# A question about a group presentation

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. : (

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
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.