Proof. It is easy to check that inverse images of basis elements of the form ($\star$) are open.

Proof. Observe that

\begin{equation*}\mathcal{S}(G)=\{K\in\mathcal{K}(G):\ K\cdot K^{-1}=K\}.\end{equation*}

It is easy to check that this set is closed. (Use Lemma 2.3.)

Proof of Proposition 2.15

Fix an algebraically closed group H. By Theorem 2.18 there is a countably infinite elementary subgroup G of H. We claim that G is algebraically closed.

Pick any finite system (E, I) of equations and inequations with parameters from G such that (E, I) has a solution in some bigger group $K\geq G$. Clearly, we may choose K so that $G=H\cap K$. By Theorem 2.16 there is a group M such that $H,K\leq M$. Clearly, (E, I) has a solution in M but contains only parameters from G. Since H is algebraically closed, (E, I) has a solution in H as well. Moreover, (E, I) has a solution in G because G is an elementary subgroup of H, which concludes the proof.

observation 3.5.

The above-mentioned standard clopen basis (3.1) of ${\mathbb{N}}^{{\mathbb{N}}\times{\mathbb{N}}}$ induces a clopen basis for $\mathcal{G}$: sets of the form

(3.2)\begin{equation} \{G\in\mathcal{G}:\ \forall i\leq l\ (n_i\cdot m_i=k_i)\} \end{equation}

with $l\in{\mathbb{N}}$ and $n_i,m_i,k_i\in{\mathbb{N}}$ for all $i\leq l$ constitute a basis. For practical reasons we introduce another clopen basis for $\mathcal{G}$.

Proof. This family extends (3.2), so it suffices to show that its elements are clopen. Since clopen sets form an algebra, it suffices to consider only one word W and only the case of equation. The proof is a straightforward induction on the length of W and uses the same argument as Remark 3.4.

observation 3.7.

An important observation is that permutations induce homeomorphisms. Let $\varphi:{\mathbb{N}}\to{\mathbb{N}}$ be a bijection that fixes 1. Then the induced homeomorphism $h_\varphi:\mathcal{G}\to\mathcal{G}$ is defined as follows. Intuitively, we define $h_\varphi(G)$ by pushing forward the structure of G via φ. More precisely, for any $G\in\mathcal{G}$ the multiplication table $h_\varphi(G)$ is defined by the equations $i\cdot j:=\varphi(G(\varphi^{-1}(i),\varphi^{-1}(j)))$ for all $i,j\in{\mathbb{N}}$. Thus φ is an isomorphism between $\widetilde G$ and $\widetilde{h_\varphi(G)}$. It is an easy exercise to verify that h_φ is indeed a homeomorphism.

Notation 3.10.

If $\mathcal{B}$ is a clopen set of the form $\{G\in\mathcal{G}:\ \forall i,j\leq k\ (i\cdot j=m_{i,j})\}$ with $k\in{\mathbb{N}}$ and $m_{i,j}\in{\mathbb{N}}$ for all $i,j\leq k$, let $supp\mathcal{B}:=\{1,\dots,k\}\cup\{m_{i,j}:\ i,j\leq k\}$.

Proof. Let F be the free group generated by $X=\{x_1,x_2\dots\}$. Let $G\in\mathcal{G}$. By definition, $\widetilde G$ is algebraically closed if and only if the following holds. For every finite system (E, I) of equations and inequations with elements from $F\star\widetilde G$ either (E, I) is inconsistent with $\widetilde G$ or it has a solution in $\widetilde G$ (see $\S$ 2.3 for the definitions). Note that elements of $F\star \widetilde G$ are words with letters from X and ${\mathbb{N}}$. Clearly, there are countably many such words. Thus, it suffices to prove that for any finite system (E, I) the set

\begin{equation*}\mathcal C(E,I):=\{G\in\mathcal G:\;(E,I)\text{ is inconsistent with }\widetilde G\text{ or}\,\,(E,I)\,\,\text{has a solution in}\,\,\widetilde G\}\end{equation*}

contains a dense open set. Fix

\begin{equation*}E=\{U_1(x_1,\dots,x_n),\dots,U_k(x_1,\dots,x_n)\}\;\text{and}\;I=\{V_1(x_1,\dots,x_n),\dots,V_l(x_1,\dots,x_n)\}\end{equation*}

where $x_1,\dots,x_n$ are the variables occurring in elements of E and I.

Fix a nonempty basic clopen set $\mathcal{B}=\{G\in\mathcal{G}:\ \forall i,j\leq k\ (i\cdot j=m_{i,j})\}$ with $k\in{\mathbb{N}}$ and $m_{i,j}\in{\mathbb{N}}$ for all $i,j\leq k$. We need to show that $\mathcal{B}\cap\mathcal{C}(E,I)$ contains a nonempty open set. There are two cases.

Case 1. There exists $H\in\mathcal{B}$ such that (E, I) has a solution in $\widetilde H$. That is, for some natural numbers $a_1,\dots,a_n$ we have $\bigwedge_{i=1}^k U_i(a_1,\dots,a_n)=1$ and $\bigwedge_{j=1}^l V_j(a_1,\dots,a_n)\neq 1$ in $\widetilde H$. Then

\begin{equation*}\mathcal{U}:=\left\{G\in\mathcal{G}:\ \bigwedge_{i=1}^k U_i(a_1,\dots,a_n)=1\land \bigwedge_{j=1}^l V_j(a_1,\dots,a_n)\neq 1\right\},\end{equation*}

is clopen in $\mathcal{G}$ by Proposition 3.6. Now $\mathcal{U}\subseteq\mathcal{C}(E,I)$ and $H\in\mathcal{U}\cap\mathcal{B}$, hence $\mathcal{U}\cap\mathcal{B}$ is a nonempty open subset of $\mathcal{B}\cap\mathcal{C}(E,I)$, which completes Case 1.

Case 2. For every $G\in\mathcal{B}$ the finite system (E, I) is unsolvable in $\widetilde G$. It suffices to prove that for every $G\in\mathcal{B}$ the finite system (E, I) is inconsistent with $\widetilde G$.

