Canonical images and ambient conjugacy of split extensions in automorphism groups of Cartesian wreath products

Problem

Describe the images of Merzlyakov’s canonical embeddings of semidirect products, and determine when two such images are conjugate in the ambient group.

B\rtimes_\alpha A\hookrightarrow A\times\operatorname{Aut}(B^C\rtimes C),\qquad C\ne1.

This is the Cartesian regular wreath-product formulation, with the coordinate and action conventions specified below.

The canonical images

Let $A,B,C$ be arbitrary groups, with $C\ne1$ . We use left conjugation $\mathrm c_x(y)=xyx^{-1}$ and composition as multiplication of automorphisms. Put

\tag{1} F=B^C,\qquad W=F\rtimes C,\qquad (f,c)(h,k)=\bigl(f\,T_ch,ck\bigr),\qquad (T_ch)(x)=h(c^{-1}x).

Thus $F$ is the full function group, including when $C$ is infinite. Write $e_c=(1,c)$ and identify $F$ with $\{(f,1):f\in B^C\}$ . For $b\in B$ , let

\delta(b)(x)= \begin{cases}b,&x=1,\\1,&x\ne1,\end{cases} \qquad u_b=\mathrm c_{(\delta(b),1)}\in\operatorname{Aut}(W).

For $\beta\in\operatorname{Aut}(B)$ , define

\tag{3} d_\beta(f,c)=(\beta\circ f,c).

Put

\tag{4} T=\operatorname{Aut}(W),\qquad U=\{u_b:b\in B\},\qquad D=\{d_\beta:\beta\in\operatorname{Aut}(B)\}, \qquad L=UD.

Lemma 3 proves that $L$ is a subgroup and that its displayed factorization is unique.

For an action $\alpha:A\to\operatorname{Aut}(B)$ , use the convention

(b,a)(b',a')=(b\alpha_a(b'),aa') \quad\text{in }B\rtimes_\alpha A.

The canonical map and its image are

\tag{6} \Phi_\alpha(b,a)=(a,u_bd_{\alpha_a}),\qquad H_\alpha=\Phi_\alpha(B\rtimes_\alpha A)\le A\times T.

Merzlyakov’s unified embedding construction gives these maps; see (2, §3, Theorem 4). Problem 4.40 of the Kourovka Notebook asks how the resulting subgroups are situated in the ambient direct product (1). We consider its Cartesian wreath product version, with the coordinate and complement conventions specified in (1)–(6).

Theorem 1. A subgroup $H\le A\times T$ has the form $H=H_\alpha$ for a unique action $\alpha:A\to\operatorname{Aut}(B)$ if and only if

\tag{7} H\le A\times L,\qquad p_A(H)=A,\qquad H\cap(\{1\}\times T)=\{1\}\times U.

The subgroups $U,D,L$ depend only on $B$ and $C$ .

Theorem 2. For actions $\alpha,\gamma:A\to\operatorname{Aut}(B)$ ,

\tag{8} \begin{aligned} H_\alpha\text{ and }H_\gamma\text{ are conjugate in }A\times T \quad\Longleftrightarrow\quad &\exists\rho\in\operatorname{Aut}(B)\\[-2pt] &\forall a\in A,\quad\gamma_a=\rho\alpha_a\rho^{-1}. \end{aligned}

When these conditions hold, an element of $\{1\}\times D$ conjugates $H_\alpha$ to $H_\gamma$ .

The faithful wreath construction

Lemma 3. The maps $b\mapsto u_b$ and $\beta\mapsto d_\beta$ are injective homomorphisms, and

\tag{9} d_\beta u_b d_\beta^{-1}=u_{\beta(b)}.

The group $L=UD$ is isomorphic to $B\rtimes\operatorname{Aut}(B)$ via $(b,\beta)\mapsto u_bd_\beta$ .

Proof. The map $b\mapsto(\delta(b),1)$ is a homomorphism, so conjugation gives a homomorphism $b\mapsto u_b$ . The maps $d_\beta$ are automorphisms by (1), and $d_\beta d_\eta=d_{\beta\eta}$ . Moreover,

d_\beta(\delta(b),1)=(\delta(\beta(b)),1),

which gives (9).

