Finitely generated groups are not closed under $\aleph_1$-cofiltered limits

Proof. We have $\ker(p) = 1 \times G \cong G$. Clearly, $p \circ s = \id_B$, and $s(B) = B \times 1$ commutes with $1 \times G$. $\square$

Proof. We have $q \circ p \circ s \circ t = q \circ t = \id_C$.

Next, we compute the kernel of $q \circ p$. By the internal direct product decomposition $A = s(B) \times \ker(p)$, every $a \in A$ can be written uniquely as $a = s(b) k$ with $b \in B$ and $k \in \ker(p)$, namely $b = p(a)$. Then $q(p(a)) = q(b)$, so $a \in \ker(q \circ p)$ if and only if $b \in \ker(q)$. This shows $\ker(q \circ p) = s(\ker(q)) \cdot \ker(p)$. Since $s(\ker(q)) \subseteq s(B)$, the two factors commute elementwise and intersect trivially. Therefore, $$\ker(q \circ p) \cong s(\ker(q)) \times \ker(p) \cong \ker(q) \times \ker(p) \cong G \times G \cong G.$$

Finally, we show that $s(t(C))$ commutes elementwise with $\ker(q \circ p) = s(\ker(q)) \cdot \ker(p)$. Since $t(C)$ commutes elementwise with $\ker(q)$, applying $s$ shows that $s(t(C))$ commutes elementwise with $s(\ker(q))$. And since $s(t(C)) \subseteq s(B)$, it commutes elementwise with $\ker(p)$. $\square$

Proof. Let $C \coloneqq \ker(\pi)$ and $K \coloneqq \ker(p)$. Since $A \cong G$ and $C \cong G$, we may choose an isomorphism $c : A \to C$. We have the internal direct product decompositions $A = s(B) \times K$ and $X = \iota(B) \times C$. Applying $c$ to the former, we get $C = c(K) \times c(s(B))$, and therefore $$X = \iota(B) \times c(K) \times c(s(B))$$ as an internal direct product of three subgroups. That is, every element of $X$ can be written uniquely as $\iota(b) \, c(k) \, c(s(b'))$ with $b,b' \in B$ and $k \in K$. We define the homomorphisms $\iota' : A \to X$ and $\pi' : X \to A$ by $$\iota'(s(b) k) \coloneqq \iota(b) \, c(k), \qquad \pi'\bigl(\iota(b) \, c(k) \, c(s(b'))\bigr) \coloneqq s(b) k.$$ They are well-defined homomorphisms, since they are defined factor by factor on internal direct products. We now verify the required properties.

• We have $\pi' \circ \iota' = \id_A$ by construction.
• We have $p\bigl(\pi'(\iota(b) \, c(k) \, c(s(b')))\bigr) = p(s(b) k) = b$. This equals $\pi\bigl(\iota(b) \, c(k) \, c(s(b'))\bigr)$, since $\pi \circ \iota = \id_B$ and $c(A) = C = \ker(\pi)$. Hence, $p \circ \pi' = \pi$.
• We have $\iota'(s(b)) = \iota(b)$, i.e. $\iota' \circ s = \iota$.
• The kernel of $\pi'$ is $c(s(B)) \cong B \cong G$.
• The image $\iota'(A) = \iota(B) \cdot c(K)$ commutes elementwise with $\ker(\pi') = c(s(B))$. Indeed, $\iota(B)$ commutes elementwise with $C$, and $c(K)$ commutes elementwise with $c(s(B))$ because $K$ commutes elementwise with $s(B)$. $\square$

Proof. We construct $\Delta_\beta$ and the homomorphisms $\pi_{\alpha\beta},\iota_{\alpha\beta}$ for $\alpha \leq \beta$ by transfinite recursion on $\beta$.

We start with $\Delta_0 \coloneqq G$ and $\pi_{00} \coloneqq \iota_{00} \coloneqq \id_G$.

