Decision Problems for Finitely Presented Groups - Examples

Cancel the adjacent pair $b{b}^{-1}$ in $w$, then cancel the adjacent pair $a^{-1}a$. The resulting reduced word is $ab{a}^{-1}$.

The reduced word $ab{a}^{-1}$ is not empty, so [L1] implies that the original word $w$ is nontrivial.

In the cyclic group of order $6$, exponents are taken modulo $6$, so $x^{14}=x^2$.

Because $2\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}6)$, the element $x^2$ is not the identity. This is the abelian coordinate check promised by [L1].

The word $b{a}^{-1}{b}^{-1}a$ is obtained from $ab{a}^{-1}{b}^{-1}$ by moving the initial letter $a$ to the end. So it is a cyclic permutation of the first word.

Both words are cyclically reduced, and [L1] therefore makes them conjugate in the free group.

In a finite quotient obtained from a partial Todd-Coxeter table, if the input word acts nontrivially on some enumerated coset, then the word is certainly nontrivial in the presented group, because every quotient homomorphism preserves equality with the identity.

This shows how a partial enumeration can settle some instances by exhibiting a separating finite quotient. It does not claim that every enumeration terminates or every instance is decided.

The word $a^6$ is the product $(a^3)(a^3)$, so [L1] gives $\mathrm{Area}(a^6)\le 2$.

It is not $1$, because a single conjugate of the only relator or its inverse is just $a^3$ or $a^{-3}$ in this cyclic presentation. Thus [L2] forces the minimum to be exactly $2$.

The presentation $⟨a\mid \mathrm{\varnothing }⟩$ presents the free group on $a$. The word $a$ is reduced and nonempty, so [L2] says that it is nontrivial.

By [L1], enumerating relator consequences lists precisely the trivial words. Since $a$ is nontrivial by step 1.1, it never appears, and waiting for it gives no halting certificate of nontriviality.

Hence the asserted decision procedure is false, and the claimed principle is refuted.