Fix $c\ne1$ in $C$ . Direct multiplication yields

\tag{11} u_b(e_c)=\bigl(\delta(b)T_c\delta(b^{-1}),c\bigr), \qquad \bigl(u_b(e_c)\bigr)_{\mathrm{base}}(1)=b.

Since every $d_\beta$ fixes $e_c$ , an equality $u_bd_\beta=u_{b'}d_{\beta'}$ implies $b=b'$ by (11). Cancellation gives $d_\beta=d_{\beta'}$ ; applying these automorphisms to $(\delta(x),1)$ and evaluating at $1$ gives $\beta(x)=\beta'(x)$ for all $x\in B$ .

Thus the factors in $UD$ are unique. Equation (9) gives

(u_bd_\beta)(u_{b'}d_{\beta'}) =u_{b\beta(b')}d_{\beta\beta'},

so the map from the holomorph is a homomorphism. Its injectivity was just proved, and its image is $L$ . ◻

Proposition 4. The map $\Phi_\alpha$ in (6) is an injective homomorphism. For every $a\in A$ ,

\tag{13} H_\alpha\cap(\{a\}\times T)=\{a\}\times Ud_{\alpha_a}.

Proof. By (9),

\begin{aligned} \Phi_\alpha(b,a)\Phi_\alpha(b',a') &=\bigl(aa',u_b d_{\alpha_a}u_{b'}d_{\alpha_{a'}}\bigr)\\ &=\bigl(aa',u_{b\alpha_a(b')}d_{\alpha_{aa'}}\bigr) =\Phi_\alpha\bigl(b\alpha_a(b'),aa'\bigr). \end{aligned}

If two images are equal, their first coordinates are equal; uniqueness of $UD$ factors then gives equality of their $B$ coordinates. This proves injectivity. Formula (13) follows from (6). ◻

Recognition and intersections

Proof of Theorem 1. Formula (13) gives all three conditions in (7). Conversely, assume those conditions. Given $a\in A$ , choose $(a,t)\in H$ . Write $t=u_bd_\beta$ using $t\in L$ . The kernel condition gives $(1,u_b^{-1})\in H$ , and hence

(a,d_\beta)=(1,u_b^{-1})(a,u_bd_\beta)\in H.

There is only one such $\beta$ . If $(a,d_\beta)$ and $(a,d_\eta)$ both belong to $H$ , then

(1,d_{\beta^{-1}\eta})\in H, \qquad d_{\beta^{-1}\eta}\in U\cap D=1.

Thus $\beta=\eta$ .

Define $\alpha_a$ to be this unique diagonal representative. The identity of $H$ gives $\alpha_1=1$ , while

(a,d_{\alpha_a})(a',d_{\alpha_{a'}}) =(aa',d_{\alpha_a\alpha_{a'}})\in H

gives $\alpha_{aa'}=\alpha_a\alpha_{a'}$ . Hence $\alpha$ is an action. The prescribed kernel now shows that the entire fibre of $H$ over $a$ is $\{a\}\times Ud_{\alpha_a}$ . Therefore $H=H_\alpha$ . Uniqueness of the diagonal representative also proves uniqueness of $\alpha$ . ◻

Proposition 5. For actions $\alpha,\gamma:A\to\operatorname{Aut}(B)$ ,

\begin{align*} H_\alpha\cap(A\times\{1\}) &=(\ker\alpha)\times\{1\},\tag{18a}\\ H_\alpha\cap(\{1\}\times T) &=\{1\}\times U,\tag{18b}\\ H_\alpha\cap(A\times D) &=\{(a,d_{\alpha_a}):a\in A\},\tag{18c}\\ H_\alpha\cap H_\gamma &=\{(a,u_bd_{\alpha_a}):\alpha_a=\gamma_a,\ b\in B\}. \tag{18d} \end{align*}

Furthermore,

\tag{19} H_\alpha\le H_\gamma\quad\Longleftrightarrow\quad\alpha=\gamma.