For the successor step, let $\Delta_{\beta+1} \coloneqq \Delta_\beta \times G$, and let $(\pi_{\beta,\beta+1},\iota_{\beta,\beta+1})$ be the good pair from Lemma 1. For $\alpha < \beta$ we define $$\pi_{\alpha,\beta+1} \coloneqq \pi_{\alpha\beta} \circ \pi_{\beta,\beta+1}, \qquad \iota_{\alpha,\beta+1} \coloneqq \iota_{\beta,\beta+1} \circ \iota_{\alpha\beta}.$$ Then (1) clearly holds, and (2) follows from Lemma 2.

For the limit step, let $\lambda < \omega_1$ be a limit ordinal. Since $\lambda$ is countable, we may choose a cofinal sequence $\alpha_0 < \alpha_1 < \cdots$ in $\lambda$. First, we define good pairs $(\pi_{\alpha_n\lambda},\iota_{\alpha_n\lambda}) : \Delta_\lambda \rightleftarrows \Delta_{\alpha_n}$ by recursion on $n$. Let $\Delta_\lambda \coloneqq \Delta_{\alpha_0} \times G$, and let $(\pi_{\alpha_0\lambda},\iota_{\alpha_0\lambda})$ be the good pair from Lemma 1. If $(\pi_{\alpha_n\lambda},\iota_{\alpha_n\lambda})$ has been defined, we apply Lemma 3 to this good pair and to the good pair $(\pi_{\alpha_n\alpha_{n+1}},\iota_{\alpha_n\alpha_{n+1}}) : \Delta_{\alpha_{n+1}} \rightleftarrows \Delta_{\alpha_n}$. This gives a good pair $(\pi_{\alpha_{n+1}\lambda},\iota_{\alpha_{n+1}\lambda}) : \Delta_\lambda \rightleftarrows \Delta_{\alpha_{n+1}}$ with $$\pi_{\alpha_n\alpha_{n+1}} \circ \pi_{\alpha_{n+1}\lambda} = \pi_{\alpha_n\lambda}, \qquad \iota_{\alpha_{n+1}\lambda} \circ \iota_{\alpha_n\alpha_{n+1}} = \iota_{\alpha_n\lambda}.$$ By induction, using (1) below $\lambda$, we get $$\pi_{\alpha_n\alpha_m} \circ \pi_{\alpha_m\lambda} = \pi_{\alpha_n\lambda}, \qquad \iota_{\alpha_m\lambda} \circ \iota_{\alpha_n\alpha_m} = \iota_{\alpha_n\lambda}$$ for all $n \leq m$.

Now, for an arbitrary $\alpha < \lambda$, choose some $n$ with $\alpha < \alpha_n$ and define $$\pi_{\alpha\lambda} \coloneqq \pi_{\alpha\alpha_n} \circ \pi_{\alpha_n\lambda}, \qquad \iota_{\alpha\lambda} \coloneqq \iota_{\alpha_n\lambda} \circ \iota_{\alpha\alpha_n}.$$ Also, let $\pi_{\lambda\lambda} \coloneqq \iota_{\lambda\lambda} \coloneqq \id$. By the previous equations and (1) below $\lambda$, this definition does not depend on the choice of $n$, and it agrees with the previous definition when $\alpha = \alpha_m$. For (1), let $\alpha \leq \beta < \lambda$ and choose $n$ with $\beta < \alpha_n$. Then $$\pi_{\alpha\beta} \circ \pi_{\beta\lambda} = \pi_{\alpha\beta} \circ \pi_{\beta\alpha_n} \circ \pi_{\alpha_n\lambda} = \pi_{\alpha\alpha_n} \circ \pi_{\alpha_n\lambda} = \pi_{\alpha\lambda},$$ and similarly $\iota_{\beta\lambda} \circ \iota_{\alpha\beta} = \iota_{\alpha\lambda}$. For (2), the pair $(\pi_{\alpha\lambda},\iota_{\alpha\lambda})$ is the composite of the good pairs $(\pi_{\alpha_n\lambda},\iota_{\alpha_n\lambda})$ and $(\pi_{\alpha\alpha_n},\iota_{\alpha\alpha_n})$ in the sense of Lemma 2, hence it is good. $\square$

