C prime one sixth torsion elements come from relator roots

Choose a shortest conjugacy representative using [F2]. By [F1], a rotation $v$ of this representative and a relator $r$ satisfy $v=c^a$, $r=c^b$ for a nonempty word $c$ and positive integers $a,b$. Rotation is conjugation in the free group, since $pq$ rotates to $qp=p^{-1}(pq)p$. Therefore $g$ is conjugate in the quotient to the image of $c^a$.

Among words $d$ with $c=d^k$ literally for some positive $k$, choose one of shortest length. The finite set of candidate lengths is nonempty because $d=c,k=1$ is allowed. If $d=e^j$ with $j\ge 2$, then $c=e^{jk}$, contradicting the shorter length of $e$. Hence $d$ is not a proper power, $r=d^{kb}$, and $v=d^{ka}$. Thus $d$ is a root in [F2] and $g$ is conjugate to a power of its image, as claimed. If $kb=1$, the relator itself is the root and is trivial in the quotient, which would contradict $g\mathrm{\ne }1$; this endpoint simply cannot occur for the given element.