid a^2c^2\rangle$$ are isomorphic to each other (see ch. 23 "The Seifert- Van Kampen theorem: I Generators", pages 178-179), and I'm not doubting the proof since it seems to be correct, and this claim is also found in other sources (online or not).

But... how CAN it be true? As far as I know, $\langle a, b \mid baba^{-1}\rangle$ is the fundamental group of the Klein bottle. An exercise found in another chapter of Kosniowski's very book even challenges the reader to prove that this group is isomorphic to a certain group $G$ generated by two transformations $$a(z) = z + i$$ and $$b(z) = \overline{z} + \frac{1}{2} + i$$ of the complex plane, and by looking at it geometrically it's pretty obvious that all elements $a^m(z)b^n(z)\in G$ are distinct from each other.

I'm actually trying to show that this group is indeed isomorphic to $\langle a, b \mid baba^{-1}\rangle$ right now... and not only can't I do it because there's no guarantee that $a^hb^i \neq a^jb^k$ whenever $h\neq j$ or $i\neq k$ (not as far as I know, at least), but now I've even learnt that this group is supposed to be the same as $\langle a, c \mid a^2c^2\rangle$, where elements like $a^4c^4$ and $a^2c^2$ are clearly the same!

I'm really clueless as to what's going on here. Did I somehow misunderstand how group presentations even work, or what?

Labba
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  • Also relevant are this, this and this. – Xander Henderson Feb 16 '20 at 00:40
  • And the answer to this question seems like it might give a bit more geometric intuition. – Xander Henderson Feb 16 '20 at 00:44
  • @XanderHenderson They all show how to prove they're isomorphic, but none of those questions give me an answer as to how it is possible that they're isomorphic to each other when one group (apparently) contains all elements of the form $a^mb^n$ and they're all distinct from each other while the other one clearly doesn't. – Labba Feb 16 '20 at 00:45
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    I don't understand your confusion. It is possible because you can prove that it is possible. You claim that you don't doubt the proof, but then you express doubts about the result. Is there some part of the proof which you do not understand? – Xander Henderson Feb 16 '20 at 00:49
  • I'm not trying to prove all of mathematics wrong, I know that proof must be correct but I just don't get the final results. How can two group presentations be isomorphic if one of them lets you end up with elements of the form $a^mb^n$ that are all distinct from each other, whereas in the other one you end up with stuff like $a^6c^6 = a^4c^4 = a2^c^2 = \dots$? How can they even be the same group? I know that they must be the same group because there's a proof that shows it, but then what's up with this? – Labba Feb 16 '20 at 00:52
  • Which exercise in Koniewski are you referencing regarding two transformations of the complex plane? – Semiclassical Feb 16 '20 at 01:12
  • @Semiclassical 17.9 (e) (page 167), later referenced in 19.5 (e) (page 177) and 23.1 (e) (page 205). – Labba Feb 16 '20 at 01:14
  • Okay. One point to clarify, then: In that exercise, the group relation is $ba=a^{-1}b\implies abab^{-1}=e$. This, rather annoyingly, is the opposite of what you want and what Koniewski does on page 177 ($a$ instead of $b$). So to avoid confusion one should probably reverse the labels on the $a(z)$ and $b(z)$ from that exercise. – Semiclassical Feb 16 '20 at 01:30
  • @Semiclassical Yep, I know, but that's the least of my worries. Regardless of how you label the letters, I already managed to prove that the elements of that group of complex functions can be written as $a^m(z)b^n(z)$ and that they're all distinct from each other, so I really only need to prove that the same can be said for that group presentation to show that they're equal. But I'm unable to do so, and the fact that this group is apparently the same as that other one with the $a^2c^2$ relation in which is this fact is clearly false is really confusing me. – Labba Feb 16 '20 at 01:36
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    It is possible, with any group, to generate the group by different sets. For example, the group $\mathbb Z$ has a presentation $\langle a,b \mid b=2a\rangle$, and it also has presentation $\langle a,c \mid c=3a\rangle$. Would your objections apply to that situation? – Lee Mosher Feb 16 '20 at 01:46
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    If you carry out Koniewski's isomorphism ($a\leftrightarrow a$, $ba^{-1}\leftrightarrow c$) on $b(z)=z+i$, $a(z)=\bar{z}+1/2+i$, then one obtains $c(z)=\bar{z}-1/2+2i$. One may then verify that $(a^2)(z) = z+1$ and $(c^2)(z) = z-1$, and hence $(a^2 c^2) (z) =z$ as required. So the fact that $(a^m b^n)(z)\neq z$ is irrelevant. – Semiclassical Feb 16 '20 at 02:14
  • You need to find a canonical form for the elements and an isomorphism that respects that. The obvious is $a^nb^n$ for the 1st group and $a\leftrightarrow d^{-1}$ with $b\leftrightarrow cd$ is the isomorphism so $d^{-n}(cd)^m$ is the corresponding canonical form for the 2nd. You need to check that these are indeed canonical forms. – Somos Feb 16 '20 at 03:48

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