Products and extensions of finite simple groups

September 26, 2026

Products and extensions of finite simple groups

In a previous post on finite simple groups I mentioned that one motivation for understanding these was various classifying results, and alluded to this in another post on finitely generated profinite groups. In this post I would like to discuss a couple of results of the flavour any set of finite simple groups can appear as the answer to a natural classification question. Much of this is taken from a survey of Segal and the two papers that cover the results I mention.

Upper composition factors

The first is the following. Recall that an upper composition factor of a group means a composition factor of some finite image of the group, and a group is just-infinite if every proper quotient group is finite. Segal proved the following, illustrating that the set of upper composition factors can be very wild:

Theorem 1: Let $\mathcal{P}$ be any non-empty collection of non-abelian finite simple groups. Then there exists a 63-generator just-infinite group whose upper composition factors comprise exactly the set $\mathcal{P}$.

If we are concerned only with the non-abelian upper composition factors, then in fact three generators will suffice. The proof strategy will be to build some very large group out of the desired set of composition factors and find some finitely generated subgroup which 'remembers' the ambient group, in the same way that lattices in higher rank semisimple Lie groups 'remember' their ambient group (by Margulis' superrigidity theorem).

We begin with a sequence of finite permutation groups $ S_n $ ($ n \geq 0 $), each of which satisfies the following hypothesis, which we call hypothesis H:

\[
S_n \text{ is transitive and distinct points have distinct stabilisers in } S_n.\]

Note that every finite non-nilpotent group $ S_n $ has a non-trivial permutation representation with this property, namely its action by right multiplication on the right cosets of a non-normal maximal subgroup. Now put $ W_0 = S_0 $, for $ n \geq 1 $ let $ W_n $ be the permutational wreath product

\[
W_n = S_n \wr W_{n-1},
\]

and form the inverse limit $ W = \varprojlim W_n $. Thus $ W $ is a profinite group, whose finite (continuous) images are just the images of the groups $ W_n $. Let $ K_n $ denote the kernel of the natural map of $ W $ onto $ W_n $. Then the family $\{K_n: n \geq 1\}$ is a base for the neighbourhoods of 1 in $ W $, and a subgroup $ G $ of $ W $ is dense if and only if $ K_nG = W $ for every $ n $. One says that $ G $ has the congruence subgroup property (CSP) if for each normal subgroup $ N $ of finite index in $ G $ there exists $ n $ such that $ K_n \cap G \leq N $. I shall say that $ G $ has the weak CSP if for each normal subgroup $ N $ of finite index in $ G $ there exists $ n $ such that $(K_n \cap G)' \leq N$, where $(K_n \cap G)'$ denotes the derived group of $ K_n \cap G $. Thus if $ G $ is dense and has the CSP then the finite images of $ G $ are exactly the images of the various groups $ W_n $, and the profinite completion $\hat{G}$ of $ G $ is isomorphic to $ W $; while if $ G $ is dense and merely has the weak CSP, then each finite image of $ G $ is an extension of some abelian normal subgroup by some image of one of the $ W_n $, and the natural map from $\hat{G}$ onto $ W $ has abelian kernel.

We will use branch groups, to be defined in the next section, to prove the following theorem:

Theorem 2: Suppose that each of the permutation groups $ S_n $ is non-abelian, simple and satisfies (H). Then $ W $ contains a dense 3-generator subgroup that has the weak congruence subgroup property.

Branch groups

Throughout, we keep fixed a sequence $(l_n)_{n \geq 0}$ of integers greater than 3. The spherically homogeneous rooted tree of type $(l_n)$ is a tree $T$ with a distinguished vertex $v_0$ (the root) of valency $l_0$, such that every vertex at distance $n \geq 1$ from $v_0$ has valency $1 + l_n$.

The distance from $v_0$ to a vertex $v$ is called the level of $v$. If we picture the tree as growing downwards, we see that each vertex $v$ of level $n \geq 1$ hangs from a unique vertex of level $n - 1$, and has hanging from it a spherically homogeneous rooted tree $T_v$ of type $(l_n, l_{n+1}, \dots)$.

Let $T[n]$ denote the finite rooted subtree, with root $v_0$, whose vertices are all the vertices of $T$ of level at most $n$. It is evident that the automorphism group of $T[n]$ (automorphisms fixing the root) is the iterated permutational wreath product

\[
\operatorname{Aut}(T[n]) = \operatorname{Sym}(l_{n-1}) \wr \cdots \wr \operatorname{Sym}(l_0),
\]

and that $\operatorname{Aut}(T)$ is the inverse limit of these as $n \to \infty$.

