All-arity identities of two-generated metabelian groups via $(n,I)$ in $\mathbb Z[X^{\pm1},Y^{\pm1}]$

Proof. Reduce the equation modulo $X-1$. The quotient is $\mathbb Z[Y^{\pm1}]$, an integral domain, and $Y-1$ is nonzero in it. Hence $q=(X-1)f$ for some $f\in R$. Substitution and cancellation of $X-1$ give $p=(1-Y)f$. Cancellation also gives uniqueness. The reverse implication follows by multiplication. ◻

Proof. The kernel assertion is the rank-two case of the Magnus embedding theorem, equivalently the kernel theorem for the winding invariant (1, Theorem 14). We describe the image explicitly. The powers of the generators give

For any triple satisfying (12) with this linear coordinate, division by $a^r b^s$ on the left gives a triple with linear coordinate $1$. Lemma 4 identifies it uniquely with $c^f$.

The elements in (14) belong to the subgroup generated by $a,b$: integer multiples and sums of Laurent monomials correspond to products of conjugates of $c$ and their inverses. Thus every compatible triple is in the image. The linear coordinate is abelian, so $M'$ is contained in its kernel. Conversely, all $c^f$ lie in $M'$, giving (14). Injectivity of $(r,s)\mapsto X^rY^s$ and Lemma 4 give uniqueness of the normal form. ◻

Proof. The normal form gives the stated images, and the universal property of $M$ allows every such choice. Formula (10) gives

Applying this first to $a^s b^t,a^u b^v$ gives $c^{D_p}$. Adding the two commutator factors gives

Conjugation by $\phi(a)$ and $\phi(b)$ multiplies Laurent coordinates by $\alpha_p$ and $\beta_p$. Extend from Laurent monomials to their integral linear combinations using (16) to obtain (18). ◻

Proof. If the conditions hold, formula (18) places every $\phi(c^f)$ in $c^I$. Conversely, take $(h,k)=(0,0),(1,0),(0,1)$ and subtract the resulting Laurent coefficients. Since $\alpha_p$ and $\beta_p$ are units, one obtains all three memberships in (6).

For fixed $p$, each multiplier in (6) is fixed, and Laurent evaluation is a ring homomorphism. If a condition holds for $f_1,f_2$, it holds for $f_1+f_2$ and for $rf_1$, $r\in R$. The polynomials satisfying all conditions therefore form an ideal. This proves the assertion about generators. ◻

Proof. Let $\tau_i$ denote translation by the $i$th coordinate vector. The monic polynomial

annihilates $E$ under $T=\tau_i$. Its constant coefficient is a unit. From

we obtain $P_i(\tau_i)F=0$ by cancellation in $A$. Thus $F$ satisfies a recurrence of length $d$ in each coordinate, with unit first and last coefficients. Membership of $d$ consecutive values in $J$ propagates forward using the leading coefficient and backward using the constant coefficient. Apply this successively in all coordinates. If $d=0$, then $E=0$ and cancellation gives $F=0$ directly. ◻

Proof of Theorem 3. The ring $R$ is Noetherian: it is the image of the polynomial ring $\mathbb Z[U_1,U_2,U_3,U_4]$ under

Choose a finite ideal generating set $E$. Proposition 7 reduces (6) to these generators.

Fix $f\in E$ and write $m=|\mathrm{supp}(f)|$. Each monomial of $f(\alpha_p,\beta_p)$ is an exponential term in the four integer coordinates $s,t,u,v$. There are $m$ such terms. Multiplication by $\alpha_p-1$ or $\beta_p-1$ gives a sum of $2m$ exponential terms, all with unit rates.

