Proof Consider the LHS spectral sequence

$$ \begin{align*}E_{i,j}^2 = H_i(C_0, H_j(A, \mathbb{F}_p)) \text{ that converges to }H_{i+ j} (B_0, \mathbb{F}_p).\end{align*} $$

Similarly there is an LHS spectral sequence

$$ \begin{align*}\widehat{E}_{i,j}^2 = H_i(C, H_j(A, \mathbb{F}_p)) \text{ that converges to }H_{i+ j} (B, \mathbb{F}_p).\end{align*} $$

Since A is of type $FP_n$ , we have that $H_j(A, \mathbb {F}_p)$ is finite for all $j \leq n$ . Then there is a pro-p subgroup $C_1$ of finite index in C such that $C_1$ acts trivially on $H_j(A, \mathbb {F}_p)$ for every $j \leq n$ . Since C is of type $FP_{n+1}$ , we have that $C_1$ is of type $FP_{n+1}$ . Then

$$ \begin{align*}H_i(C_1, H_j(A, \mathbb{F}_p)) \simeq \oplus H_i(C_1, \mathbb{F}_p)\text{ is finite for }j \leq n, i \leq n+1,\end{align*} $$

where we have $\dim _{\mathbb {F}_p} H_j(A, \mathbb {F}_p)$ direct summands in the right-hand side of the above isomorphism. Since $C_1$ has finite index in C, we deduce that

$$ \begin{align*}\widehat{E}_{i,j}^2 = H_i(C, H_j(A, \mathbb{F}_p))\text{ is finite for }j \leq n, i \leq n+1,\end{align*} $$

hence by the convergence of the second spectral sequence, we obtain that

$$ \begin{align*}H_k(B, \mathbb{F}_p) \text{ is finite for }k \leq n.\end{align*} $$

Note that we have shown that if $i + j = n+1 , i \not = 0$ , then $\widehat {E}_{i,j}^2$ is finite, and hence $\widehat {E}_{i,j}^{\infty }$ is finite. By the convergence of the spectral sequence, there is a filtration of $H_{n+1}(B, \mathbb {F}_p)$

$$ \begin{align*}0 = F_{-1}(H_{n+1}(B, \mathbb{F}_p)) \subseteq \cdots \subseteq F_{i}(H_{n+1}(B, \mathbb{F}_p)) \subseteq F_{i+1}(H_{n+1}(B, \mathbb{F}_p))\end{align*} $$

$$ \begin{align*}\subseteq \cdots \subseteq F_{n+1}(H_{n+1}(B, \mathbb{F}_p)) = H_{n+1}(B, \mathbb{F}_p), \end{align*} $$

where $ F_{i}(H_{n+1}(B, \mathbb {F}_p)) / F_{i-1}(H_{n+1}(B, \mathbb {F}_p)) \simeq \widehat {E}_{i, n+1 - i}^{ \infty }$ . Thus,

$$ \begin{align*}H_{n+1}(B, \mathbb{F}_p) \text{ is finite if and only if }\widehat{E}_{0,n+1}^{\infty} \text{ is finite.}\end{align*} $$

Note that since any differential that comes out from $\widehat {E}^r_{0, n+1}$ is zero, we have that $\widehat {E}_{0,n+1}^{\infty }$ is a quotient of $\widehat {E}^2_{0, n+1} = H_0(C, H_{n+1}(A, \mathbb {F}_p))$ , and thus there is a map

$$ \begin{align*}\mu : H_0(C, H_{n+1}(A, \mathbb{F}_p)) \to H_{n+1}(B, \mathbb{F}_p)\end{align*} $$

with image that equals $\widehat {E}^{ \infty }_{0, n+1}$ . Thus,

$$ \begin{align*}B \text{ is of type }FP_{n+1}\text{ if and only if }\mathrm{Im} (\mu)\text{ is finite.}\end{align*} $$

Similarly, there is a map

$$ \begin{align*}\mu_0 : H_0(C_0, H_{n+1}(A, \mathbb{F}_p)) \to H_{n+1}(B_0, \mathbb{F}_p)\end{align*} $$

with image that equals ${E}^{ \infty }_{0, n+1}$ and such that $B_0$ is of type $FP_{n+1}$ if and only if $\mathrm{Im} (\mu _0)$ is finite. Since $B_0$ is of type $FP_{n+1}$ , we conclude that $\mathrm{Im} (\mu _0)$ is finite.

The naturality of the LHS spectral sequence implies that we have the commutative diagram

where the maps $\rho $ and $\rho _0$ are induced by $\nu : C_0 \to C$ and by $\theta $ .

Recall that the action of $B_0$ on A via conjugation induces an action of $B_0$ on $H_{n+1}(A, \mathbb {F}_p)$ where A acts trivially and this induces the action of $C_0$ on $H_{n+1}(A, \mathbb {F}_p)$ that is used to define $H_0(C_0, H_{n+1}(A, \mathbb {F}_p))$ . Similarly, the action of B on A via conjugation induces an action of B on $H_{n+1}(A, \mathbb {F}_p)$ where A acts trivially and this induces the action of C on $H_{n+1}(A, \mathbb {F}_p)$ that is used to define $H_0(C, H_{n+1}(A, \mathbb {F}_p))$ . If $\nu $ is surjective, then $\rho $ is an isomorphism; if $\nu $ is injective, then $\rho $ is surjective. Since every homomorphism $\nu $ is a composition of one epimorphism followed by one monomorphism, we conclude that $\rho $ is always surjective. Then

$$ \begin{align*}\mathrm{Im} (\mu) = \mathrm{Im} (\mu \circ \rho) = \mathrm{Im} (\rho_0 \circ \mu_0) \text{ is a quotient of } \mathrm{Im} (\mu_0).\end{align*} $$

Since $\mathrm{Im} (\mu _0)$ is finite, we conclude that $\mathrm{Im} (\mu )$ is finite. Hence, B is of type $FP_{n+1}$ as required.

Proof We induct on $m \geq 1$ . The case $m = 1$ is obvious since a pro-p group is of type $FP_1$ if and only if the group is finitely generated (as a pro-p group).

