Finitely generated groups do not have $\aleph_1$-filtered colimits

Proof. Let $I$ be the set of finitely generated subgroups of $H$, partially ordered by inclusion, and let $D : I \to \Grp_\fg$ be the diagram which maps each $U \in I$ to $U$ and each inclusion $U \subseteq U'$ to the inclusion homomorphism $U \hookrightarrow U'$. The poset $I$ is $\aleph_1$-filtered by (1): every countable subset of $I$ has an upper bound in $I$, since the union of countably many finitely generated subgroups is countable.

Assume that $D$ has a colimit $C$ in $\Grp_\fg$ with colimit cocone $\lambda_U : U \to C$ for $U \in I$. Since $H$ is the directed union of all $U \in I$, these homomorphisms glue to a homomorphism $f : H \to C$ with $f|_U = \lambda_U$. For each $n$, the restrictions $\pi_n|_U : U \to H_n$ of the projection $\pi_n : H \to H_n$ form a cocone of $D$ with tip $H_n$, so there is a homomorphism $h_n : C \to H_n$ with $h_n \circ \lambda_U = \pi_n|_U$ for all $U \in I$, i.e. $h_n \circ f = \pi_n$. By (2), the $\pi_n$ are jointly injective, so $f$ is injective. But $H$ is uncountable, whereas $C$ is countable as a finitely generated group. $\square$

Proof. Since $V^{\IN}$ is uncountable, so is $H$. The map $(z,g) \mapsto ((z_n,g))_{n \in \IN}$ embeds $H$ into $(V \rtimes G)^{\IN}$, and $V \rtimes G$ is generated by $u,r,\varphi,s,e_{(0,0)}$, since $u^i r^j \cdot e_{(0,0)} = e_{(i,j)}$. $\square$

Proof. For $d = 0$ this holds since $P$ is the identity on $V_0$. For $d \neq 0$, we have $u^d \cdot x \in V_d \subseteq W$, and $P$ vanishes on $W$. $\square$

Proof. Write $v = x + y$ with $x \in V_0$ and $y \in W$, so that $P(v) = x$. Since $\varphi \cdot x \in W$ and $\varphi \cdot y \in V_0$, we have $P(\varphi \cdot v) = \varphi \cdot y$, and hence $\varphi \cdot P(\varphi \cdot v) = \varphi^2 \cdot y = y$. Therefore, $P(v) + \varphi \cdot P(\varphi \cdot v) = x + y = v$. $\square$

Proof. By definition of the semidirect product, conjugation with $g \in G$ in $H$ acts on $V^{\IN}$ by $z \mapsto g \cdot z$. Hence, $\Gamma \cap V^{\IN}$ is closed under the action of $G$. For $v = x + y$ with $x \in V_0$ and $y \in W$, we have $$s \cdot v - v = (-x + y) - (x + y) = -2x = x = P(v),$$ since $-2 = 1$ in $\IF_3$. Hence, $P(z) = s \cdot z - z$ for $z \in V^{\IN}$, so $\Gamma \cap V^{\IN}$ is also closed under $P$. $\square$

Proof. Let $X \subseteq H$ be countable. Since every element of $H$ is a product of an element of $V^{\IN}$ and an element of $G$, there are $w_0,w_1,\dotsc \in V^{\IN}$ such that $X$ is contained in the subgroup generated by $G$ and $w_0,w_1,\dotsc$. Write $w_k = (w_{k,n})_{n \in \IN}$ and define $b,c \in V^{\IN}$ by $$\textstyle b_n \coloneqq u^n \cdot e_{(0,0)}, \qquad c_n \coloneqq \sum_{k=0}^{n} \bigl(u^{2k+1} \cdot P(w_{k,n}) + u^{2k+2} \cdot P(\varphi \cdot w_{k,n})\bigr).$$ Let $\Gamma \subseteq H$ be the finitely generated subgroup generated by $u,r,\varphi,s,b,c$, i.e. by $G$, $b$ and $c$. By Lemma 5, $\Gamma \cap V^{\IN}$ is a subgroup of $V^{\IN}$ containing $b$ and $c$ which is closed under the action of $G$ and under $P$. We show that $w_k \in \Gamma$ for all $k$, which implies $X \subseteq \Gamma$.

First, $\Gamma$ contains every $z \in V^{\IN}$ with $z_n = 0$ for almost all $n$. Indeed, let $n \in \IN$. The coordinate of $P(u^{-n} \cdot b) \in \Gamma$ at $n'$ is $P(u^{n'-n} \cdot e_{(0,0)})$, which is $e_{(0,0)}$ for $n' = n$ and $0$ otherwise. Acting on this element with $u^i r^j$ replaces $e_{(0,0)}$ by $e_{(i,j)}$, and these elements generate all $z \in V^{\IN}$ with $z_{n'} = 0$ for $n' \neq n$.

Second, let $k \in \IN$ and $n \geq k$. All vectors $P(w_{k',n})$ and $P(\varphi \cdot w_{k',n})$ lie in $V_0$. If we apply the linear map $P \circ u^{-(2k+1)}$ to $c_n$, Lemma 3 shows that only the summand with $u^{2k+1}$ survives, so that $P(u^{-(2k+1)} \cdot c_n) = P(w_{k,n})$. Similarly, $P(u^{-(2k+2)} \cdot c_n) = P(\varphi \cdot w_{k,n})$. Therefore, by Lemma 4, the element $$y_k \coloneqq P(u^{-(2k+1)} \cdot c) + \varphi \cdot P(u^{-(2k+2)} \cdot c) \in \Gamma$$ has coordinate $P(w_{k,n}) + \varphi \cdot P(\varphi \cdot w_{k,n}) = w_{k,n}$ for all $n \geq k$. Hence, $w_k - y_k$ has only finitely many non-zero coordinates, so it lies in $\Gamma$ by the first step, and therefore $w_k \in \Gamma$. $\square$

Proof. Apply Lemma 1 to the group $H$, using Lemma 2 and Proposition 6. $\square$