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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 $\langle\ a,b\ \vert\ a=b\ \rangle$, ie. 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

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