Assume that $m> 1$ and that the result holds for $m-1$ ; hence, C is of type $FP_{m-1}$ . Since A is $FP_{m-1}$ , we have that $H_j(A, \mathbb {F}_p)$ is finite for $0 \leq j \leq m-1$ . Consider the LHS spectral sequence

$$ \begin{align*}E_{i,j}^2 = H_i(C, H_j(A, \mathbb{F}_p)) \text{ that converges to }H_{i+j}(B, \mathbb{F}_p).\end{align*} $$

By substituting if necessary C with a subgroup of finite index, we can assume that C acts trivially on $ H_j(A, \mathbb {F}_p)$ for $ 0 \leq j \leq m-1$ . Then

$$ \begin{align*}E_{i,j}^2 = H_i(C, H_j(A, \mathbb{F}_p)) \simeq H_i (C, \mathbb{F}_p) \widehat{\otimes} H_j(A, \mathbb{F}_p)\end{align*} $$

is finite for $0 \leq i,j \leq m-1$ . Then, for $r \geq 2$ , consider the differential

$$ \begin{align*}d_{m,0}^r : E_{m,0}^r \to E_{m-r, r- 1}^r\end{align*} $$

and note that either $m - r < 0$ ; hence, $ E_{m-r, r- 1}^2 = 0$ or $m-r\leq m-1, r- 1 \leq m-1$ . In all cases, $E_{m-r, r-1}^2 $ is finite, and hence $E_{m-r, r- 1}^r$ is finite and so $E_{m,0}^{r+1}= \mathrm{Ker}(d_{m,0}^r)$ is finite if and only if $E_{m,0}^r$ is finite. Thus,

$$ \begin{align*}E_{m,0}^{\infty} \text{ is finite if and only if }E_{m,0}^2 = H_m(C, \mathbb{F}_p) \text{ is finite.}\end{align*} $$

Finally, since B is $FP_m$ , we have that $H_m(B, \mathbb {F}_p)$ is finite and by the convergence of the spectral sequence $E_{m,0}^{\infty }$ is finite. Thus, we conclude that $ H_m(C, \mathbb {F}_p)$ is finite. And, this together with C is of type $FP_{m-1}$ implies that C is $FP_m$ .

Proof By [Reference Ribes and Zalesskii19, Theorem 4.1], the proper pro-p HNN extension gives rise to the exact sequence of $\mathbb {F}_p[[G]]$ -modules

(3.1) $$ \begin{align} 0 \to \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p \to \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p \to \mathbb{F}_p \to 0. \end{align} $$

Note that since $K \not \subseteq M$ , we have that $M \backslash G / K = G/ MK$ is a proper pro-p quotient of $G/ M \simeq \mathbb {Z}_p$ , hence is finite. Similarly, $M \backslash G / A = G/ MA$ is finite.

Note that there is an isomorphism of (left) $\mathbb {F}_p[[M]]$ -modules

$$ \begin{align*}\mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p \simeq (\oplus_{t \in M \backslash G / K} \mathbb{F}_p[[M]] t \mathbb{F}_p[[K]]) \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p \simeq\end{align*} $$

$$ \begin{align*}\oplus_{t \in M \backslash G / K} \mathbb{F}_p[[M]] \otimes_{\mathbb{F}_p[[ M \cap t K t^{ -1}]]} \mathbb{F}_p. \end{align*} $$

Similarly, there is an isomorphism of (left) $\mathbb {F}_p[[M]]$ -modules

$$ \begin{align*}\mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p \simeq (\oplus_{t \in M \backslash G / A} \mathbb{F}_p[[M]] t \mathbb{F}_p[[A]]) \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p \simeq\end{align*} $$

$$ \begin{align*}\oplus_{t \in M \backslash G / A} \mathbb{F}_p[[M]] \otimes_{\mathbb{F}_p[[M \cap t A t^{ -1}]]} \mathbb{F}_p. \end{align*} $$

The short exact sequence (3.1) gives rise to a long exact sequence in pro-p homology

$$ \begin{align*}\cdots \to H_{m+1}(M, \mathbb{F}_p) \to H_m(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p ) \to\end{align*} $$

$$ \begin{align*}H_m(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p) \to H_m (M, \mathbb{F}_p) \to H_{m-1}(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p )\end{align*} $$

$$ \begin{align*}\to \cdots \to H_1(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p) \to H_1(M, \mathbb{F}_p) \to\end{align*} $$

$$ \begin{align*}H_0(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p ) \to H_0(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p ) \to H_0(M, \mathbb{F}_p) \to 0. \end{align*} $$

Note that

$$ \begin{align*}H_i(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[K]]} \mathbb{F}_p ) \simeq H_i (M, \oplus_{t \in M \backslash G / K} \mathbb{F}_p[[M]] \otimes_{\mathbb{F}_p[[M \cap t K t^{ -1}]]} \mathbb{F}_p) \simeq\end{align*} $$

$$ \begin{align*}\oplus_{t \in M \backslash G / K} H_i (M, \mathbb{F}_p[[M]] \otimes_{\mathbb{F}_p[[M \cap t K t^{ -1}]]} \mathbb{F}_p) \simeq \oplus_{t \in M \backslash G / K} H_i (M \cap t K t^{ -1}, \mathbb{F}_p) =\end{align*} $$

$$ \begin{align*}\oplus_{t \in M \backslash G / K} H_i (t( M \cap K) t^{ -1}, \mathbb{F}_p) \simeq \oplus_{t \in M \backslash G / K} H_i (M \cap K, \mathbb{F}_p). \end{align*} $$

Similarly,

$$ \begin{align*}H_i(M, \mathbb{F}_p[[G]] \otimes_{\mathbb{F}_p[[A]]} \mathbb{F}_p ) \simeq \oplus_{t \in M \backslash G / A} H_i (M \cap A, \mathbb{F}_p). \end{align*} $$

Then the long exact sequence could be rewritten as

$$ \begin{align*}\cdots \to H_{m+1}(M, \mathbb{F}_p) \to \oplus_{t \in M \backslash G / K} H_m(M \cap K, \mathbb{F}_p)\to \oplus_{t \in M \backslash G / A}H_m(M \cap A, \mathbb{F}_p) \to\end{align*} $$

$$ \begin{align*}H_m (M, \mathbb{F}_p) \to \oplus_{t \in M \backslash G / K} H_{m-1}(M \cap K, \mathbb{F}_p) \to \cdots \to \oplus_{t \in M \backslash G / A} H_1(M \cap A, \mathbb{F}_p) \to\end{align*} $$

$$ \begin{align*}H_1(M, \mathbb{F}_p) \to \oplus_{t \in M \backslash G / K} H_0(M \cap K, \mathbb{F}_p) \to\end{align*} $$

$$ \begin{align*}\oplus_{t \in M \backslash G / A} H_0(M \cap A, \mathbb{F}_p) \to H_0(M, \mathbb{F}_p) \to 0. \end{align*} $$

Since $M \cap A$ is of type $FP_m$ , we have that $H_i(M \cap A, \mathbb {F}_p)$ is finite for $i \leq m$ . Combining with $M \backslash G / A$ is finite, we conclude that

$$ \begin{align*}\oplus_{t \in M \backslash G / A} H_i(M \cap A, \mathbb{F}_p) \text{ is finite for }i \leq m.\end{align*} $$

a) Note that M is of type $FP_m$ if and only if $H_i(M, \mathbb {F}_p)$ is finite for $i \leq m$ . By the above long exact sequence together with the fact that $M \backslash G / K$ is finite, $H_i(M, \mathbb {F}_p)$ is finite for $i \leq m$ if and only if $ \oplus _{t \in M \backslash G / K} H_i(M \cap K, \mathbb {F}_p)$ is finite for $i \leq m-1$ , i.e., $M \cap K$ is of type $FP_{m-1}$ .

b) If M is of type $FP_{m+1}$ , then $H_{m+1}(M, \mathbb {F}_p)$ is finite and since $H_m(M \cap A, \mathbb {Z}_p)$ is finite by the long exact sequence $H_m(M \cap K, \mathbb {F}_p)$ is finite. We already know by a) that $M \cap K$ is of type $FP_{m-1}$ , and hence $M \cap K$ is of type $FP_m$ .

