TITLE: Is $\langle\ a,b\ \vert\ aba=bab,\ abab=baba\ \rangle$ a presentation of the free group on a single generator?
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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.