Suppose that there is some $H\in\mathcal{B}$ and a group $L\geq \widetilde H$ such that (E, I) has a solution $a_1,\dots,a_n$ in L. Clearly, we may assume that L is countable since otherwise we could replace it by one of its countably generated subgroups. Let M be the set of natural numbers occurring in elements of E or I. We choose any bijection $\varphi:L\to {\mathbb{N}}$ that extends the identity of the finite set $M\cup(supp\mathcal{B})\cup\{1\}$ (recall Notation 3.10). We define the multiplication table K on ${\mathbb{N}}$ by pushing forward the structure of L via φ. Then $\widetilde K\in\mathcal{B}$ and $\varphi(a_1),\dots,\varphi(a_n)$ is a solution of (E, I) in $\widetilde K$, a contradiction.

Proof. This follows from Remark 2.23 and Theorem 3.11.

Proof. Let H be a group such that Φ holds in H. By the definition of algebraically closed groups and Remark 3.13, it suffices to show that for any algebraically closed group G the sentence Φ holds in G × H. Since $\{1\}\times H$ and H are isomorphic, Φ holds in $\{1\}\times H$. Hence Φ holds in G × H because Φ is existential.

Proof. Fix a nonempty basic clopen $\mathcal{B}=\{G\in\mathcal{G}:\ \forall i,j\leq k\ (i\cdot j=m_{i,j})\}$ with $k\in{\mathbb{N}}$ and $m_{i,j}\in{\mathbb{N}}$ for all $i,j\leq k$. Fix any multiplication table $A\in\mathcal{G}$ with $\widetilde A$ algebraically closed. Pick any $H\in\mathcal{B}$. We associate variables x_i to i for each $i\leq k$ and $x_{i,j}$ to (i, j) for each $i,j\leq k$. For each $i,j,l\leq k$ let

\begin{equation*} \varphi_{i,j,l}= \begin{cases} x_{i,j}=x_l\qquad\text {if}\quad m_{i,j}=l, \\ x_{i,j}\neq x_l\qquad\text{if}\quad m_{i,j}\neq l, \end{cases} \end{equation*}

and for each $i,j,r,s\leq k$ let

\begin{equation*} \varphi_{i,j,r,s}= \begin{cases} x_{i,j}=x_{r,s}\qquad\text {if}\quad m_{i,j}=m_{r,s}, \\ x_{i,j}\neq x_{r,s}\qquad\text{if}\quad m_{i,j}\neq m_{r,s}. \end{cases} \end{equation*}

Now let Φ denote the following formula:

\begin{equation*} \begin{array}{cc} \exists x_1,\ldots,x_k,x_{1,1},\ldots,x_{1,k},x_{2,1},\ldots,x_{2,k},\ldots,x_{k,1},\ldots,x_{k,k} \\ \displaystyle\left(\left(\bigwedge_{i,j\leq k} x_i\cdot x_j=x_{i,j}\right)\land\left(\bigwedge_{\substack{i,j\leq k\\ i\neq j}}x_i\neq x_j\right)\land\left(\bigwedge_{i,j,l\leq k}\varphi_{i,j,l}\right)\land\left(\bigwedge_{i,j,r,s\leq k}\varphi_{i,j,r,s}\right)\right). \end{array} \end{equation*}

Clearly, Φ holds in $\widetilde H$. Thus Φ holds in $\widetilde A$ as well by lemma 3.14. That is, there are numbers $a_i, n_{i,j}\in {\mathbb{N}}$ such that the equations $a_i\cdot a_j=n_{i,j}$ hold in $\widetilde A$ for each $i,j\leq k$ and two of them equal if and only if the corresponding elements are equal in H. Note that $a_1=1$ since both $a_1\cdot a_1=n_{1,1}$ and $n_{1,1}=a_1$ hold. Let $\alpha:{\mathbb{N}}\to{\mathbb{N}}$ be a bijection that extends the finite map $a_i\mapsto i$, $n_{i,j}\mapsto m_{i,j}$ for each $i,j\leq k$. Then $h_\alpha(A)\in\mathcal{B}$ and $\widetilde{h_\alpha(A)}$ is isomorphic to $\widetilde A$ (recall from observation 3.7 that h_α is the induced homeomorphism).

Proof. Fix a finite generating set $\{a_1,\dots,a_n\}$ for H. Let $G\in\mathcal{G}$ be arbitrary.

Claim. The group H is embeddable into $\widetilde G$ if and only if there are $b_1,\dots,b_n\in{\mathbb{N}}$ such that for every word W in n variables $W(a_1,\dots,a_n)=1$ in H $\iff$ $W(b_1,\dots,b_n)=1$ in $\widetilde G$.

If there is an embedding $i:H\hookrightarrow \widetilde G$, then clearly for every word W in n variables $W(a_1,\dots,a_n)$ maps to $W(i(a_1),\dots,i(a_n))$, therefore $b_j=i(a_j)$ for each $1\leq j\leq n$ is suitable. On the other hand, if for some $b_1,\dots,b_n\in \widetilde G$ we have $W(a_1,\dots,a_n)=1$ in H $\iff$ $W(b_1,\dots,b_n)=1$ in $\widetilde G$ for every word W in n variables, then H and $\langle b_1,\dots,b_n\rangle_{\widetilde G}$ have the same presentation, hence they are isomorphic. This proves the claim.

Thus we may write $\mathcal{E}_H$ as

\begin{equation*}\bigcup_{b_1,\dots,b_n\in{\mathbb{N}}}\ \bigcap_{\substack{W\text{is a word}\\ \text{in }n\text{variables}}}\underbrace{\{G\in\mathcal{G}:\ W(a_1,\dots,a_n)=1\text{in }H \iff W(b_1,\dots,b_n)=1\text{in }\widetilde G\}}_{\text{clopen by Proposition }3.6},\end{equation*}

which proves the lemma.

Proof. (2)$\implies$(3) follows from Theorem 2.14 and Theorem 2.22.

(3)$\implies$(2): Suppose (3). By Theorem 2.22 every finitely generated subgroup of H can be embedded into every algebraically closed group. Since every algebraically closed group is homogeneous by Proposition 2.21, (2) follows from Lemma 2.20.

(2)$\implies$(1) is immediate from Theorem 3.11.