For the remaining multiplier, put $\delta=(X-1)(Y-1)\ne0$. Direct expansion of (3) yields

Consequently, $\delta D_p f(\alpha_p,\beta_p)$ is a sum of $6m$ exponential terms. Apply Lemma 8 with $A=R$ and $J=I$ to $D_p f(\alpha_p,\beta_p)$. Apply it with $\delta=1$ to the other two expressions. The box of side $6m$ includes the needed boxes of side $2m$, proving sufficiency. Necessity is immediate. The uniqueness assertion follows from Theorem 2. ◻

Proof. Apply the retractions $(a,b)\mapsto(a,1)$ and $(a,b)\mapsto(1,a)$ to an element of $N$ in normal form. They kill $M'$ and give $a^r,a^s\in N$. Hence $n\mid r,s$. Substitution also gives $b^s\in N$, and cancellation in $N$ gives $c^f\in N$. Conversely, if $n\mid r,s$ and $c^f\in N$, the two power factors belong to $N$ by (27), so their product with $c^f$ belongs to $N$. ◻

Proof. The commutator formulas follow from the affine coordinates of $a^n,b^n$. For the product power, define over a commutative ring

Induction on $n$ gives

Now $(ab)^n=(X^nY^n;S_n(XY),X S_n(XY))$. Since $C_n(X,Y)=T_n$, the two displayed identities identify this with $c^{-XT_n}a^nb^n$. ◻

Proof. Apply (27) to $c$: since $c^n=c^{n\cdot1}$, we obtain $n\in I$. The first two formulas in (29), together with $a^n,b^n\in N$, give $S_n(X),S_n(Y)\in I$. The third gives $-XT_n\in I$ because $(ab)^n,a^n,b^n\in N$. As $-X$ is a unit, $T_n\in I$. Full invariance of $c^I$ supplies the substitution conditions. ◻

Proof. First form $Q=M/c^I$. By Proposition 7, $c^I$ is fully invariant. The conditions $S_n(X),S_n(Y)\in I$ and Lemma 10 show that $a^n,b^n$ are central in $Q$. Let $C=\langle a^n,b^n\rangle\le Q$.

The linear-coordinate homomorphism $Q\to R^\times$ sends

Its restriction to $C$ is injective. For $n>0$, a trivial image forces $r=s=0$; for $n=0$, $C=1$. Every element of $Q'$ has trivial linear coordinate. Therefore $C\cap Q'=1$, and quotienting by $C$ introduces no additional relations in the image of $M'$. This proves $N(n,I)\cap M'=c^I$ and the derived-subgroup assertion of Theorem 2.

Since $T_n\in I$, the last formula in (29) gives $(ab)^n=a^nb^n$ in $Q$. Full invariance of $c^I$ permits substitution of arbitrary elements of $M$ for $a,b$. Hence

Thus $x\mapsto x^n$ is a homomorphism on $Q$. After passage to $Q/C$, it kills both generators and is therefore trivial. It follows that $g^n\in N(n,I)$ for all $g\in M$.

An endomorphism of $M$ preserves $c^I$ and sends each of $a^n,b^n$ into $N(n,I)$. It therefore preserves their normal closure, proving full invariance. Finally, if $a^z\in N(n,I)$, then its image in $Q$ lies in $C$. Comparing linear coordinates in (34) gives $n\mid z$. The reverse implication is immediate. Thus the image of $a$ in $G(n,I)=Q/C$ has order exactly $n$, with the stated convention at zero. ◻

Proof. Every value in $M$ of an identity of $M/N$ belongs to $N$, proving one inclusion. For the reverse inclusion, take $g\in N$ and a word $w(a,b)$ representing it. Given any two elements of $M/N$, lift them to $x,y\in M$. The endomorphism sending $a,b$ to $x,y$ sends $g$ into $N$. Consequently $w$ is a two-variable identity of $M/N$, and its value $g$ belongs to the left side of (36). ◻

Proof of Theorem 2. Let $H$ be a two-generated metabelian group and choose an epimorphism $M\to H$. Put $N=V_{\mathrm{Id}(H)}(M)$. The quotient $M/N$ satisfies $\mathrm{Id}(H)$ by construction. The epimorphism to $H$ factors through $M/N$, so every identity of $M/N$ is also an identity of $H$. Hence

The subgroup $N$ is fully invariant, so it equals $N(n,I)$ for its unique admissible pair. This proves existence.

If two admissible pairs give the same identities, Lemma 13 shows that their fully invariant kernels in $M$ coincide. Recovering the exponent and the ideal gives equality of the pairs. Conversely, equal pairs define the same quotient. Every such quotient is metabelian and two-generated. The exact exponent and the isomorphism $G(n,I)'\cong(R/I,+)$ were proved in Proposition 12. ◻