Proof By Theorem 2.1, if P is a pro-p subgroup of Q, then A is finitely generated as pro- $p\ \mathbb {Z}_p[[P]]$ -module if and only if $T(Q, P) \cap \Delta (A) = \{1 \}.$

Let

$$ \begin{align*}J = \mathrm{ann}_{\mathbb{Z}_p[[Q]]}(A).\end{align*} $$

Since A is finitely generated as a pro- $p\ \mathbb {Z}_p[[H]]$ -module for every $\chi \in T(Q, H) \setminus \{ 1 \}$ , we have that $J \not \subseteq \mathrm{Ker} (\overline {\chi })$ . Since $\mathbb {Z}_p[[Q]]$ is a Noetherian ring, there is a finite subset $\Lambda $ of J that generates J as an ideal (abstractly or topologically is the same).

We aim to show that for sufficiently big j, we have $T(Q, H_j) \cap \Delta (A) = \{1 \}.$ Let $\mu _j \in T(Q, H_j) \setminus \{ 1 \}$ ; thus, we aim to show that $\mu _j \not \in \Delta (A)$ . Then, by Theorem 2.1, A is finitely generated as a pro- $p\ \mathbb {Z}_p[[H_j]]$ -module.

Let

$$ \begin{align*}\overline{\mu}_j : \mathbb{Z}_p[[Q]] \to \mathbb{F}[[t]]\end{align*} $$

be the continuous ring homomorphism induced by $\mu _j$ . Since $\overline {\mu }_j(H_j) = 1$ , we have $\mu _j(x) = \mu _j(y^{p^j}) \not = 1.$ Let

$$ \begin{align*}\chi \in T(Q, H) \setminus \{ 1 \}\text{ be such that }\chi(y) = \mu_j(y), \chi(x) = 1\end{align*} $$

and

$$ \begin{align*}\overline{\chi} : \mathbb{Z}_p[[Q]] \to \mathbb{F}[[t]]\end{align*} $$

be the continuous ring homomorphism induced by $\chi $ . Let

$$ \begin{align*}\lambda \in \mathbb{Z}_p[[Q]] = \mathbb{Z}_p[[ t_1, t_2]], \text{ where }x = 1 + t_1, y = 1 + t_2.\end{align*} $$

Then, since $\chi (y) = \mu _j(y)$ , $\chi (x) = 1$ , we have $\chi (t_2) = 0$ , and hence

$$ \begin{align*}\overline{\chi} (\lambda) = \overline{\chi} (\lambda |_{t_1 = 0}) = \overline{\mu}_j(\lambda |_{t_1 =0}),\end{align*} $$

where $\lambda = \sum _{i,k \geq 0} z_{i,k} t_1^i t_2^k$ , $z_{i,k} \in \mathbb {Z}_p$ and $\lambda |_{t_1 = 0} = \sum _{k \geq 0} z_{0,k} t_2^k$ . Note that

$$ \begin{align*}\overline{\mu}_j(t_2) = \overline{\mu}_j(1 + t_2) - \overline{\mu}_j(1) = \overline{\mu}_j(y) - 1 \in 1 + t \mathbb{F}[[t]] - 1 = t \mathbb{F}[[t]].\end{align*} $$

Note that since $\mathbb {F}$ has characteristic $p> 0$ , we have

$$ \begin{align*}1 + \overline{\mu}_j( t_1) = \overline{\mu}_j( 1 + t_1) = \overline{\mu}_j(x) = \mu_j(x) =\end{align*} $$

$$ \begin{align*}\mu_j(y^{p^j}) = \overline{\mu}_j( (1 + t_2)^{p^j}) = \overline{\mu}_j( 1 + t_2^{p^j}) = 1 + \overline{\mu}_j( t_2^{p^j}).\end{align*} $$

Consider

$$ \begin{align*}\lambda |_{t_1 = t_2^{p^j}}: = \sum_{i,k \geq 0} z_{i,k} t_2^{i p^j + k} = \sum_{s \geq 0} \left(\sum_{ i p^j + k = s} z_{i,k}\right) t_2^{s}.\end{align*} $$

Then, using that $\overline {\mu }_j( t_1) = \overline {\mu }_j( t_2^{p^j})$ , we conclude that

(3.2) $$ \begin{align} \overline{\mu}_j(\lambda) &= \overline{\mu}_j(\lambda |_{t_1 = t_2^{p^j}}) = \overline{\mu}_j\left( \sum_{i,k \geq 0} z_{i,k} t_2^{i p^j + k}\right) = \overline{\mu}_j( \sum_{k \geq 0} z_{0,k} t_2^{k}) + \overline{\mu}_j\left( \sum_{i \geq 1,k \geq 0} z_{i,k} t_2^{i p^j + k}\right)\nonumber \\& = \overline{\mu}_j(\lambda |_{t_1 =0})+ \overline{\mu}_j\left( \sum_{i \geq 1,k \geq 0} z_{i,k} t_2^{i p^j + k}\right) = \overline{\chi} (\lambda) + \overline{\mu}_j\left( \sum_{i \geq 1,k \geq 0} z_{i,k} t_2^{i p^j + k}\right). \end{align} $$

Suppose now that $\mu _j \in \Delta (A)$ . Then $\overline {\mu }_j(J) = 0$ , in particular $\overline {\mu }_j(\Lambda ) = 0.$ On the other hand, $\chi \not \in \Delta (A)$ ; hence, $\overline {\chi }(J) \not = 0$ . This is equivalent with $\overline {\chi }(\Lambda ) \not = 0$ . So there is $\lambda _0 \in \Lambda $ such that

$$ \begin{align*}\overline{\chi}(\lambda_0) \not= 0 = \overline{\mu}_j(\lambda_0) .\end{align*} $$

Write as before $\lambda _0 = \sum _{i,k \geq 0} z_{i,k} t_1^i t_2^k$ where $z_{i,k} \in \mathbb {Z}_p$ . Then, by (3.2),

$$ \begin{align*}0 = \overline{\mu}_j(\lambda_0) = \overline{\chi} (\lambda_0) + \overline{\mu}_j\left( \sum_{i \geq 1,k \geq 0} z_{i,k} t_2^{i p^j + k}\right). \end{align*} $$