(1)$\implies$(2): Fix a generically embeddable group H. By Lemma 2.20 it suffices to prove that any finitely generated subgroup K of H can be embedded into every algebraically closed group. Fix K. Note that $\mathcal{E}_K$ is comeager because it contains $\mathcal{E}_H$. On the other hand, $\mathcal{E}_K$ is F_σ by Lemma 3.16, hence it has nonempty interior by Remark 2.10. However, isomorphism classes of algebraically closed groups are dense in $\mathcal{G}$ by Theorem 3.15; therefore K can be embedded into every countably infinite algebraically closed group. Now Proposition 2.15 completes the proof.

Proof. By definition,

\begin{equation*}\mathcal{D}=\{G\in\mathcal{A}:\ \forall n,k\in{\mathbb{N}}\ \exists m\in{\mathbb{N}}\ (m^k=n)\}=\bigcap_{n,k\in{\mathbb{N}}}\ \bigcup_{m\in{\mathbb{N}}}\ \underbrace{\{G\in\mathcal{A}:\ m^k=n\}}_{\text{clopen}},\end{equation*}

which is G_δ in $\mathcal{A}$. Again, by definition,

\begin{equation*}\mathcal{T}=\{G\in\mathcal{A}:\ \forall n\in{\mathbb{N}}\ \exists k\in{\mathbb{N}}\ (n^k=1)\}=\bigcap_{n\in{\mathbb{N}}}\ \bigcup_{k\in{\mathbb{N}}}\ \underbrace{\{G\in\mathcal{A}:\ n^k=1\}}_{\text{clopen}},\end{equation*}

which is G_δ in $\mathcal{A}$. For $\mathcal{F}$ note that there are countably infinitely many finite abelian groups up to isomorphism, hence it suffices to prove that for any fixed finite abelian group H the set $\mathcal{A}\cap\mathcal{E}_H$ is G_δ. Let $\{h_1,\dots,h_n\}$ be the underlying set of H. For a $G\in\mathcal{A}$ the group H is embeddable into $\widetilde G$ if and only if there exist pairwise distinct numbers $g_1,\dots,g_n\in {\mathbb{N}}$ such that $h_i\cdot h_j=h_l$ in H implies $g_i\cdot g_j=g_l$ in $\widetilde G$ for all $i,j,l\leq n$. That is,

\begin{equation*}\mathcal{E}_H\cap\mathcal{A}=\bigcup_{\substack{g_1,\dots,g_n\in{\mathbb{N}}\\ \text{p.~distinct}}}\ \bigcap_{\substack{h_i,h_j,h_l\in H,\\ h_i\cdot h_j=h_l}}\ \{G\in\mathcal{A}:\ g_i\cdot g_j=g_l\},\end{equation*}

which is open in $\mathcal{A}$.

Proof. Fix any nonempty basic clopen $\mathcal{B}=\{G\in\mathcal{A}:\ \forall i,j\leq k\ (i\cdot j=m_{i,j})\}$ with $k\in{\mathbb{N}}$ and $m_{i,j}\in{\mathbb{N}}$ for all $i,j\leq k$. Pick some $H\in\mathcal{B}$ and a divisible abelian group K with an embedding $\varphi:\widetilde H\hookrightarrow K$; such a K exists by Remark 3.21. Clearly, K can be chosen to be countable. We write K in the form

\begin{equation*}\left(\bigoplus_{p\in\mathbf{P}}\mathbb{Z}[p^\infty]^{(I_{K,p})}\right)\oplus\mathbb{Q}^{(I_K)}.\end{equation*}

Consider the finite subset $\varphi(supp\mathcal{B})\subseteq K$ (recall notation 3.10). Each $x\in\varphi(supp\mathcal{B})$ has finitely many nonzero coordinates x_i with $i\in I_K$. Let n be a natural number greater than $2\cdot\max\{|x_i|:\ x\in\varphi(supp\mathcal{B}), i\in I_K\}$. Let N be the subgroup of K generated by elements of the form $((0,0,\dots),(0,\dots,0,\underset{r\text{th}}{n},0,\dots))$ for every $r\in I_K$. Let $\psi:K\to K/N$ be the quotient map. Clearly, $K/N$ is

\begin{equation*}\left(\bigoplus_{p\in\mathbf{P}}\mathbb{Z}[p^\infty]^{(I_{K,p})}\right)\oplus(\mathbb{Q}/n\mathbb{Z})^{(I_K)}.\end{equation*}

This is a countable abelian torsion group, thus there is an embedding $\nu: K/N\hookrightarrow \widetilde A$ by Remark 3.21. Notice that $\nu\circ\psi\circ\varphi:\ \widetilde H\to \widetilde A$ is homomorphism that is injective on the set $supp\mathcal{B}\subseteq\widetilde H$ by the choices of n and N. Thus, there is a bijection $\vartheta:{\mathbb{N}}\to{\mathbb{N}}$ that extends the finite bijection $(\nu\circ\psi\circ\varphi|_{supp\mathcal{B}})^{-1}$. For such a ϑ we have $h_\vartheta(A)\in\mathcal{B}$ because H and $h_\vartheta(A)$ coincide on $\{1,\dots,k\}\times\{1,\dots,k\}$ (recall from Observation 3.7 that h_ϑ is the induced homeomorphism). Since $\widetilde{h_\vartheta(A)}\cong\widetilde A$, we conclude that $\mathcal{I}_A$ is dense in $\mathcal{A}$.

Notation 4.4.

Let $\mathcal{F}$ denote the set of finite groups in $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$ with finite support.

Proof. Fix some ɛ > 0 and $K\in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}).$ Since $\pi_{[n]}(K)\overset{d_H}{\to}K$ (as $n\to\infty$, recall Notation 2.2), we may assume $K\leq{\mathbb{T}}^N\times\{0\}\times\{0\}\times\ldots$ for some $N\in{\mathbb{N}}$. We abuse notation and use simply ${\mathbb{T}}^N$ instead of ${\mathbb{T}}^N\times\{0\}\times\{0\}\times\ldots$. We work with the compatible metric $d_{{\mathbb{T}}^N}(x,y):=\sum_{i=1}^N d_{\mathbb{T}}(x(i),y(i))$.

Let $x_1,x_2\ldots x_k$ be a finite $\frac{\varepsilon}{2}$-net in K, and let us fix an integer M such that $M \gt \frac{2Nk}{\varepsilon}$.

