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Is it true that the group presented by $\langle a, b \mid ab = 1\rangle$ is isomorphic to the free group generated by $a$?

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Yes: $\,b=a^{-1}\,$ , in fact. –  DonAntonio Jan 14 '13 at 3:30
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up vote 3 down vote accepted

All you are looking for is summerized in @Don's leading comment. Let $H=\langle a\rangle$. Since $$ab=1\to aba^{-1}=a^{-1}\to H\unlhd G$$ Now we can have the presentation quotient group $G/\langle a\rangle $ as $$G/\langle a\rangle\cong\langle a,b\mid ab=1, a=1\rangle\cong\langle a,b\mid a=1,b=1\rangle\cong \{1\}$$ This what you wanted.

Edit: After being remarked by @user1729, I have found out that there is a hole in my proof above. If fact as $a=b^{-1}$ we have also $$b^{-1}ab=b^{-1}=a\in H$$ and $$bab^{-1}=a^{-1}ab^{-1}=b^{-1}=a\in H$$ These points all together prove $H$ is normal in $G$.

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That's not quite right - you need to show that $bab^{-1}\in H$ and $b^{-1}ab\in H$. You have shown $aba^{-1}\in H$! It is a nice idea though. As $ab=1$ and $a=b^{-1}$ we have $b^{-1}ab=b^{-1}=a\in H$ while as $ba=b(ab)b^{-1}=1$ and $b^{-1}=a$ we have $bab^{-1}=b^{-1}=a\in H$. –  user1729 Jan 14 '13 at 10:39
    
@user1729: It is fixed. I don't know what happened to me in that time not to consider that elementary requisites. Thanks a lot. –  B. S. Jan 14 '13 at 11:08
    
I don't think your edit plugs the hole... –  user1729 Jan 14 '13 at 11:19
    
@user1729: Am I losing more? Sorry... –  B. S. Jan 14 '13 at 14:32
    
It is better now, although you do not need the $ba=b(ab)b^{-1}$ bit. I was just doing that to point out that $ba=1$ so $bab^{-1}=b^{-1}=a\in H$. You have proven this in a different way. (Also, I should point out, your last line of working should read $bab^{-1}=a^{-1}ab^{-1}=b^{-1}=a\in H$...you have a $b$ which should be a $b^{-1}$.) –  user1729 Jan 14 '13 at 15:14
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Because Tietze transformations don't change the isomorphy class of the group, $$\langle a,b|b=a^{-1} \rangle \simeq \langle a | \ \rangle \simeq \mathbb{Z}$$

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