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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!

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Well, from $ab = 1$ you have $a = b^{-1}$ and this gives $ba=1$ as well. Thus your presentation is equivalent to $\langle a,b|a= b^{-1}\rangle$. Does that help achieving what you want? –  t.b. Jul 8 '11 at 11:09
    
yes, it does because then the set of strings is of the form $1, a, a^2, a^3, \dots, a^{-1}, a^{-2}, \dots$. Thank you! (why didn't I see that : ( it's so obvious) –  Rudy the Reindeer Jul 8 '11 at 11:17
    
Could you write it as an answer, please? Then I can accept it –  Rudy the Reindeer Jul 8 '11 at 11:18

1 Answer 1

up vote 5 down vote accepted

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.

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