We need a simultaneous version of Dirichlet’s Approximation Theorem [Reference Schmidt37, page 27].

We apply this theorem with the parameters d = kN and Q = M to the real numbers $x_1(1),\ldots x_1(N),\ldots x_k(1),\ldots x_k(N)$. Thus we get integers $p_1^1,\ldots p_N^1,\ldots p_1^k,\ldots p_N^k$ and q with $1\leq q\leq M^{kN}$ such that

\begin{equation*}\left|x_i(j)-\frac{p_j^i}{q}\right|\leq\frac{1}{qM} \lt \frac{\varepsilon}{2qkN}.\end{equation*}

The last inequality follows from our assumption on M.

This means that $d_{\mathbb{T}}\left(x_i(j),\frac{p_j^i}{q}\right) \lt \frac{\varepsilon}{2qkN}$ holds on the circle for each $1\leq i\leq k$ and $1\leq j\leq N$. Let L be the generated subgroup $\left\langle\left( \frac{p_1^1}{q},\ldots,\frac{p_N^1}{q}\right),\ldots,\left( \frac{p_1^k}{q},\ldots,\frac{p_N^k}{q}\right) \right\rangle\leq{\mathbb{T}}^N$. Note that L is finite.

We claim that $d_H(L,K) \lt \varepsilon$. First we prove that $L_\varepsilon\supseteq K$ (recall that L_ɛ is the ɛ-neighborhood of L). This holds since $x_1,x_2,\ldots, x_k$ was an $\frac{\varepsilon}{2}$-net in K and

\begin{equation*}d_{{\mathbb{T}}^N}\left(x_i,\left( \frac{p_1^i}{q},\ldots,\frac{p_N^i}{q}\right)\right)=\sum_{j=1}^N d_{{\mathbb{T}}}\left(x_i(j), \frac{p_j^i}{q}\right) \lt \frac{\varepsilon}{2qkN} \cdot N=\frac{\varepsilon}{2qk}\leq\frac{\varepsilon}{2}.\end{equation*}

For $K_\varepsilon\supseteq L$ notice that every element of L is of the form

\begin{equation*}\sum_{i=1}^k l_i\cdot\left( \frac{p_1^i}{q},\ldots,\frac{p_N^i}{q}\right),\end{equation*}

with $l_i\in\mathbb{Z}$, $0\leq l_i \lt q$ for each $1\leq i\leq k$, thus

\begin{equation*}d_{{\mathbb{T}}^N}{\left(\sum_{i=1}^k l_i x_i, \sum_{i=1}^k l_i\cdot\left( \frac{p_1^i}{q},\ldots,\frac{p_N^i}{q}\right)\right)}\leq\sum_{i=1}^k\sum_{j=1}^N l_i\cdot d_{{\mathbb{T}}}{\left(x_i(j), \frac{p_j^i}{q}\right)} \lt k\cdot N\cdot q\cdot\frac{\varepsilon}{2qkN} =\frac{\varepsilon}{2},\end{equation*}

which proves the claim since $\sum_{i=1}^k l_i x_i\in K$.

Proof. Let

\begin{equation*}\mathcal{B}_1:=\{K \in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}): \text{every finite abelian group}\,\, A\,\, \text{is a quotient of}\,\, K\}.\end{equation*}

Recall that for compact groups quotients and continuous homomorphic images coincide, hence

\begin{equation*}\mathcal{B}_1=\bigcap_{A \text{ is finite abelian}}\{K\in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}):\ \exists\ \varphi:K \to A\text{ surjective continuous homomorphism}\}.\end{equation*}

Since there are countably many finite abelian groups up to isomorphism, it sufficies to show that for a fixed finite abelian group A the following subset of $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$ is dense open:

\begin{equation*}\mathcal{B}_A:=\{K\in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}):\ \exists\ \varphi:K \to A\text{surjective continuous homomorphism}\}.\end{equation*}

Claim 1. The set $\mathcal{B}_A$ is open.

Let $\{a_1,a_2,\ldots, a_n\}$ be the underlying set of A. For a fixed $K\in \mathcal{B}_A$ let $\varphi:K\to A$ be a surjective continuous homomorphism and let $H:=Ker\varphi$. Then we have a partition $K=\bigcup_{i=1}^n (k_i+H)$, where $k_i\in K$ with $\varphi(k_i)=a_i$. Note that $k_i+H$ is compact for each i and let

\begin{equation*}\delta:= \min\{dist(k_i+H, k_j+H):\ 1\leq i,j\leq n,\ i\neq j\} \gt 0.\end{equation*}

We claim that $B_{d_H}{\left(K,\frac{\delta}{4}\right)}\subseteq \mathcal{B}_A$. To show this, for any $L\in B_{d_H}{\left(K,\frac{\delta}{4}\right)}$ we have to find a surjective continuous homomorphism $\psi:L\to A$. By the choice of δ, if we define $L_i:=L\cap (k_i+H)_{\frac{\delta}{4}}$, then $\bigcup_{i=1}^n L_i$ is a decomposition of L into disjoint nonempty open sets. Let us define ψ as $\psi(x)=a_i$ if and only if $x\in L_i$. The continuity of ψ is clear and the surjectivity follows from the fact that each L_i is nonempty. Now we have to show that ψ is a homomorphism.

Pick any $l_i\in L_i$ and $l_j\in L_j$. Then there exist $x_i\in k_i+H$ and $x_j\in k_j+H$ such that $d_{{\mathbb{T}}^{\mathbb{N}}}(x_i,l_i) \lt \frac{\delta}{4}$, $d_{{\mathbb{T}}^{\mathbb{N}}}(x_j,l_j) \lt \frac{\delta}{4}$, $\psi(l_i)=\varphi(x_i)=a_i$ and $\psi(l_j)=\varphi(x_j)=a_j$. Note that

\begin{equation*}d_{{\mathbb{T}}^{\mathbb{N}}}(x_i+x_j, l_i+l_j)\leq d_{{\mathbb{T}}^{\mathbb{N}}}(x_i+x_j,x_i+l_j)+d_{{\mathbb{T}}^{\mathbb{N}}}(x_i+l_j,l_i+l_j)=d_{{\mathbb{T}}^{\mathbb{N}}}(x_j,l_j)+d_{{\mathbb{T}}^{\mathbb{N}}}(x_i,l_i) \lt \frac{\delta}{2},\end{equation*}

therefore $l_i+l_j\in (x_i+x_j+H)_{\frac{\delta}{2}}$. Now it follows from $L\subseteq K_{\frac{\delta}{4}}$ and the choice of δ that $l_i+l_j\in (x_i+x_j+H)_{\frac{\delta}{4}}$. Thus, by the definition of ψ, we have $\psi(l_i+l_j)=\varphi(x_i+x_j)=a_i+a_j$. We conclude that ψ is a homomorphism, which completes the proof of Claim 1.

