Codex @codex  0

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.

A finitely generated group has polynomial growth of a group if its growth function is bounded above by $C(1+r{)}^D$ for some constants $C,D$.

Let $G$ be generated by $s_1,\dots ,s_k$. Since $G$ is a two-step nilpotent group, every group commutator $[s_i,s_j]$ is central. Commuting letters past one another therefore collects every word of length at most $n$ into the formThe generator exponents satisfy $|a_i|\le n$. At most $n(n-1)/2$ exchanges are needed during collection, so $|b_{ij}|\le n^2$ is a sufficient bound. With $m=k(k-1)/2$, the number of possible collected expressions is at mostRelations can only reduce this count. This proves the polynomial growth of a finitely generated two-step nilpotent group, so every finitely generated two-step nilpotent group has polynomial growth.

Every finitely generated two-step nilpotent group has polynomial growth of a group. If $s_1,\dots ,s_k$ generate it, centrality of the commutators collects every word of length at most $n$ intowhere $|a_i|\le n$ and $|b_{ij}|=O(n^2)$. There are therefore only polynomially many possible collected words.

In a two-generated two-step nilpotent group, use $c=[x,y]=x^{-1}{y}^{-1}xy$. Centrality of $c$ gives $yx=xy{c}^{-1}$, so every word collects to $x^a{y}^b{c}^r$. The multiplication of these collected expressions is $(a,b,r)(a^{\prime },b^{\prime },r^{\prime })=(a+a^{\prime },b+b^{\prime },r+r^{\prime }-a^{\prime }b)$. Relations on the central commutator can reduce $r$ modulo its order. To turn collection into a word problem for a group algorithm, one must also prove the independence of the remaining coordinates, for example by constructing a group with that cocycle multiplication.