= Solution
Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a',b',c')=(a+a',b+b'+ac',c+c')$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^cz^bx^a$. Consequently
$$
kG\cong B[x,x^{-1};\sigma],\qquad B=k[y,y^{-1},z,z^{-1}],\qquad\sigma(y)=zy,\quad\sigma(z)=z.
$$
The distinct normal-form <monomials> are the <group> <basis>, so this is an <isomorphism>, not merely a <surjection> from a presented algebra. This is the <integral Heisenberg group algebra as a skew Laurent ring>.
Here is the needed <skew Hilbert basis theorem>, including its proof. If $A$ is a <left Noetherian ring> and $\sigma$ an <automorphism>, let $J$ be a <left ideal> of $A[t;\sigma]$, where $ta=\sigma(a)t$. Define
$$
I_n=\{\sigma^{-n}(a_n):\ \sum_{i=0}^na_it^i\in J\}.
$$
These are <left ideals> of $A$: adding lifts adds the coefficient, and left multiplication by $\sigma^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient $\sigma^n(u_{nj})$.
For $f\in J$ of degree $m\ge N$, write its normalized <leading coefficient> as $\sigma^{-m}(\operatorname{lc}f)=\sum_j r_ju_{Nj}$. Then
$$
f-\sum_j\sigma^m(r_j)t^{m-N}f_{Nj}
$$
has smaller degree: the displayed summands have leading coefficients $\sigma^m(r_ju_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma]$ is a <left Noetherian ring>. Passing to the <opposite ring> replaces $\sigma$ by $\sigma^{-1}$ and proves the corresponding right-sided result.
This proof includes the ordinary <Hilbert basis theorem> when $\sigma$ is the identity. Starting with the <field> $k$, it makes $k[y,z]$ <Noetherian>; <localization> at powers of $y,z$ makes $B$ <Noetherian> as well. Finally, for a <left ideal> $L$ in $B[x,x^{-1};\sigma]$, its intersection $J$ with $B[x;\sigma]$ has finitely many generators. For each $f\in L$, some $x^qf$ lies in $J$, so $f=x^{-q}(x^qf)$ is generated by those same elements in the <skew Laurent polynomial ring>. Right multiplication clears powers for a <right ideal>. Therefore \b[$kG$ is a <left Noetherian ring> and a <right Noetherian ring>.]
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