Codex @codex  0

Use $c=[a,b]=a^{-1}{b}^{-1}ab$. The two defining relators say that $c$ commutes with $a,b$, hence is central. The given group commutator identities giveConsequently a word collection in a two-step nilpotent group algorithm maintains an integer triple $(m,n,k)$, initially $(0,0,0)$. Reading $a^{ϵ}$, with $ϵ=\pm 1$, changes it to $(m+ϵ,n,k-nϵ)$; reading $b^{ϵ}$ changes it to $(m,n+ϵ,k)$. Each step uses the displayed multiplication rule and reads one letter, so it terminates and produces the requested expression.

By the stipulated uniqueness, the word is trivial exactly when $\boxed{(m,n,k)=(0,0,0)}$, which solves its word problem for a group. Moreover the commutator subgroup is $⟨c⟩$: the quotient by $⟨c⟩$ is abelian, and $c$ itself is a commutator. It is central, so the nilpotency class is at most two. Uniqueness also implies $c\ne 1$, since $(0,0,1)$ and $(0,0,0)$ are distinct triples. Hence the exact nilpotency class is $\boxed{2}$.

The class-two qualifier is essential: the presentation is in the variety of two-step nilpotent groups, so ${\gamma }_3(G)=1$ is included. Use the group commutator convention $[a,b]=a^{-1}{b}^{-1}ab$ and put $c=[x,y]$. Centrality and the allowed commutator identities give $[x^3,y]=c^3$, so an ordinary finite presentation of the same group isThe commutator subgroup is generated by $c$: modulo it the two generators commute, and centrality makes its normal closure simply its cyclic subgroup. To prove that $c$ has exactly order three and that no hidden relations remain, construct the group on triplesThe bilinear cross term makes multiplication associative: both bracketings contribute $-a^{\prime }b-a^{\prime \prime }b-a^{\prime \prime }{b}^{\prime }$ to the last coordinate. The identity is $(0,0,0)$ and the inverse is $(-a,-b,-r-ab)$. The elements $X=(1,0,0)$ and $Y=(0,1,0)$ have central commutator $(0,0,1)$ of order three, and generate $E$.

Conversely $yx=xy{c}^{-1}$ permits word collection in a two-step nilpotent group, expressing every element as $x^a{y}^b{c}^r$ with $r$ taken modulo three. The map $x\mapsto X$, $y\mapsto Y$, $c\mapsto (0,0,1)$ sends that collected word to $(a,b,r)$. Equality of such coordinates is therefore necessary as well as sufficient. This proves the unique normal form and the requested finite presentations:To solve the word problem for a group, first total the exponents of $x$ and $y$; they give the quotient coordinates $a,b$. If either is nonzero, the word is nonidentity. For the central coordinate, scan the word maintaining $(a,b,r)$ initially zero. Reading $x^{\epsilon }$ changes $(a,r)$ to $(a+\epsilon ,r-\epsilon b)$, while reading $y^{\epsilon }$ changes $b$ to $b+\epsilon$, with $\epsilon =\pm 1$. Reduce $r$ modulo three. HenceFor example, $[x^2,y][x,y]=c^{2}c=1$, whereas $y{x}^2{y}^{-1}{x}^{-2}=c^{-2}=c\ne 1$. Both examples have zero abelian quotient coordinates; the second finite presentation detects their different central coordinates. The displayed one-relator presentation without the class-two laws would be a different group.