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I have calculated the fundamental group of the annulus and got the following group presentation:

$ \langle a, b | ab = ba = 1 \rangle$

This is the set of strings of the form: $1, a, a^2, a^3, \dots , b , b^2 , \dots$.

Is this equivalent to $\langle a | \rangle = \mathbb{Z}$? If yes, how do I see that?

Edit I think it's not equivalent. : (

Many thanks for your help!

1 Answers 1

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On Matt's request I'm posting my comment as an answer:

From $ab=1$ we have $a = b^{-1}$ and thus also $ba = 1$. This means that your presentation is equivalent to $\langle a,b\mid a = b^{-1}\rangle$ and thus your group is isomorphic to $\mathbb{Z} \cong \langle a \mid \;\rangle$ as you wanted.