So, for

$$ \begin{align*}\lambda_0 |_{t_1 = 0} = \sum_{i \geq 0} z_i t_2^i,\end{align*} $$

where $z_i \in \mathbb {Z}_p$ and using that $ \overline {\chi } (\lambda _0) = \overline {\chi } (\lambda _0 |_{t_1 = 0})$ , we have

(3.3) $$ \begin{align} - \overline{\chi} ( \sum_{i \geq 0} z_i t_2^i) = \overline{\mu}_j\left( \sum_{i \geq 1,k \geq 0} z_{i,k} t_2^{i p^j + k}\right).\end{align} $$

Let

$$ \begin{align*}f_0 : = \overline{\mu}_j(y) -1 = \overline{\mu}_j(t_2 + 1) - 1 = \overline{\mu}_j(t_2) = \mu_j(t_2) \in t \mathbb{F} [[t]] \setminus \{ 0 \};\end{align*} $$

hence,

$$ \begin{align*}\overline{\chi} (t_2) = \overline{\chi} (t_2 + 1) - 1 = \overline{\chi} (y) -1 = \chi(y) -1 = \mu_j(y) -1 = f_0.\end{align*} $$

Then, by (3.3), we conclude that

(3.4) $$ \begin{align} 0 \not= - \sum_{i \geq 0} \overline{z}_i f_0^i = \sum_{i \geq 1,k \geq 0} \overline{z}_{i,k} f_0^{i p^j + k}, \end{align} $$

where for $z \in \mathbb {Z}_p$ we denote by $\overline {z}$ the image of z in $\mathbb {F}_p = \mathbb {Z}_p/ p \mathbb {Z}_p$ .

Consider the map

$$ \begin{align*}o : \mathbb{F}[[t]] \setminus \{ 0 \} \to \mathbb{Z}_{\geq 0}\end{align*} $$

that sends $\sum _{i \geq 0} a_i t^i$ to the smallest $i_0$ such that $a_{i_0} \not = 0$ , where each $a_i \in \mathbb {F}$ . Define

$$ \begin{align*}d = o(f_0) \geq 1 \text{ and } k_0 = o(\sum_{i} \overline{z}_i t^i) \geq 0.\end{align*} $$

Then

(3.5) $$ \begin{align} o\left(- \sum_{i \geq 0} \overline{z}_i f_0^i\right) = d k_0. \end{align} $$

By (3.4) and (3.5), there is $\overline {z}_{i,k} \not = 0$ for some $i \geq 1, k \geq 0$ such that

$$ \begin{align*}d p^j \leq d ( i p^j + k) = o ( f_0^{i p^j + k}) \leq dk_0, \text{ hence } p^j \leq k_0.\end{align*} $$

Note that $k_0$ depends only on $\lambda _0 \in \Lambda $ , where $\Lambda $ is a finite set, hence it does not depend on j. From the very beginning, we can choose $j_0 \in \mathbb {Z}_{>0}$ such that

$$ \begin{align*}p^{j_0}> max \{ \widetilde{k} = o( \sum_i \overline{m}_i t^i) \ | \ \widetilde{\lambda} |_{t_1 = 0} = \sum_i m_i t_2^i \not= 0, \widetilde{\lambda} \in \Lambda \} \geq k_0.\end{align*} $$

Then, for $j \geq j_0$ , we get a contradiction, so $\mu _j \not \in \Delta (A)$ as required.

Proof Note that since S is a pro-p group of type $FP_m$ and $G/ S\simeq \mathbb {Z}_p$ is a pro-p group of type $FP_{\infty }$ , hence of type $FP_m$ , we can conclude that G is a pro-p group of type $FP_m$ . Set

$$ \begin{align*}Q = G/ G_0 = \langle x, y \rangle \text{ and }H = S/ G_0 = \langle x \rangle.\end{align*} $$

Since $Q = G/ G_0$ is a finitely generated abelian pro-p group, hence $\mathbb {Z}_p[[Q]]$ is left and right Noetherian and G is of type $FP_m$ , we conclude that

$$ \begin{align*}A_i = H_i(G_0, \mathbb{Z}_p) \text{ is finitely generated as a pro-}p \ \mathbb{Z}_p[[Q]]\text{-module for }i \leq m.\end{align*} $$

Since S is a pro-p group of type $FP_m$ , we conclude that

$$ \begin{align*}A_i\text{ is finitely generated as a pro-}p \ \mathbb{Z}_p[[H]]\text{-module for } i \leq m.\end{align*} $$

Then, by Proposition 3.4, for sufficiently big j, we have that $A_i$ is finitely generated as a pro- $p\ \mathbb {Z}_p[[H_j]]$ -module, where $H_j = \langle x y^{ - p^j} \rangle \leq Q$ , for every $ i \leq m$ . Then we define $S_0$ as the preimage in G of one such $H_j$ .

Proof of Theorem 1.1

Consider a commutative diagram

where the lines are short exact sequences of pro-p groups, $F_n$ is the free pro-p group with a free basis $s_1, \ldots , s_n$ , and the vertical maps are surjective homomorphisms of pro-p groups with the most left map being the identity map.

Define

$$ \begin{align*}\Pi = \Pi_1 \coprod_K \Pi_2 \coprod_K \ldots \coprod_K \Pi_n,\end{align*} $$

where $\coprod _K$ is the amalgamated free product with amalgam K in the category of pro-p groups, and each

$$ \begin{align*}\Pi_i = K \rtimes \langle s_i \rangle, \ \langle s_i \rangle \simeq \mathbb{Z}_p.\end{align*} $$

Note that since K is normal in $\Pi $ and $\Pi / K \simeq \Pi _1/ K \coprod \Pi _2/ K \coprod \cdots \coprod \Pi _n/ K$ is a free pro-p product, we conclude that $\Pi _1 \coprod _K \Pi _2 \coprod _K \cdots \coprod _K \Pi _i$ embeds in $\Pi $ for every $1 \leq i \leq n$ .

Recall that $\Gamma ^{ab}$ is infinite; hence, the image in $\Gamma ^{ab}$ of at least one $\pi (s_i)$ has infinite order. Without loss of generality, we can assume that the image of $\pi (s_1)$ in $\Gamma ^{ab}$ has infinite order. In particular, the restriction map

$$ \begin{align*}\pi |_{\Pi_1} : \Pi_1 \to \pi ({\Pi_1})\end{align*} $$

is an isomorphism.

Note that $[K, s_1] \subseteq G' \cap K \subseteq N$ , hence $\Pi _1'\subseteq N$ . We have $N \subseteq K \subseteq \Pi _1$ where $K/ N \simeq \mathbb {Z}_p, \Pi _1 / K \simeq \mathbb {Z}_p$ , this together with the inclusion $\Pi _1'\subseteq N$ implies that $\Pi _1 / N \simeq \mathbb {Z}_p^2$ .

