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?