Proof. The limit is the subgroup $$\textstyle L \coloneqq \bigl\{x \in \prod_{\beta < \omega_1} \Delta_\beta : \pi_{\beta\gamma}(x_\gamma) = x_\beta \text{ for all } \beta \leq \gamma\bigr\}.$$ For $\alpha < \omega_1$ we define a homomorphism $j_\alpha : \Delta_\alpha \to \prod_{\beta < \omega_1} \Delta_\beta$ by $$j_\alpha(x)_\beta \coloneqq \begin{cases} \iota_{\alpha\beta}(x) & \beta \geq \alpha, \\ \pi_{\beta\alpha}(x) & \beta \leq \alpha. \end{cases}$$ For $\beta = \alpha$, both cases give $x$, so this is well-defined. We claim that $j_\alpha$ maps into $L$, i.e. that for all $x \in \Delta_\alpha$ and $\beta \leq \gamma$ we have $$\pi_{\beta\gamma}(j_\alpha(x)_\gamma) = j_\alpha(x)_\beta.$$ This follows from (1) and $\pi_{\alpha\beta} \circ \iota_{\alpha\beta} = \id$, distinguishing three cases:

$$\begin{align*} \pi_{\beta\gamma}(j_\alpha(x)_\gamma) & = \pi_{\beta\gamma}(\iota_{\alpha\gamma}(x)) = \pi_{\beta\gamma}(\iota_{\beta\gamma}(\iota_{\alpha\beta}(x))) = \iota_{\alpha\beta}(x) = j_\alpha(x)_\beta && \text{if } \alpha \leq \beta, \\ \pi_{\beta\gamma}(j_\alpha(x)_\gamma) & = \pi_{\beta\gamma}(\iota_{\alpha\gamma}(x)) = \pi_{\beta\alpha}(\pi_{\alpha\gamma}(\iota_{\alpha\gamma}(x))) = \pi_{\beta\alpha}(x) = j_\alpha(x)_\beta && \text{if } \beta \leq \alpha \leq \gamma, \\ \pi_{\beta\gamma}(j_\alpha(x)_\gamma) & = \pi_{\beta\gamma}(\pi_{\gamma\alpha}(x)) = \pi_{\beta\alpha}(x) = j_\alpha(x)_\beta && \text{if } \gamma \leq \alpha. \end{align*}$$

So we get a homomorphism $j_\alpha : \Delta_\alpha \to L$. A coordinatewise check of the same kind shows $j_\beta \circ \iota_{\alpha\beta} = j_\alpha$ for $\alpha \leq \beta$. Hence, the subgroups $L_\alpha \coloneqq j_\alpha(\Delta_\alpha)$ of $L$ satisfy $$L_\alpha \subseteq L_\beta$$ for $\alpha \leq \beta$. Next, we show that $L_\alpha \neq L_{\alpha+1}$. Since $G$ is non-trivial and $\ker(\pi_{\alpha,\alpha+1}) \cong G$ by (2), there is some element $1 \neq k \in \ker(\pi_{\alpha,\alpha+1})$. Assume that $j_{\alpha+1}(k) = j_\alpha(x)$ for some $x \in \Delta_\alpha$. Comparing the $(\alpha+1)$-coordinates gives $k = \iota_{\alpha,\alpha+1}(x)$. Applying $\pi_{\alpha,\alpha+1}$ gives $1 = x$, and hence $k = 1$, a contradiction. Thus, $j_{\alpha+1}(k) \in L_{\alpha+1} \setminus L_\alpha$.

Now choose an element $y_\alpha \in L_{\alpha+1} \setminus L_\alpha$ for every $\alpha < \omega_1$. These elements are pairwise distinct: if $\alpha < \beta$, then $y_\alpha \in L_{\alpha+1} \subseteq L_\beta$, but $y_\beta \notin L_\beta$. Therefore, $L$ has at least $\aleph_1$ elements. $\square$

Proof. The diagram $\Delta : \omega_1^{\op} \to \Grp$ is $\aleph_1$-cofiltered, and each $\Delta_\alpha \cong G$ is finitely generated. But its limit is uncountable by Proposition 5, whereas every finitely generated group is countable. $\square$