By assumption, K is of type $FP_{n_0}$ . By Proposition 3.5, there is $S_0$ a normal pro-p subgroup of $\Pi _1$ such that

$$ \begin{align*}N \subseteq S_0, S_0\text{ is of type }FP_{n_0}, S_0 \not= K \text{ and }\Pi_1/ S_0 \simeq \mathbb{Z}_p.\end{align*} $$

Let

$$ \begin{align*}\mu : G \to \mathbb{Z}_p\end{align*} $$

be a homomorphism of pro-p groups such that

$$ \begin{align*}\mathrm{Ker} (\mu \circ \pi) \cap \Pi_1 = S_0, \text{ i.e., }\mathrm{Ker}(\mu) \cap \pi( \Pi_1) = \pi(S_0).\end{align*} $$

This is possible since $\Pi _1 / N \simeq \mathbb {Z}_p^2$ is abelian and $G' \cap K \subseteq N \subseteq S_0$ . Note that $K \not \subseteq S_0$ , hence $\mu (K) \not = 0$ .

Consider the epimorphism of pro-p groups

$$ \begin{align*}\chi = \mu \circ \pi : \Pi \to \mathbb{Z}_p.\end{align*} $$

Note that

$$ \begin{align*}\chi(K) \not= 0, \mathrm{Ker}(\chi) \cap \Pi_1 = S_0 \text{ is of type }FP_{n_0}\end{align*} $$

and

$$ \begin{align*}\mathrm{Ker}(\chi) \cap K = S_0 \cap K = N \text{ is of type }FP_{n_0 - 1}.\end{align*} $$

Then we view $\Pi _1 \coprod _K \Pi _2$ as a proper HNN extension

$$ \begin{align*}\langle \Pi_1, s_2 \ | \ K^{s_2} = K \rangle\end{align*} $$

with a pro-p base group $\Pi _1$ , associated pro-p subgroup K and stable letter $s_2$ . Then, by Lemma 3.3(a),

$$ \begin{align*}\mathrm{Ker}(\chi) \cap ( \Pi_1 \coprod_K \Pi_2)\text{ is of type }FP_{n_0}.\end{align*} $$

We view $\Pi _1 \coprod _K \Pi _2 \coprod _K \Pi _3$ as a proper HNN extension with a base pro-p group $\Pi _1 \coprod _K \Pi _2$ , associated pro-p subgroup K and stable letter $s_3$ then by Lemma 3.3(a)

$$ \begin{align*}\mathrm{Ker}(\chi) \cap \left( \Pi_1 \coprod_K \Pi_2 \coprod_K \Pi_3\right)\text{ is of type }FP_{n_0}.\end{align*} $$

Then, repeating this argument several times, we deduce that $\mathrm{Ker}(\chi )$ is of type $FP_{n_0}$ .

By construction, $\mathrm{Ker}(\mu )$ is a quotient of $\mathrm{Ker}(\chi )$ . If $n_0 = 1$ , then $\mathrm{Ker}(\chi )$ is finitely generated (as a pro-p group), then any pro-p quotient of $\mathrm{Ker}(\chi )$ is finitely generated (as a pro-p group). In particular, $\mathrm{Ker}(\mu )$ is finitely generated (as a pro-p group).

Now, for the general case, i.e., $n_0 \geq 2$ , we will apply Lemma 3.1. Write $\widetilde {\mathrm{Ker} (\chi )}$ for the image of $\mathrm{Ker} (\chi )$ in $F_n$ and $\widetilde {\mathrm{Ker}(\mu )}$ for the image of $\mathrm{Ker}(\mu )$ in $\Gamma $ . By construction,

$$ \begin{align*}\mathrm{Ker}(\chi) \cap K = N= \mathrm{Ker}(\mu) \cap K.\end{align*} $$

By assumption, N is of type $FP_{n_0-1}$ and we have already shown that $\mathrm{Ker}(\chi )$ is of type $FP_{n_0}$ . By construction, $\mu (K) \not = 0$ , hence $K. \mathrm{Ker} (\mu ) \not = \mathrm{Ker}(\mu )$ and since $G/ \mathrm{Ker}(\mu ) \simeq \mathbb {Z}_p$ , we deduce that $K. \mathrm{Ker}(\mu )$ has finite index in G and so $\widetilde {\mathrm{Ker}(\mu )}$ has finite index in $\Gamma $ .

By Lemma 3.2, since in the short exact sequence of pro-p groups

$$ \begin{align*}1 \to K \to G \to \Gamma \to 1\end{align*} $$

the pro-p groups G and K are of type $FP_{n_0}$ (it suffices that K is of type $FP_{n_0-1}$ ), we deduce that $\Gamma $ is of type $FP_{n_0}$ . Then $\widetilde {\mathrm{Ker}(\mu )}$ is a pro-p group of type $FP_{n_0}$ . Then we can apply Lemma 3.1 for the commutative diagram

to deduce that $\mathrm{Ker}(\mu )$ is a pro-p group of type $FP_{n_0}$ . Finally, we set $M = \mathrm{Ker}(\mu )$ .

Proof of Corollary 1.2

We define M as in the proof of Theorem 1.1 for $\Gamma = F_n$ and $\pi $ the identity map, $\mu = \chi $ . Thus, $M = \mathrm{Ker}(\chi ) = \mathrm{Ker}(\mu )$ is a normal pro-p subgroup of G, $G/ M \simeq \mathbb {Z}_p$ and M is of type $FP_{n_0}$ . We view

$$ \begin{align*}G = \Pi = \Pi_1 \coprod_K \Pi_2 \coprod_K \ldots \coprod_K \Pi_n\end{align*} $$

as a proper HNN extension with a base pro-p subgroup

$$ \begin{align*}A =\Pi_1 \coprod_K \Pi_2 \coprod_K \ldots \coprod_K \Pi_{n-1},\end{align*} $$

associated pro-p subgroup K and stable letter $s_n$ . By the proof of Theorem 1.1,

$$ \begin{align*}A \cap M = A \cap \mathrm{Ker} (\chi) \text{ is of type }FP_{n_0}.\end{align*} $$

Suppose that M is of type $FP_{n_0+1}$ . By Lemma 3.3(b), $N = M \cap K$ is of type $FP_{n_0}$ , a contradiction. Hence, M is not of type $FP_{n_0 + 1}$ . This completes the proof of the corollary.