Claim 2. The set $\mathcal{B}_A$ is dense in $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$.

By Proposition 4.5 it sufficies to approximate elements of $\mathcal{F}$. Fix $F\in \mathcal{F}$ and ɛ > 0. Let $N\in{\mathbb{N}}$ be such that $\frac{1}{2^N} \lt \varepsilon$ and $\{1,\ldots,N\}\supseteq supp(F)$. Since A is a CMA group, there is an embedding

\begin{equation*}\varphi:A\hookrightarrow\underbrace{\{0\}\times\ldots\times\{0\}}_{N\text{times}}\times {\mathbb{T}}\times {\mathbb{T}}\ldots.\end{equation*}

Now the subgroup H of ${\mathbb{T}}^{\mathbb{N}}$ generated by F and $\varphi(A)$ is isomorphic to F × A. Therefore, we have a natural surjective continuous homomorphism H → A and also $d_H(H,F) \lt \varepsilon$ because $\pi_{[N]}(H)=\pi_{[N]}(F)$ and $\frac{1}{2^N} \lt \varepsilon$.

Proof. Let

\begin{equation*}\mathcal{B}_2:=\{K \in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}): K \text{is torsion-free} \}.\end{equation*}

We will show that $\mathcal{B}_2$ is dense G_δ in $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$.

Claim 1. The set $\mathcal{B}_2$ is G_δ in $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$.

Let us define $f_n:{\mathbb{T}}^{\mathbb{N}}\rightarrow {\mathbb{T}}^{\mathbb{N}}$ as $f_n: x\mapsto n\cdot x$. Clearly, f_n is continuous and

\begin{equation*}\mathcal{B}_2=\left\{K\in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}):\ \forall n,k\in {\mathbb{N}}\ \left(0\notin f_n{\left[K\setminus B{\left(0,\tfrac{1}{k}\right)}\right]}\right)\right\}=\end{equation*}

\begin{equation*}=\bigcap_{n,k\in{\mathbb{N}}} \underbrace{\left\{K\in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}):\ K\cap \underbrace{B{\left(0,\tfrac{1}{k}\right)}^c\cap f_n^{-1}(\{0\})}_{\text{closed}}=\emptyset\right\}}_{\text{open}}.\end{equation*}

Claim 2. The set $\mathcal{B}_2$ is dense in $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$.

By Proposition 4.5 it suffices to approximate elements of $\mathcal{F}$. Fix any $F\in \mathcal{F}$ and ɛ > 0. Let $N\in{\mathbb{N}}$ be such that $\frac{1}{2^N} \lt \varepsilon$ and $\{1,\ldots,N\}\supseteq supp(F)$. As in Theorem 4.7, let $Z=\prod_{p\in P}(Z_p)^{\mathbb{N}}$, which is torsion-free (recall Remark 4.8). Since Z is a CMA group, there is an embedding

\begin{equation*}\varphi:Z\hookrightarrow\underbrace{\{0\}\times\ldots\times\{0\}}_{N\text{times}}\times {\mathbb{T}}\times {\mathbb{T}}\times\ldots.\end{equation*}

As we have already observed in Remark 4.8, every finite abelian group is a quotient of Z. Since Z and $\varphi(Z)$ are isomorphic, there is a surjective continuous homomorphism $\theta:\varphi(Z)\to F$. Now consider the following map:

\begin{equation*}\Phi: F\times \varphi(Z)\to F\times F,\quad \Phi(f,x):=(f,\theta(x)).\end{equation*}

(More precisely, instead of $F\times \varphi(Z)$ we should consider the subgroup of ${\mathbb{T}}^{\mathbb{N}}$ generated by F and $\varphi(Z)$ as in the end of the proof of Lemma 4.9. Here we avoid such rigor for notational simplicity.) Clearly, Φ is a surjective continuous homomorphism. Then for the diagonal subgroup $\tilde{F}=\{(f,f):f \in F\}\leq F\times F$ we have $K:=\Phi^{-1}(\tilde{F})\in\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$. We claim that K is torsion-free and $d_H(K,F) \lt \varepsilon$.

Assume that $(f,x)\in K\setminus\{(0,0)\}$ (with $f\in F$ and $x\in\varphi(Z)$) and $n(f,x)=(nf,nx)=(0,0)$ with some $n\in{\mathbb{N}}$. Since $x\in \varphi(Z)$, and $\varphi(Z)$ is torsion-free, we get x = 0. On the other hand, $(f,x)\in K$ means $\theta(x)=f$, hence we have f = 0. Thus K is torsion-free.

Note that $\pi_{[N]}(K)=\pi_{[N]}(F)$, therefore $d_H(F,K) \lt \varepsilon$. This completes the proof of Claim 2.

Proof. Let

\begin{equation*}\mathcal{B}_3:=\{K \in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}):\ K \text{is totally disconnected} \}.\end{equation*}

We need to show that for the generic $K\in\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$ the singleton $\{0\}$ is a connected component. Notice that whenever $\varphi:K\to F$ is a continuous homomorphism to a finite abelian group, the connected component of 0 is a subset of $Ker\varphi$. Let us define

\begin{equation*}\mathcal{B}_3^{'}:=\bigcap_{n\in{\mathbb{N}}}\underbrace{\left\{K \in \mathcal{S}({\mathbb{T}}^{\mathbb{N}}):\ \begin{array}{ll} \text{there exists a finite abelian group}\,\, F_n\,\, \text{and a continuous} \\ \text{homomorphism}\,\, \varphi_n:K\to F_n\,\, \text{with}\,\, diam(Ker(\varphi_n)) \lt \frac{1}{n} \end{array} \right\}}_{U_n}.\end{equation*}

Now we have $\mathcal{B}_3^{'}\subseteq\mathcal{B}_{3}$, hence it suffices to prove that $\mathcal{B}_3^{'}$ is dense G_δ in $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$. The denseness follows directly from Proposition 4.5 and the fact that every finite group is in $\mathcal{B}_3^{'}$.