Proof. In (13), equality $u_bd_{\alpha_a}=1$ forces $b=1$ and $\alpha_a=1$ , proving (18a). Setting $a=1$ gives (18b). If $u_bd_{\alpha_a}=d_\beta$ , uniqueness of the factors gives $b=1$ and $\beta=\alpha_a$ ; this proves (18c).

If $u_bd_{\alpha_a}=u_{b'}d_{\gamma_a}$ , the same uniqueness gives $b=b'$ and $\alpha_a=\gamma_a$ , proving (18d). If $H_\alpha\le H_\gamma$ , apply that formula to $(a,d_{\alpha_a})\in H_\alpha$ for each $a$ . Then $\alpha_a=\gamma_a$ for every $a$ , so $\alpha=\gamma$ . The converse is immediate. ◻

Preservation of the Cartesian base

Normalizers in this section are taken in $T=\operatorname{Aut}(W)$ .

Lemma 6. Suppose $B\ne1$ . If $\theta\in N_T(U)$ , then $\theta(F)=F$ .

Proof. For $w=(f,c)\in W$ ,

\tag{20} \bigl(\mathrm c_w(\delta(b),1)\bigr)_{\mathrm{base}}(x) =f(x)\delta(b)(c^{-1}x)f(x)^{-1}.

First, if $\mathrm c_w\in U$ , then $c=1$ . Indeed, suppose $c\ne1$ and choose $b\ne1$ in $B$ . Write $\mathrm c_w=u_d$ . Evaluating both sides on $(\delta(b),1)$ at coordinate $c$ , the left side is $f(c)bf(c)^{-1}$ by (20); the right side is $1$ . This contradicts $b\ne1$ . Naturality of inner conjugation,

\tag{21} \theta\,\mathrm c_w\,\theta^{-1}=\mathrm c_{\theta(w)},

therefore shows that

\tag{22} \theta(\delta(b),1)\in F\qquad(b\in B),

since $\theta u_b\theta^{-1}\in U$ .

Second, suppose $C$ has at least three elements. Then

\tag{23} \mathrm c_w\in N_T(U)\quad\Longleftrightarrow\quad w\in F.

For $w=(f,1)$ , formula (20) gives

\mathrm c_w(\delta(b),1) =(\delta(f(1)bf(1)^{-1}),1),

so $\mathrm c_w$ normalizes $U$ . Conversely, suppose $\mathrm c_w\in N_T(U)$ and $c\ne1$ . Choose $b\ne1$ and $k\in C\setminus\{1,c\}$ . There is $d\in B$ such that

\mathrm c_{\mathrm c_w(\delta(b),1)} =\mathrm c_wu_b\mathrm c_w^{-1}=u_d.

The element $\mathrm c_w(\delta(b),1)$ is supported at coordinate $c$ , where its value is $f(c)bf(c)^{-1}$ . Evaluate the last equality on $e_k$ at coordinate $c$ . On the left the value is $f(c)bf(c)^{-1}$ , since $k\ne1$ ; on the right it is $1$ , since $c\ne1$ and $c\ne k$ . Again $b=1$ , a contradiction. This proves (23).

If $C$ has at least three elements and $w\in F$ , then $\mathrm c_w\in N_T(U)$ . Since $\theta\in N_T(U)$ , (21) gives $\mathrm c_{\theta(w)}\in N_T(U)$ . Equation (23) yields $\theta(w)\in F$ .

It remains to treat $C=\{1,c\}$ . Every $(f,1)\in F$ has the decomposition

\tag{26} (f,1)=(\delta(f(1)),1)\, e_c(\delta(f(c)),1)e_c^{-1}.

By (22), both coordinate elements have images in $F$ . The group $F$ is normal in $W$ , so applying $\theta$ to (26) shows $\theta(f,1)\in F$ .

Thus $\theta(F)\le F$ in both cases. The same argument applies to $\theta^{-1}\in N_T(U)$ and gives equality. ◻

Corollary 7. Under the hypotheses of Lemma 6, if $c\ne1$ and $\theta(e_c)=(f,k)$ , then $k\ne1$ .

