Coordinate retractions and the image of the Magnus–Shmelkin embedding

Problem

Which elements lie in the image of the Magnus–Shmelkin embedding?

F/\mathcal V(R)\hookrightarrow B\rtimes(F/R), \qquad F=F(X),\quad R\trianglelefteq F.

Here $\mathcal V$ is an arbitrary group variety and $B$ is the corresponding relatively free verbal base. The requested criterion is structural; an algorithm requires additional effective hypotheses.

Setup and result

Let $F=F(X)$ be free on $X=\{x_i:i\in I\}$ , let $R\trianglelefteq F$ , and write

Q=F/R,\qquad \pi:F\longrightarrow Q,\qquad a_i=\pi(x_i).

Fix a variety $\mathcal V$ of groups. Its verbal subgroup in a group $H$ is denoted by $\mathcal V(H)$ . Let

B=F_{\mathcal V}(\{e_{g,i}:g\in Q,\ i\in I\}), \qquad h\cdot e_{g,i}=e_{hg,i}.

The verbal wreath product has the model $W=B\rtimes Q$ , with multiplication

\tag{3} (b,q)(c,r)=(b(q\cdot c),qr).

The assignment $x_i\mapsto(e_{1,i},a_i)$ defines a homomorphism

\tag{4} \eta_0:F\longrightarrow W,\qquad \eta_0(f)=(d(f),\pi(f)).

The Magnus–Shmelkin theorem identifies its kernel with $\mathcal V(R)$ . We shall include a proof in Section “The exact embedding kernel”.

Problem 7.58 of the Kourovka Notebook (3) asks which elements of $W$ belong to the image of the induced embedding. McCool (1, pp. 139–140) reduces this question to membership in the base image of $R/\mathcal V(R)$ . His equation (3.2) is necessary in general and sufficient for the abelian variety; his Corollary 5 gives an algorithm under word-problem hypotheses. Romanovskii (2, Theorem 2) gives a criterion for abelian varieties in a more general free-product setting.

The construction below uses a retraction onto the base image of $R$ . McCool’s free-factor decomposition (1, Theorem 1(4)) already provides the underlying structural antecedent. Here we give the retraction on the original generators of $B$ and prove the resulting equality criterion directly. The proof uses only the universal property of relatively free groups and the cocycle identity for $d$ .

Choose a set-theoretic section $s:Q\to F$ satisfying

\tag{5} \pi(s(g))=g,\qquad s(1)=1,

and put $\tau_g=d(s(g))$ . Since $B\in\mathcal V$ , the assignment

\tag{6} \rho_s(e_{g,i})=\tau_g e_{g,i}\tau_{ga_i}^{-1}

extends uniquely to an endomorphism $\rho_s:B\to B$ .

Theorem 1. The map $\eta_0$ induces an embedding

\eta:F/\mathcal V(R)\hookrightarrow B\rtimes Q, \qquad x_i\mathcal V(R)\longmapsto(e_{1,i},a_i).

For every section satisfying (5),

\tag{8} (b,q)\in\eta(F/\mathcal V(R)) \quad\Longleftrightarrow\quad \rho_s(b)=b\tau_q^{-1}.

Moreover, $\rho_s^2=\rho_s$ and $\rho_s(B)=d(R)$ .

The criterion is an equality in the given relatively free base. Neither the definition of $\rho_s$ nor that of $\tau_g$ uses a membership test for the embedded subgroup. No effective section or decidable equality is assumed, so the theorem asserts no unrestricted decision procedure.

The coordinate retraction

The homomorphism (4) gives

\begin{align*} d(1)&=1,& d(x_i)&=e_{1,i},\tag{9a}\\ d(fh)&=d(f)(\pi(f)\cdot d(h)),& d(f^{-1})&=\pi(f)^{-1}\cdot d(f)^{-1}.\tag{9b} \end{align*}

In particular, $d|_R:R\to B$ is a homomorphism and $\tau_1=1$ .

Lemma 2. For $f\in F$ and $g\in Q$ ,

\tag{10} \rho_s(g\cdot d(f)) =\tau_g(g\cdot d(f))\tau_{g\pi(f)}^{-1}.

Consequently,

\tag{11} \rho_s(d(f))=d(f)\tau_{\pi(f)}^{-1}, \qquad \rho_s(\tau_g)=1.

Proof. For $f=x_i$ , equation (10) is (6). For $f=x_i^{-1}$ , the inverse identity in (9b) gives

g\cdot d(x_i^{-1})=e_{ga_i^{-1},i}^{-1},

whose image under $\rho_s$ is

(\tau_{ga_i^{-1}}e_{ga_i^{-1},i}\tau_g^{-1})^{-1} =\tau_g e_{ga_i^{-1},i}^{-1}\tau_{ga_i^{-1}}^{-1}.