Claim. The set U_n is open.

Pick some $K\in U_n$. Fix a finite abelian group F_n and a continuous homomorphism $\varphi_n:K\to F_n$ with $diam(Ker(\varphi_n)) \lt \frac{1}{n}$. As in the proof of Claim 1 of Lemma 4.9, for any given ɛ > 0 if $L\in\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$ is suitably close to K, then φ_n gives rise to a continuous homomorphism $\psi_n:L\rightarrow F_n$ with $Ker(\psi_n)\subseteq \left(Ker(\varphi_n)\right)_\varepsilon$. Then $\varepsilon:=\frac{1}{2}\left(\frac{1}{n}-diam(Ker(\varphi_n))\right)$ witnesses that U_n is open.

Notation 4.13.

Recall that $\mathbb{Z}^{({\mathbb{N}})}$ is the countably infinite direct sum of $\mathbb{Z}$ with itself. Let $\mathcal{A}'$ denote the set of all subgroups of $\mathbb{Z}^{({\mathbb{N}})}$ with the subspace topology inherited from the Cantor space $2^{\mathbb{Z}^{({\mathbb{N}})}}$.

Sketch of proof.

Notation 4.16.

Let H be a fixed locally compact abelian group, and $\widehat{H}$ its Pontryagin dual. If G is a closed subgroup of H, then the annihilator of G is defined as:

\begin{equation*}Ann(G):=\{\chi\in\widehat{H}:\ \forall a \in G\ (\chi(a)=0)\}.\end{equation*}

Notice that Ann(G) is subgroup of $\widehat{H}$.

Recall that any CMA group G can be embedded into ${\mathbb{T}}^{\mathbb{N}}$, hence the annihilator of G is a subgroup of $\mathbb{Z}^{({\mathbb{N}})}$, that is, $Ann(G)\in \mathcal{A}'$. Thus the canonical map, which turns out to be a homeomorphism between $\mathcal{A}'$ and $\mathcal{S}({\mathbb{T}}^{\mathbb{N}})$, is Ann itself.

Notation 4.17.

For $x\in\mathbb{Z}^{({\mathbb{N}})}$ let

\begin{equation*}supp (x):=\{n\in{\mathbb{N}}:\ x(n)\neq 0\}.\end{equation*}

Notation 4.18.

Let e_i be the following element of $\mathbb{Z}^{({\mathbb{N}})}$: $e_i(k)=1$ if k = i and $e_i(k)=0$ otherwise.

Proof. Injectivity. Pick distinct elements $A,B\in \mathcal{A}$. If we use the original notations introduced in the beginning of Section 3.1, this means that there exist $k,l,m\in{\mathbb{N}}$ such that $m=A(k,l)\neq B(k,l)$. By the definition of Φ it follows that $e_k+e_l-e_m\in \Phi(A)$ and $e_k+e_l-e_m\notin\Phi(B)$.

Continuity and openness. Sets of the form

\begin{equation*}U_x=\{A'\in \mathcal{A}': x\in A'\}\quad \text{and }\quad V_x=\{A'\in \mathcal{A}': x\notin A'\}\end{equation*}

with $x\in\mathbb{Z}^{({\mathbb{N}})}$ constitute a subbasis in $\mathcal{A}'$. By the definition of Φ we have

\begin{equation*}\Phi^{-1}(U_x)=\left\{A\in\mathcal A:\;\prod_{n\in\mathbb{N}}n^{x(n)}=1\;\text{in }\widetilde A\right\},\end{equation*}

and

\begin{equation*}\Phi^{-1}(V_x)=\left\{A\in\mathcal A:\;\prod_{n\in\mathbb{N}}n^{x(n)}\neq1\;\text{in }\widetilde A\right\}.\end{equation*}

By Proposition 3.6 these sets are open, hence Φ is continuous. Again, by Proposition 3.6 the set $\{\Phi^{-1}(U_x):\ x\in\mathbb{Z}^{({\mathbb{N}})}\}$ is a subbasis for $\mathcal{A}$. Thus $\Phi(\Phi^{-1}(U_x))= U_x\cap\Phi(\mathcal{A})$ and the injectivity of Φ shows that Φ is open (onto its range).

Proof. Observe that $e_i\notin A'$ if $i\geq 2$ and $A'\in \Phi(\mathcal{A})$, therefore we have $\Phi(\mathcal{A})\subset \bigcap_{i=2}^\infty V_{e_i}$. Then it suffices to prove that $\bigcup_{i=2}^\infty U_{e_i}$ is dense in $\mathcal{A}'$. Pick any nonempty basic clopen set $U:= U_{x_1}\cap\ldots\cap U_{x_n}\cap V_{y_1}\cap\ldots\cap V_{y_k}$ in $\mathcal{A}'$. Let $N\in{\mathbb{N}}$ be such that $N\notin(\bigcup_{i=1}^n supp(x_i))\cup(\bigcup_{j=1}^ksupp(y_j))\cup\{1\}$. Now it is easy to check that $U\cap U_{e_N}\neq\emptyset$.

observation 5.1.

Let $G\in\mathcal{S}(\mathbb{SU})$ and $G_n\in\mathcal{S}(\mathbb{SU})$ for every $n\in{\mathbb{N}}$. Notice that if $\pi_{[n]}(G_n)=\pi_{[n]}(G)$ for every $n\in{\mathbb{N}}$, then $G_n\overset{d_H}{\longrightarrow} G$.

Proof of Lemma 5.2

The trivial group SU(1) has no proper subgroup, hence we may assume $n\geq 2$. Suppose that $(K_i)_{i\in {\mathbb{N}}}\leq SU(n)$ is a sequence of proper compact subgroups such that $K_i\overset{d_{n,H}}{\longrightarrow} SU(n)$. Notice that $(K_i)^\circ$ is closed and therefore, by Theorem 5.3, it is a connected Lie subgroup of SU(n) for every $i\in{\mathbb{N}}$.