Proof. If $k=1$ , then $\theta(e_c)\in F$ . Applying $\theta^{-1}(F)=F$ gives $e_c\in F$ , contrary to $c\ne1$ . ◻

Full ambient conjugacy

Proof of Theorem 2. If $B=1$ , there is just one action and one image, so the result holds. Assume $B\ne1$ , and suppose

\tag{27} (g,\theta)H_\alpha(g,\theta)^{-1}=H_\gamma.

Intersecting with the second direct factor and using (18b) gives $\theta U\theta^{-1}=U$ . Since $b\mapsto u_b$ is an isomorphism $B\to U$ , conjugation by $\theta$ induces $\beta\in\operatorname{Aut}(B)$ with

\tag{28} \theta u_b\theta^{-1}=u_{\beta(b)}\qquad(b\in B).

For $a\in A$ , set $\varphi_a=\gamma_{gag^{-1}}$ . The image of $(a,d_{\alpha_a})$ in (27) lies over $gag^{-1}$ . By (13), there is a unique $t(a)\in B$ such that

\tag{29} \theta d_{\alpha_a}\theta^{-1}=u_{t(a)}d_{\varphi_a}.

Taking the induced conjugation actions on $U$ , equations (9) and (28) give

\tag{30} \beta\alpha_a\beta^{-1}=\mathrm c_{t(a)}\varphi_a \quad\text{in }\operatorname{Aut}(B).

Indeed, for $b\in B$ we have

\beta(\alpha_a(b)) =t(a)\varphi_a(\beta(b))t(a)^{-1},

obtained by conjugating $u_b$ with both sides of (29) and using injectivity of $b\mapsto u_b$ .

We claim that $t(a)=q\varphi_a(q)^{-1}$ for a fixed $q\in B$ . Fix $c\ne1$ and write

\theta(e_c)=(f,k),\qquad q=f(1).

Corollary 7 gives $k\ne1$ . Every $d_{\alpha_a}$ fixes $e_c$ , so (29) implies

u_{t(a)}d_{\varphi_a}(f,k)=(f,k).

The base component on the left is

\delta(t(a))\,(\varphi_a\circ f)\,T_k\delta(t(a)^{-1}).

At coordinate $1$ , the last factor is $1$ because $k\ne1$ . Therefore

\tag{35} q=t(a)\varphi_a(q),\qquad t(a)=q\varphi_a(q)^{-1}\quad(a\in A).

In particular, $q$ is independent of $a$ .

For any automorphism $\varphi$ of $B$ ,

\mathrm c_{q\varphi(q)^{-1}}\varphi =\mathrm c_q\varphi\mathrm c_q^{-1}.

Substituting (35) into (30) consequently gives

(\mathrm c_q^{-1}\beta)\alpha_a(\mathrm c_q^{-1}\beta)^{-1} =\varphi_a=\gamma_g\gamma_a\gamma_g^{-1}.

Thus

\tag{38} \rho=\gamma_g^{-1}\mathrm c_q^{-1}\beta \quad\text{satisfies}\quad \gamma_a=\rho\alpha_a\rho^{-1}\quad(a\in A).

Conversely, suppose the right side of (8) holds. Equation (9) gives

\begin{aligned} d_\rho(u_bd_{\alpha_a})d_\rho^{-1} &=u_{\rho(b)}d_{\rho\alpha_a\rho^{-1}}\\ &=u_{\rho(b)}d_{\gamma_a}. \end{aligned}

Since $\rho$ is bijective, this maps the entire fibre over $a$ onto the corresponding fibre of $H_\gamma$ . Hence $(1,d_\rho)H_\alpha(1,d_\rho)^{-1}=H_\gamma$ , proving both assertions. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., 2026, Problem 4.40. arXiv:1401.0300.

Yu. I. Merzlyakov, Integral representation of holomorphs of polycyclic groups, Algebra and Logic 9 (1970), 326–337. Russian original: Algebra i Logika 9, no. 5 (1970), 539–558. doi:10.1007/BF02321896.