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Is the following a presentation of the free group generated by a single element?

$\langle\ a,b\ \vert\ aba=bab,\ abab=baba\ \rangle.$

My thinking is the following:

$abab = baba=b(bab)=b^2ab$ by substituting the first relation into the second. Simplifying, we get $a=b$. Since these steps give equivalent statements, the above presentation is in fact $\langle\ a,b\ \vert\ a=b\ \rangle$, i.e., the free group on one generator.

Is this correct?

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    $\begingroup$ Yes , What you have said its correct $\endgroup$ – Chirantan Chowdhury Feb 5 '17 at 18:52
  • $\begingroup$ Please don't forget to accept answers using the checkmark $\checkmark$ symbol :) $\endgroup$ – Shaun Aug 18 at 20:58
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Yes, this is correct.

One thing I would leave out is the $b^2ab$ step, for $a=b$ follows from $abab=b(bab)$ by cancelling $bab$ on the right.

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