Consider the sequence of normalizers $N_i:= N_{SU(n)}((K_i)^\circ)$. By Theorem 5.4 we may assume that the N_i are mutually conjugate. Since $(K_i)^\circ\triangleleft K_i$, we have $K_i\leq N_i\leq SU(n)$. Thus $N_i\overset{d_{n,H}}{\longrightarrow}SU(n)$ as well. On the other hand, d_n is biinvariant, hence the distance $d_H(SU(n), N_i)$ is the same for each $i\in{\mathbb{N}}$. We conclude that $N_i=SU(n)$ for every $i\in{\mathbb{N}}$. In other words, every $(K_i)^\circ$ is normal in SU(n).

The Lie algebra of a connected normal subgroup of SU(n) is an ideal of the Lie algebra $\mathfrak{su}(n)$ (see [Reference Helgason14, page 128]). Since $\mathfrak{su}(n)$ is simple (see [Reference Pfeifer34, page 13]), we conclude that every $(K_i)^\circ$ is the trivial group. Therefore, K_i is finite for every $i\in{\mathbb{N}}$.

By Theorem 5.5 this contradicts the fact that SU(n) is a connected nonabelian group (see [Reference Hall13, Proposition 13.11] for the connectedness).

Question 5.8.

Which compact Lie groups can be approximated by proper subgroups in the Hausdorff metric?

Question 5.9.

Let G be a Lie group. Which compact subgroups K of G can be approximated by other compact subgroups of G that are non-conjugate to K? (It follows easily that in a compact connected Lie group G any nonnormal compact subgroup can be approximated by its conjugates.)

Proof. Fix $n\in{\mathbb{N}}$. By Theorem 5.11 there is a neighbourhood U of the identity of SU(n) that contains no nontrivial subgroups of SU(n). By Theorem 5.12 and the continuity of π_n we can pick a compact open subgroup H of G such that $\pi_n(H)\subseteq U$ and thus $\pi_n(H)=\{1_{SU(n)}\}$. Also notice that the index of H is finite since H is open, therefore $\pi_n(G)$ is finite.

Proof. Let

\begin{equation*}\mathcal{U}:=\{G\in \mathcal{S}(\mathbb{SU}):\ \forall k\in {\mathbb{N}}\ \exists n\geq k \ (\pi_n(G)=SU(n))\}=\end{equation*}

\begin{equation*}=\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty\underbrace{\{G\in\mathcal{S}(\mathbb{SU}):\ \pi_n(G)=SU(n)\}}_{\mathcal{U}_n}.\end{equation*}

It follows from Lemma 2.3 and Lemma 5.2 that each $\mathcal{U}_n$ is open in $\mathcal{S}(\mathbb{SU})$, hence $\mathcal{U}$ is G_δ. Thus it suffices to prove that $\mathcal{U}$ is dense. Pick any group $G\in \mathcal{S}(\mathbb{SU})$. Then, with a slight abuse of notation,

\begin{equation*}(\pi_{[m]}(G)\times SU(m+1)\times SU(m+2)\times\ldots)\overset{d_H}{\longrightarrow} G\end{equation*}

shows that G can be approximated by elements of $\mathcal{U}$.

Proof. We have shown in the proof of Theorem 5.14 that for any $k\in{\mathbb{N}}$ the set $\bigcup_{n=k}^\infty\mathcal{U}_n$ is dense open. On the other hand, by Lemma 5.13 it does not contain totally disconnected groups for $k\geq 2$.

Proof. Since $G/G^\circ$ is a totally disconnected metrizable (see [Reference Haagerup and Przybyszewska12, Theorem 4.5]) compact group, it suffices to show that for the generic $G\in\mathcal{S}(\mathbb{SU})$ it is not finite. Let n be a fixed positive integer. Let

\begin{equation*}\mathcal{U}_n:=\{G\in\mathcal{S}(\mathbb{SU}):\ |G:G^\circ|\geq n \}.\end{equation*}

We will show that $\mathcal{U}_n$ is open and dense, which completes the proof.

Claim 1. The set $\mathcal{U}_n$ is dense.

Note that SU(m) contains the cyclic group C_m for each $m\in{\mathbb{N}}$ (up to isomorphism). Also for any $G\in\mathcal{S}(\mathbb{SU})$ we have, with an abuse of notation,

\begin{equation*}G_m:=\pi_{[m]}(G)\times C_{m+1}\times \{1\}\times \{1\}\times\ldots \overset{d_H}{\to} G,\end{equation*}

by Observation 5.1. Here G_m has at least m + 1 connected components, hence $\mathcal{U}_n$ is dense.

Claim 2. The set $\mathcal{U}_n$ is open.

Pick any group $G\in\mathcal{U}_n$. First we show that G can be partitioned into finitely many but at least n clopen cosets. It follows from Theorem 5.12 that the group $G/G^\circ$ can be partitioned into finitely many but at least n clopen cosets. Since the natural homomorphism $G\to G/G^\circ$ is continuous, the inverse images of these clopen cosets form a suitable partition of G.

Now let δ be the minimal distance occurring between these clopen cosets of G. Then it is easy to see that any $L\in\mathcal{S}(\mathbb{SU})$ with $d_H(L,K) \lt \frac{\delta}{2}$ has at least n connected components.

Proof. By [Reference Hofmann and Morris20, Corollary 10.38.] for every compact group G the group $G^\circ\times G/G^\circ$ is homeomorphic to G, hence the corollary follows from Theorem 5.17.

Proof. Let $\varphi:G/G^\circ\to H/\Phi(G^\circ)$ be the natural continuous surjection, that is, $\varphi(gG^\circ):=\Phi(g)\Phi(G^\circ)$. We need to prove that $H/\Phi(G^\circ)$ is trivial.

Suppose that there is an open neighbourhood V of the identity of $H/\Phi(G^\circ)$ such that $V\neq H/\Phi(G^\circ)$. By Theorem 5.12 and the continuity of φ there is a clopen subgroup $U\leq G/G^\circ$ such that $\varphi(U)\subseteq V$.

Recall that φ is an open map because it is a surjective continuous homomorphism between compact groups. Therefore, $\varphi(U)\neq H/\Phi(G^\circ)$ is clopen, which contradicts the connectedness of $H/\Phi(G^\circ)$.

