Sc toolkit commuting positive words have a common root

Prove the claim by recursion on the positive integer $\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }$. When $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }y\mathrm{\mid }$, the prefixes of that length in the equality $xy=yx$ give $x=y$; take $c=x$ and $a=b=1$. This includes the smallest possible total length two.

If $\mathrm{\mid }x\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }$, prefix comparison gives $x=yu$ with $u$ nonempty. Substitute into $xy=yx$ to obtain $yuy=yyu$, and cancel the first literal copy of $y$, yielding $uy=yu$. The pair $(u,y)$ has smaller total length, so the recursive assertion gives $u=c^a$ and $y=c^b$ with $a,b>0$. Then $x=yu=c^{a+b}$. If $\mathrm{\mid }y\mathrm{\mid }>\mathrm{\mid }x\mathrm{\mid }$, interchange $x,y$ and the identical reduction gives the assertion. Each reduction strictly lowers a positive integer, so it terminates at the equal-length case. All cancellations are prefix cancellations of strings, not group reductions.