Proof We claim that there is a finitely generated non-procyclic pro-p subgroup $\Gamma _0$ of $\Gamma $ such that $\Gamma _0$ acts trivially on the abilianization $K^{ab} = K/ K'$ via conjugation. Indeed, let $T = \mathrm{tor}(K/ K')$ be the torsion pro-p subgroup of $K^{ab}$ . Then $V = K^{ab}/ T \simeq \mathbb {Z}_p^d$ , where $d \geq 1$ . Note that the conjugation action of $\Gamma $ on $V \simeq \mathbb {Z}_p^d$ induces a homomorphism

$$ \begin{align*}\rho : \Gamma \to GL_d(\mathbb{Z}_p).\end{align*} $$

Note that $\mathrm{Im} (\rho )$ is a pro-p subgroup of $GL_d(\mathbb {Z}_p)$ , hence is p-adic analytic and there is an upper bound on the number of generators of any finitely generated pro-p subgroup of $\mathrm{Im} (\rho )$ [Reference Dixon, Du Sautoy, Mann and Segal7]. Hence, $\rho $ is not injective. Alternatively, we can use the main result of [Reference Barnea and Larsen1] to deduce that $\rho $ is not injective. Thus, $\mathrm{Ker}(\rho )$ is a nontrivial normal pro-p subgroup of $\Gamma $ and we can choose $\Gamma _0$ any non-procyclic finitely generated pro-p subgroup of $\mathrm{Ker}(\rho )$ .

Set $G_0 = K \rtimes \Gamma _0$ . Then, by Corollary 1.2, there is a normal pro-p subgroup M of $G_0$ such that $G_0/ M \simeq \mathbb {Z}_p$ , $M \cap K = N$ , and M is $FP_1$ but not $FP_2$ , i.e., it is finitely generated as a pro-p group, but is not finitely presented as a pro-p group. Thus, $G_0$ and hence G are incoherent (in the category of pro-p groups). This completes the proof.

Proof of Corollary 1.3

Let $\Gamma _0$ be a finitely generated, free non-abelian pro-p subgroup of $\Gamma $ . Consider the preimage $G_0$ of $\Gamma _0$ in G, i.e., there is a short exact sequence

$$ \begin{align*}1 \to K \to G_0 \to \Gamma_0 \to 1.\end{align*} $$

Note that $G_0 = K \rtimes \Gamma _0$ , then by Theorem 3.6, $G_0$ is not coherent.

Proof of Corollary 1.4

By Proposition 4.1, $K^{ab}$ is infinite. Let N be a normal pro-p subgroup of K such that $K/ N \simeq \mathbb {Z}_p$ . By part (4) from the main theorem of [Reference Kochloukova and Zalesskii14], we have that N is not finitely generated as a pro-p group. Then we can apply Corollary 1.3. This completes the proof.

Proof The classification of Demushkin groups has several cases described in Theorems 2.3–2.5. In the case of 2-generated Demushkin group, we have a 1-relation presentation with a relation of the type $[x_1, x_2]$ or $x_1^q[ x_1, x_2]$ , where q is a power of p or of the type $2^f + 2$ , $p = 2$ . In all these cases, the group is soluble, since it is $\langle x_1 \rangle \rtimes \langle x_2 \rangle $ and has zero Euler characteristic.

Proof of Corollary 1.5

We claim that $\Gamma $ has a free non-abelian pro-p subgroup F. Suppose first that $\Gamma $ is a pro-p RAAG. Let $v_1,v_2$ be vertices of the graph that defines the pro-p RAAG $\Gamma $ that are not adjacent. Then the pro-p subgroup F of $\Gamma $ generated by $v_1$ and $v_2$ , is a retract of $\Gamma $ , hence it is non-abelian free pro-p.

If $\Gamma $ is a non-soluble Demushkin group, then every pro-p subgroup of infinite index in $\Gamma $ has cohomological dimension 1, so is free pro-p [Reference Serre24, Ex. 5b), p. 44]. Furthermore, the abelianization of $\Gamma $ is infinite, so we can set $F $ to be a normal pro-p subgroup of $\Gamma $ such that $\Gamma / F \simeq \mathbb {Z}_p$ .

We claim that there is a normal pro-p subgroup N of K such that N is not finitely generated (as a pro-p group) and $K/ N \simeq \mathbb {Z}_p$ . Suppose first that K is a non-abelian pro-p RAAG. Let $w_1$ and $w_2$ be vertices of the graph that defines the pro-p RAAG K that are not adjacent. Then the pro-p subgroup $F_0$ of K generated by $w_1$ and $w_2$ is non-abelian free pro-p and it is a retract of K. Note that any normal pro-p subgroup S of $F_0$ such that $F_0/ S \simeq \mathbb {Z}_p$ is not finitely generated (as a pro-p group), hence any preimage N of S in K is not finitely generated (as a pro-p group).

Suppose that K is a non-soluble Demushkin group. Note that $K/ K'$ is infinite and let N be a normal pro-p subgroup of K such that $K/ N \simeq \mathbb {Z}_p$ . If N is finitely generated (as a pro-p subgroup), together with the fact that N is free pro-p, we conclude that N has finite Euler characteristic $\chi (N)$ , $\chi (K/N) = \chi (\mathbb {Z}_p) = 0$ and

$$ \begin{align*}0= \chi(N) \chi(K/ N) = \chi(K) = 1 - d(K) + 1 = 2 - d(K),\end{align*} $$

so $d(K) = 2$ . But by Lemma 4.2 2-generated Demushkin groups are soluble, a contradiction.

Finally, by Corollary 1.3, G is not coherent.

Proof Let H be a finitely generated, pro-p subgroup of G. We aim to prove that H is $FP_{\infty }$ , in particular is finitely presented.

Consider the short exact sequence of pro-p groups

$$ \begin{align*}1 \to K_0 \to H \to Q \to 1,\end{align*} $$

where Q is a pro-p subgroup of $\mathbb {Z}_p^m$ , so is finitely generated, abelian, and $K_0 = K \cap H$ . Recall that an infinite index subgroup in a Demushkin group is free pro-p, in particular $K_0$ is either free pro-p or Demushkin. In the latter, $K_0$ is $FP_{\infty }$ , hence H is $FP_{\infty }$ . Thus, we can assume from now on that $K_0$ is free pro-p.

Consider the LHS spectral sequence

$$ \begin{align*}E_{i,j}^2 = H_i(Q, H_j(K_0, \mathbb{F}_p)) \text{ that converges to } H_{i+j}(H, \mathbb{F}_p).\end{align*} $$

Note that $E_{i,j}^2 = 0$ for $j \geq 2$ and

$$ \begin{align*}d_{i,j}^k : E_{i,j}^k \to E_{i-k,j+k-1}^k \text{ for } k \geq 2.\end{align*} $$

Thus, $E_{i,j}^3 = E_{i,j}^{\infty }$ and

$$ \begin{align*}E_{i,0}^3 = \mathrm{Ker}(d_{i,0}^2)\text{ is finite }\end{align*} $$

since $E_{i,0}^2$ is finite for all i. Note that

$$ \begin{align*}E_{i,1}^3 = \mathrm{Coker}(d_{i+2,0}^2) = E_{i,1}^2/ \mathrm{Im}(d_{i+2,0}^2) \text{ is finite if and only if } E_{i,1}^2 \text{ is finite,}\end{align*} $$

since $E_{i+2,0}^2$ is finite for all i.