Proof. Fix $n\in{\mathbb{N}}$ arbitrarily. It suffices to prove that for the generic $G\in\mathcal{S}(\mathbb{SU})$ its identity component G ^∘ is not an n-dimensional Lie group. By Corollary 5.21, for the generic $G\in\mathcal{S}(\mathbb{SU})$ we have $\pi_k(G^\circ)=SU(k)$ for infinitely many $k\in{\mathbb{N}}$.

Suppose that G ^∘ is an n-dimensional Lie group, and fix some k > n with $\pi_k(G^\circ)=SU(k)$. It is well-known that any continuous homomorphism between Lie groups is smooth. (See [Reference Zuoqin41] or [Reference Hall13, Corollary 3.50.]. Although this corollary is stated for matrix Lie groups, the proof works for any Lie group.) Thus $\pi_k|_{G^\circ}:G^\circ\rightarrow SU(k)$ is smooth. It is also surjective, therefore, by Sard’s lemma [Reference Sard36, Theorem 4.1], the differential of $\pi_k|_{G^\circ}$ is surjective at some point, which contradicts k > n.

Proof. Follows immediately from Corollary 5.19 and Theorem 5.22.

Proof. Let G be a compact connected torsion-free group and let $\{S_i:\ i\in I\}$ be the family of simple connected compact Lie groups provided by Proposition 5.24. Also let $\varphi:S:= \prod_{i\in I}S_i\to [G,G]$ denote the surjective continuous homomorphism provided by Proposition 5.24.

If $[G,G]=1$, then G is abelian and we are done. Otherwise, I is nonempty and we have $Ker\varphi\subseteq Z(S)=\prod_{i\in I}Z(S_i) \lt S$ by Proposition 5.24. We claim that S contains a non-central torsion element. Pick $j\in I$ such that $S_j\setminus Z(S_j)\neq\emptyset$. Clearly, it suffices to prove that $S_j\setminus Z(S_j)$ contains a torsion element. We may assume that S_j is infinite. Then it contains a nontrivial torus (recall that a every infinite compact Lie group contains a nontrivial torus [Reference Hofmann and Morris20, Lemma 6.20.]) and thereby infinitely many torsion elements, while $Z(S_j)$ is finite. Now it follows that G has a nontrivial torsion element, a contradiction.

Proof. Suppose that SU(n) occurs as a quotient of a compact torsion-free group G. By Lemma 5.20 we may assume that G is connected. Then by Proposition 5.25 it must be abelian. Since SU(n) is nonabelian for $n\geq 2$, this is a contradiction.

Proof. By Theorem 5.14 for the generic $G\in\mathcal{S}(\mathbb{SU})$ some SU(k) is a quotient of G with $k\geq 2$. Since SU(k) contains a circle group, it contains elements of infinite order, and so does G. On the other hand, by Lemma 5.26, G cannot be a torsion-free group either.

Notation 5.28.

For a group G and a number $m\in{\mathbb{N}}$ let

\begin{equation*}F(m,G):=\{(g_1,...,g_m)\in G^m:\;g_1,...,g_m\;\text{freely generate a free group in }G\}.\end{equation*}

For any set A let $(A)^m:=\{(x_1,\ldots,x_m)\in A^m:\ x_i\neq x_j \text{if }i\neq j\}$.

Proof. Fix $m \in {\mathbb{N}}$ and a nonempty reduced word $W(x_1,...,x_m)$ in m variables arbitrarily. Consider the analytic map

\begin{equation*} f_W:SU(n)^n \to SU(n), \quad (g_1,...,g_m) \mapsto W(g_1,...,g_m), \end{equation*}

from a connected analytic manifold to an analytic manifold (see [Reference Epstein7]).

Claim. The compact set $f_W^{-1}(id_{SU(n)})$ has empty interior, hence it is nowhere dense.

Suppose $int(f_W^{-1}(id_{SU(n)}))\neq\emptyset$. Then the analytic map f_W is constant on a nonempty open set. Therefore, it is constant on $SU(n)^m$ (see, for example, [Reference Epstein7]). On the other hand, it is well-known (and also follows from [Reference Epstein7]) that SO(3) contains a free group of rank m, hence SU(n) also contains a free group of rank m. This contradicts that f_W is constant on $SU(n)^m$, which proves the claim.

Notice that we can write

\begin{equation*}F(m,SU(n))=\bigcap_{{}\;W\;\;\textit{is a nonempty reduced}\\\textit{word in }m\;\textit{variables}}(SU(n)^m\setminus f_W^{-1}(id_{SU(n)})).\end{equation*}

By the claim, the right-hand side is a countable intersection of dense open sets, which concludes the proof.

Proof. By Theorem 5.14, for the generic $G\in\mathcal{S}(\mathbb{SU})$ there is $n\geq 3$ such that $\pi_n(G)=SU(n)$. Fix such G and n. We claim that the generic $K\in\mathcal{K}(G)$ freely generates a free group of rank continuum in G.

Recall that by Theorem 2.6, the map $\pi_n:G\to SU(n)$ is continuous and open. It follows that the map $\pi_n^m: G^m\to SU(n)^m$, $(g_1,\ldots,g_m)\mapsto (\pi_n(g_1),\ldots,\pi_n(g_m))$ is also continuous and open for every $m\in{\mathbb{N}}$. Then, by Lemma 5.29 and [Reference Melleray27, Proposition 2.8], the inverse image $B_m:=(\pi_n^m)^{-1}(F(m,SU(n)))$ is comeager in G^m for every $m\in{\mathbb{N}}$. Observe that for every $m\in{\mathbb{N}}$ and $(g_1,\ldots,g_m)\in B_m$ the elements $g_1,\ldots,g_m$ freely generate a free group in G because, since π_n is a homomorphism, a nontrivial relation of the g_i in G would give us a nontrivial relation of the $\pi_n(g_i)$ in SU(n).

Now by Theorem 5.30, for the generic $K\in\mathcal{K}(G)$ for every $m\in{\mathbb{N}}$ we have $(K)^m\subseteq B_m$. Therefore, the generic $K\in\mathcal{K}(G)$ freely generates a free group in G. Furthermore, by [Reference Kechris21, (8.8) Exercise i)], the generic $K\in\mathcal{K}(G)$ has continuum many elements, which concludes the proof.