A strict C prime(1/6) presentation that is a free group of rank six

Write $r=x_1\cdots x_7$. Its seven positive cyclic conjugates have distinct first letters, as do its seven negative inverse conjugates; no positive first letter equals a negative one. Thus no distinct symmetrised relators share a nonempty prefix, so there are no pieces and strict $C^{\mathrm{\prime }}(1\mathrm{/}6)$ holds vacuously. The cyclic word is not a proper power: each generator occurs exactly once, whereas a repetition at least twice would repeat each occurring letter.

Add the inverse relator $r^{-1}$ as a redundant relator and delete $r$, since each is a consequence of the other. Now $r^{-1}=x_7^{-1}w$ with $w=(x_1\cdots x_6{)}^{-1}$, which contains no $x_7$. There are no other relators, so the legal inverse dictionary move deletes $x_7$ and $r^{-1}$. The resulting presentation is $⟨x_1,\dots ,x_6\mid \mathrm{\varnothing }⟩$. Hence [L2] identifies $G$ with the free group on these six generators.

By [L3] that free group has no nonidentity element of finite order. An isomorphism preserves powers and the identity, so $G$ is torsion-free.