If (10) holds for $f$ and $h$ , then (9b) and multiplicativity of $\rho_s$ give

\begin{align*} \rho_s(g\cdot d(fh)) &=\rho_s(g\cdot d(f))\rho_s(g\pi(f)\cdot d(h))\\ &=\tau_g(g\cdot d(f))\tau_{g\pi(f)}^{-1} \tau_{g\pi(f)}(g\pi(f)\cdot d(h))\tau_{g\pi(fh)}^{-1}\\ &=\tau_g(g\cdot d(fh))\tau_{g\pi(fh)}^{-1}. \end{align*}

Word induction proves the formula. Taking $g=1$ gives the first identity in (11); taking $f=s(g)$ there gives the second. ◻

Lemma 3. The endomorphism $\rho_s$ is idempotent, and

\tag{15} \rho_s(B)=d(R)=\{b\in B:\rho_s(b)=b\}.

Proof. For $g\in Q$ and $i\in I$ , define the section loop

\tag{16} r_{g,i}=s(g)x_i s(ga_i)^{-1}\in R.

Using (9b) and (5), we obtain

\tag{17} d(r_{g,i})=\tau_g e_{g,i}\tau_{ga_i}^{-1}=\rho_s(e_{g,i}).

The subgroup $d(R)$ therefore contains the images under $\rho_s$ of all generators of $B$ , and hence $\rho_s(B)\subseteq d(R)$ . For $r\in R$ , equation (11) gives $\rho_s(d(r))=d(r)$ . This proves both inclusions in (15) and shows that $\rho_s$ fixes its image. Thus $\rho_s^2=\rho_s$ . ◻

Proof of the image criterion. If $(b,q)=\eta_0(f)$ , then (11) gives $\rho_s(b)=b\tau_q^{-1}$ . Conversely, suppose this equality holds. By Lemma 3, choose $r\in R$ with $d(r)=\rho_s(b)$ . Then

d(rs(q))=d(r)d(s(q))=\rho_s(b)\tau_q=b, \qquad \pi(rs(q))=q.

Hence $(b,q)=\eta_0(rs(q))$ . This proves (8) for the image of $\eta_0$ ; the kernel calculation below identifies that image with the stated quotient. ◻

The exact embedding kernel

We write $N=\mathcal V(R)$ as a subgroup of $F$ . Verbal subgroups are fully invariant, so $N\trianglelefteq F$ . Let

\kappa:F\longrightarrow F/N, \qquad U=R/\mathcal V(R),

and let $j:U\to F/N$ be the homomorphism induced by inclusion of $R$ in $F$ . Since $U\in\mathcal V$ , the assignment

e_{g,i}\longmapsto r_{g,i}\mathcal V(R)

extends to a homomorphism $\lambda:B\to U$ . Set $L=j\lambda$ . Thus

\tag{21} L(e_{g,i})=\kappa(s(g))\kappa(x_i)\kappa(s(ga_i))^{-1}.

Lemma 4. For $g\in Q$ and $f\in F$ ,

\tag{22} L(g\cdot d(f))=\kappa(s(g))\kappa(f)\kappa(s(g\pi(f)))^{-1}.

Proof. For $f=1$ both sides are $1$ , and for $f=x_i$ the formula is (21). For an inverse generator,

\begin{align*} L(g\cdot d(x_i^{-1})) &=L(e_{ga_i^{-1},i})^{-1}\\ &=\kappa(s(g))\kappa(x_i)^{-1}\kappa(s(ga_i^{-1}))^{-1}. \end{align*}

For a product $fh$ , the cocycle identity expresses the left side as

L(g\cdot d(f))L(g\pi(f)\cdot d(h)).

The two middle factors $\kappa(s(g\pi(f)))^{-1}$ and $\kappa(s(g\pi(f)))$ cancel. Word induction gives (22). ◻

Proposition 5. The kernel of $\eta_0$ is $\mathcal V(R)$ .

Proof. The restriction $d|_R$ is a homomorphism into the group $B\in\mathcal V$ , so it kills $\mathcal V(R)$ . The second coordinate $\pi$ kills $R$ . Hence $N\subseteq\ker\eta_0$ .

Conversely, if $\eta_0(f)=1$ , then $d(f)=1$ and $\pi(f)=1$ . Taking $g=1$ in (22) and using $s(1)=1$ gives

1=L(d(f))=\kappa(f).

Thus $f\in N$ . The induced map $F/N\to W$ is injective, completing the proof of Theorem 1. ◻

References

Preprint · Lean (GitHub)

J. McCool, On the Magnus–Smelkin embedding, Proc. Edinburgh Math. Soc. (2) 30 (1987), no. 1, 133–142, doi:10.1017/S0013091500018058.

N. S. Romanovskii, On Shmel’kin embeddings for abstract and profinite groups, Algebra i Logika 38 (1999), no. 5, 598–612 (Russian); English translation, Algebra and Logic 38 (1999), 326–334, https://www.mathnet.ru/eng/al2483.

E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved problems in group theory, 21st ed., September 2026 update, Problem 7.58, arXiv:1401.0300v46.