Since H is finitely generated and by the convergence of the spectral sequence, we have that

$$ \begin{align*}E^{3}_{0,1} = E_{0,1}^{\infty} \text{ is finite,}\end{align*} $$

hence

$$ \begin{align*}E_{0,1}^2 = H_0(Q, H_1(K_0, \mathbb{F}_p)) \text{ is finite.}\end{align*} $$

Since $\mathbb {F}_p[[Q]]$ is a local ring, this implies that $H_1(K_0, \mathbb {F}_p)$ is finitely generated as a pro- $p\ \mathbb {F}_p[[Q]]$ -module. Then $E_{j,1}^2$ is finite for $j \geq 0$ .

By the convergence of the spectral sequence, there is an exact sequence for $i \geq 2$

$$ \begin{align*}0 \to E_{i-1,1}^{\infty} \to H_i(H, \mathbb{F}_p) \to E_{i,0}^{\infty} \to 0\end{align*} $$

with both $ E_{i-1,1}^{\infty } $ and $ E_{i,0}^{\infty }$ finite. Hence, $ H_i(H, \mathbb {F}_p) $ is finite for $i \geq 2$ and H is of type $FP_{\infty }$ , in particular is finitely presented as a pro-p group.

Proof of Corollary 1.6

Let F be a finitely generated free non-procyclic pro-p group that embeds as a closed subgroup of $\mathrm{Out}(K)$ . Note that $G = K \rtimes F$ is a pro-p group that embeds as a closed subgroup of $\mathrm{Aut}(K)$ and by Theorem 3.6 G is incoherent (in the category of pro-p groups). Finally, $G_0 = G \cap \mathrm{Aut}_0(G)$ is a pro-p subgroup of finite index in G, hence $G_0$ is incoherent (in the category of pro-p groups).

Proof of Corollary 1.7

We recall first some results from [Reference Lubotzky18]. Let G be a finitely generated pro-p group and $\mathrm{Aut}(G)$ denote all continuous automorphisms of G (which coincide with the abstract automorphisms of G). Denote $\mathrm{Inn}(G)$ the group of the internal automorphisms. The group $\mathrm{Aut}(G)$ is a profinite group.

Proof of Lemma 5.2

Here, we use significantly ideas introduced in [Reference Romankov21]. We fix $x_1, x_2, \ldots , x_n$ a generating set of M. Define

$$ \begin{align*}\mathrm{IAut}(M) = \{ \varphi \in \mathrm{Aut}(M) \ | \ \varphi \text{ induces on } M/ M'\text{ the identity map} \},\end{align*} $$

where $\mathrm{Aut}(M)$ denotes continuous automorphisms of M. In fact, every abstract automorphism of a finitely generated pro-p group is a continuous one. Then there is a short exact sequence of profinite groups

$$ \begin{align*}1 \to \mathrm{IAut}(M) \to \mathrm{Aut}(M) \to \mathrm{Aut}(M^{ ab}) = GL_n(\mathbb{Z}_p) \to 1.\end{align*} $$

By [Reference Romankov21], there is a Bachmut embedding $\beta $ of $\mathrm{IAut}(M)$ in $GL_n(\mathbb {Z}_p[[M^{ab}]])$ , where $M^{ab}$ is the maximal abelian pro-p quotient of M. By [Reference Romankov21], where $\mathrm{Aut}(M)$ acts on the right,

$$ \begin{align*}\beta(\varphi) = \left( \frac{\partial }{\partial x_j} (x_i^{ \varphi})\right) \text{ and } \frac{\partial}{\partial x_j} : M \to \mathbb{Z}_p[[M^{ab}]] \text{ are the Fox derivatives defined by }\end{align*} $$

$$ \begin{align*}\frac{\partial}{\partial x_j} (1) = 0, \ \frac{\partial}{\partial x_j} ( g_1 g_2) = \frac{\partial}{ \partial x_j} ( g_1) + \overline{g}_1 \frac{\partial}{\partial x_j} (g_2), \ \frac{\partial}{ \partial x_j} (x_i) = \delta_{i,j},\end{align*} $$

where $\overline {g}_1$ is the image of $g_1 \in M$ in $ M^{ ab}$ , $\delta _{i,j}$ is the Kronecker symbol.

Set $s_i$ for the image of $x_i-1$ in $\mathbb {Z}_p[[M^{ ab}]]$ ; thus,

$$ \begin{align*}\mathbb{Z}_p[[M^{ ab}]] \simeq \mathbb{Z}_p[[s_1, s_2, \ldots, s_n]].\end{align*} $$

Define

$$ \begin{align*}\mathrm{det}(\varphi) = \mathrm{det} (\beta(\varphi)).\end{align*} $$

By [Reference Romankov21],

$$ \begin{align*}\mathrm{det}(\mathrm{IAut}(M)) = 1 + \Delta =: P\end{align*} $$

is a multiplicative abelian group, where $\Delta $ is the unique maximal ideal of $\mathbb {Z}_p[[M^{ab}]]$ , and the $GL_n(\mathbb {Z}_p)$ -action via conjugation on the abelianization of $\mathrm{IAut}(M)$ induces an action on $\mathrm{det}(\mathrm{IAut}(M)) = P$ . Then we have a short exact sequence of profinite groups

$$ \begin{align*}1 \to P \to \mathrm{Aut}(M)/ \mathrm{Ker} (\mathrm{det}) \to GL_n(\mathbb{Z}_p) \to 1.\end{align*} $$

Consider the pro-p group

$$ \begin{align*}GL_n^1(\mathbb{Z}_p) = \mathrm{Ker}(GL_n(\mathbb{Z}_p) \to GL_n(\mathbb{F}_p)).\end{align*} $$

Let Q be the maximal pro-p quotient of P that has exponent p. Then there is a pro-p subgroup T of $ \mathrm{Aut}(M)/ \mathrm{Ker} (\mathrm{det}) $ and a short exact sequence of pro-p groups

$$ \begin{align*}1 \to P \to T \to GL_n^1(\mathbb{Z}_p) \to 1\end{align*} $$

and a pro-p quotient $T_0$ of T together with a short exact sequence of pro-p groups

$$ \begin{align*}1 \to Q \to T_0 \to GL_n^1(\mathbb{Z}_p) \to 1.\end{align*} $$

By [Reference Romankov22, (7)],

$$ \begin{align*}P^p \cap ( 1 + p \Delta) = 1 + p^2 \Delta,\end{align*} $$

and for $\delta \in \Delta $ using $[\delta ]$ for the image of $ 1 + p \delta $ in Q, we have that

$$ \begin{align*}[\delta_1] [\delta_2] = [ \delta_1 + \delta_2].\end{align*} $$

Thus, the multiplicative subgroup of Q generated by $\{ [\delta ] \ | \ \delta \in \Delta \}$ could be identified with the additive group $\Delta / p \Delta $ . Furthermore, using the long exact sequence in homology for the short exact sequence

$$ \begin{align*}0 \to \Delta \to \mathbb{Z}_p[[s_1, s_2, \ldots, s_n]] \to \mathbb{F}_p \to 0,\end{align*} $$

we have a long exact sequence

$$ \begin{align*}0 = \mathrm{Tor}^{\mathbb{Z}_p}_1( \mathbb{Z}_p[[s_1, s_2, \ldots, s_n]], \mathbb{F}_p) \to \mathrm{Tor}_1^{\mathbb{Z}_p}(\mathbb{F}_p, \mathbb{F}_p) \to\end{align*} $$

$$ \begin{align*}\Delta/ p \Delta \to \mathbb{F}_p[[s_1, s_2, \ldots, s_n]] \to \mathbb{F}_p \to 0.\end{align*} $$

