Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.