Note that $\mathrm{Tor}_1^{\mathbb {Z}_p}(\mathbb {F}_p, \mathbb {F}_p) \simeq \mathbb {F}_p$ and thus we have a short exact sequence of additive pro-p groups

$$ \begin{align*}0 \to \mathbb{F}_p \to \Delta/ p \Delta \xrightarrow{\nu} \Omega \to 0,\end{align*} $$

where $\Omega $ is the augmentation ideal of $ \mathbb {F}_p[[s_1, s_2, \ldots , s_n]] $ and for the canonical projection

$$ \begin{align*}\pi : \Delta \to \Delta/ p \Delta,\end{align*} $$

the composition map $\nu \circ \pi : \Delta \to \Omega $ is the restriction of the map $\mathbb {Z}_p[[s_1, s_2, \ldots , s_n]] \to \mathbb {F}_p[[s_1, s_2, \ldots , s_n]]$ that reduces coefficients mod p. Actually, $\mathrm{Ker}(\nu ) = \mathbb {F}_p$ is generated as an additive group by $p + p \Delta $ .

Consider now $\varphi _2 \in \mathrm{Aut}(M)$ given by

$$ \begin{align*}\varphi_2 = \rho^p, \text{ where }\rho (x_1) = x_1 x_2, \rho(x_k) = x_k \text{ for } 2 \leq k \leq n\end{align*} $$

and $\varphi _1 \in \mathrm{IAut}(M)$ such that

$$ \begin{align*}\mathrm{det}(\beta(\varphi_1)) = 1 + p s_1.\end{align*} $$

Note that $\varphi _1$ is not uniquely determined and that the image of $\varphi _2$ in $GL_n(\mathbb {Z}_p)$ is in $GL_n^1(\mathbb {Z}_p)$ . Hence, the profinite subgroup $\Gamma $ of $\mathrm{Aut}(M)$ generated by $\varphi _1, \varphi _2$ is in fact a pro-p group. Let

$$ \begin{align*}\Gamma_0 = \langle \psi_1, \psi_2 \rangle\end{align*} $$

be the image of $\Gamma $ in $T_0$ , where $\psi _i$ is the image of $\varphi _i$ in $T_0$ . Thus, $\Gamma _0$ is a pro-p group.

By [Reference Romankov21, Proposition 4.4], for every $\varphi \in \mathrm{IAut}(M)$ for $\varphi '= \rho ^{ -1} \varphi \rho $ , $h'= \mathrm{det} (\beta (\varphi '))$ and $h = \mathrm{det}( \beta (\varphi ))$ , we have that $h'$ is obtained from h applying the substitution $s_1 \to s_1 + s_2 + s_1 s_2$ . Then the action of $\psi _2$ on $\psi _1 = [s_1]$ by conjugations is induced by applying the substitution $s_1 \to s_1 + s_2 + s_1 s_2$ exactly p-times, i.e., we apply the substitution

$$ \begin{align*}s_1 \to (1 + s_1) ( 1 + s_2) ^p -1.\end{align*} $$

Similarly, the action of $\psi _2^k$ on $\psi _1 = [s_1]$ by conjugation is induced by applying the substitution $s_1 \to s_1 + s_2 + s_1 s_2$ exactly $pk$ -times, and thus gives the substitution $s_1 \to (1 + s_1) ( 1 + s_2) ^{pk} -1$ .

Let A be the normal pro-p subgroup of $\Gamma _0$ generated by $\psi _1$ . Thus, A can be identified with an additive subgroup of $\Delta / p \Delta $ and

$$ \begin{align*}\nu(A) \subseteq \nu(\Delta/ p \Delta) = \Omega =\end{align*} $$

$$ \begin{align*}s_1 \mathbb{F}_p[[s_1, s_2, \ldots, s_n]] + s_2 \mathbb{F}_p[[s_1, s_2, \ldots, s_n]] + \cdots + s_n \mathbb{F}_p[[s_1, s_2, \ldots, s_n]].\end{align*} $$

The previous paragraph shows that

$$ \begin{align*}\{ (1 + s_1) ( 1 + s_2) ^{p k} -1 \ | \ k \geq 0 \} \subseteq \nu(A),\end{align*} $$

in particular $\nu (A)$ and A are infinite.

Note that $\Gamma _0 \simeq A \rtimes D$ , where $D \simeq \mathbb {Z}_p $ is generated by $\psi _2$ . We view A as an $\mathbb {F}_p[[t]]$ -module via the conjugation action of $\psi _2 = 1 + t$ , where $\mathbb {F}_p[[t]] \simeq \mathbb {F}_p[[D]]$ . Furthermore, A is a pro-p cyclic $\mathbb {F}_p[[t]]$ -module, with a generator $\psi _1$ . Since every proper $\mathbb {F}_p[[t]]$ -module quotient of $\mathbb {F}_p[[t]]$ is a finite additive group and A is infinite, we deduce that $A \simeq \mathbb {F}_p[[t]]$ . Then, by the example after Theorem 2.2, $\Gamma _0$ is a metabelian pro-p group that is not finitely presented.

Note that the image W of $M \simeq \mathrm{Inn} (M)$ in $T_0$ is inside Q and since M is a finitely generated pro-p group and Q is an abelian pro-p group of finite exponent p, then W and consequently $\Gamma _0 \cap W$ are finite. Since $\Gamma _0 \cap W$ is finite, $\Gamma _0/ (\Gamma _0 \cap W)$ is not a finitely presented pro-p group. Actually examining the structure of $\Gamma _0$ , it is easy to see that any finite normal subgroup of $\Gamma _0$ is trivial, in particular $\Gamma _0 \cap W = 1$ . Finally, $\Gamma _0 \simeq \Gamma _0 / (\Gamma _0 \cap W)$ is a metabelian pro-p quotient of a 2-generated pro-p group $ H \leq \mathrm{Out}(M)$ . This completes the proof of the lemma.