Let $T_n$ be the spherically homogeneous rooted tree of type $(l_n, l_{n+1}, \dots)$. We fix an identification of $T_n$ with each of the rooted subtrees $T_v$ where $v$ runs over the vertices $v_{n,1}, \dots, v_{n,m_n}$ of level $n$. Then $\operatorname{Aut}(T_v) = \operatorname{Aut}(T_n)$ for each such $v$, and we can specify an automorphism $x$ of $T$ that fixes all vertices of level $n$ by writing

\[
x = (x_1, \dots, x_{m_n}),
\]

where $x_j \in \operatorname{Aut}(T_n)$ is the restriction of $x$ to the subtree $T_{v_{n,j}}$, identified with $T_n$, and $j$ runs from 1 to $m_n = l_0 l_1 \cdots l_{n-1}$.

For $\alpha \in \operatorname{Sym}(l_0)$, I denote by $\dot{\alpha}$ the automorphism of $T$ that permutes the vertices of level 1 according to $\alpha$, but preserves the identifications of all the subtrees $T_{v_{n,j}}$ with $T_1$; thus $\alpha \mapsto \dot{\alpha}$ identifies $\operatorname{Sym}(l_0)$ with the `top group' in the wreath product $\operatorname{Aut}(T) = \cdots \wr \operatorname{Sym}(l_1) \wr \operatorname{Sym}(l_0)$. Each automorphism of $T$ can then be written uniquely in the form

\[
x = (x_1, \dots, x_{l_0}) \cdot \dot{\alpha},
\]

where $\alpha \in \operatorname{Sym}(l_0)$ and $x_j \in \operatorname{Aut}(T_1)$ for each $j$.

For any subgroup $G$ of $\operatorname{Aut}(T)$ and $n \geq 0$, the kernel of the action of $G$ on $T[n]$ is denoted $\operatorname{St}_G(n)$. The subgroup of $G$ consisting of automorphisms that fix every vertex of $T$ outside a given subtree $T_v$ is denoted $\operatorname{top}_G(v)$. It is clear that

\[
\operatorname{St}_G(n) \cong \operatorname{top}_G(v_{n,1}) \times \cdots \times \operatorname{top}_G(v_{n,m_n}) := \operatorname{rist}_G(n).
\]

$\operatorname{rist}_G(n)$ the restricted nth level stabiliser; $G$ is a branch group if

\begin{enumerate}
\item[(i)] $G$ acts transitively on the vertices of level $n$, for each $n$, and
\item[(ii)] $\operatorname{rist}_G(n)$ has finite index in $G$ for each $n$.
\end{enumerate}

In many settings (but not our current one), we take the tree to be regular: every vertex has the same fixed number of descendants. The most famous branch group is Grigorchuk's group, the first example of a finitely generated group with either the property that every element is finite order (in fact a power of 2), or that the growth is superpolynomial but subexponential. It has inspired a huge amount of research in this direction. It would be fair to say this and related constructions were the original motivation for considering branch groups, but it turns out this idea is good for much more as we shall see.

For each $n$ let $S_n$ be a $d$-generator subgroup of $\operatorname{Sym}(l_n)$ that satisfies hypothesis (H). For each $n \geq 1$ we fix a generating set $\{\alpha(1)_n, \dots, \alpha(d)_n\}$ for $S_n$, and specify the following automorphisms of $T$:

\[
b(i) = b(i)_0 = (\alpha(i)_1, 1, \dots, 1, b(i)_1) \cdot \dot{\alpha(i)_1} \quad (i = 1, \dots, d)
\]

where

\[
b(i)_1 = (\alpha(i)_2, 1, \dots, 1, b(i)_2) \cdot \dot{\alpha(i)_2} \in \operatorname{Aut}(T_1)
\]

and in general

\[
b(i)_n = (\alpha(i)_{n+1}, 1, \ldots, 1, b(i)_{n+1}) \cdot \dot{\alpha(i)_{n+1}} \in \operatorname{Aut}(T_n)
\]

(with 1 in each unlabelled position). Each $ b(i) $ is a directed automorphism of $ T $. By a direct calculation, Segal proves the following lemmata:

Lemma 3: Let

\[
A = \langle S_0, b(1), \ldots, b(d) \rangle \leq \operatorname{Aut}(T).
\]

Suppose that $ S_1 $ is perfect. Then $ \operatorname{St}_A(1) = \operatorname{rist}_A(1) $ and, for each vertex $ v $ of level 1,

\[
\operatorname{top}_A(v)|_{T_v} = \langle S_1, b(1)_1, \ldots, b(d)_1 \rangle \leq \operatorname{Aut}(T_1).
\]

Lemma 4: Assume that $ d = 2 $ and that each of the groups $ S_n $ is perfect. Let

\[
B = \langle \dot{S}_0, c \rangle \leq \operatorname{Aut}(T).
\]

If $ S_1 $ is simple then $ \operatorname{St}_B(1) = \operatorname{rist}_B(1) $ and, for each vertex $ v $ of level 1,

\[
\operatorname{top}_B(v)|_{T_v} = \langle \dot{S}_1, c_1 \rangle \leq \operatorname{Aut}(T_1).
\]

Lemma 5: Let $ G $ be a subgroup of $ \operatorname{Aut}(T) $ that acts transitively on the vertices of each level. If $ 1 \neq N \triangleleft G $ then $ N \geq \operatorname{rist}_G(n)' $ for some $ n $.

Segal also gives a slick proof of this last lemma in the case of a finite index subgroup $N$, which is the case we need: $ G/N $ has only a finite number, say $ k $, of distinct subgroups; if $ n $ is so large that $ m_n > k $ then there exist distinct vertices $ u $ and $ v $ of level $ n $ such that

\[
N \operatorname{top}_G(u) = N \operatorname{top}_G(v).
\]

As $ \operatorname{top}_G(u) $ and $ \operatorname{top}_G(v) $ commute elementwise, it follows that

\[
\operatorname{top}_G(v)' \leq [N \operatorname{top}_G(u), N \operatorname{top}_G(v)] \leq N. \blacksquare
\]

We are now ready to prove Theorems 1 and 2. Recall that $ W_n = S_n \wr \cdots \wr S_0 $, which we identify with a subgroup of $ \operatorname{Aut}(T[n+1]) = \operatorname{Sym}(l_n) \wr \cdots \wr \operatorname{Sym}(l_0) $. Thus $ W = \varprojlim_{n \to \infty} W_n \leq \operatorname{Aut}(T) $, and $ K_n = \operatorname{St}_W(n+1) $ is the kernel of the natural map $ W \to W_n $.

Proof of Theorem 2: We first recall that every finite simple group can be generated by two elements, so we may take $ d = 2 $. Thus $ B $ is a 3-generator group. Now Lemma 4 shows that $ B = B_1 \wr S_0 $, where $ B_1 $ is the subgroup of $ \operatorname{Aut}(T_1) $ defined in the same way as $ B $ was in $ \operatorname{Aut}(T) $. It follows inductively that for each $ n $ we have (in the analogous notation)

\[
B = B_{n+1} \wr S_n \wr \cdots \wr S_0,
\]

hence $ B $ induces on $ T[n+1] $ the automorphism group $ S_n \wr \cdots \wr S_0 = W_n $. It follows that $ B $ is a dense subgroup of $ W $, and that $ B $ acts transitively on the vertices of each level.

Repeated applications of Lemma 4 show that $ \operatorname{St}_B(n) = \operatorname{rist}_B(n) $ for each $ n $; hence Lemma 5 implies that every non-trivial normal subgroup of $ B $ contains $ \operatorname{St}_B(n)' $ for some $ n $. As $ B $ is infinite and

\[
\operatorname{St}_B(n) = B \cap \operatorname{St}_W(n) = B \cap K_{n-1},
\]

this implies that $ B $ has the weak congruence subgroup property, and completes the proof. $\blacksquare$

In order to construct a dense finitely generated subgroup of $ W $ with the (full) CSP, we need to postulate a certain uniformity in the presentations of the groups $ S_n $. Recall that a group is said to be perfect if it is equal to its derived group.

Theorem 6: Suppose that each of the permutation groups $ S_n $ satisfies (H). Assume further that there exists a $ d $-generator perfect group that has $ S_n $ as an epimorphic image for every $ n \geq 1 $, and that $ S_0 $ is an $ r $-generator group. Then $ W $ contains a dense $(d+r)$-generator just-infinite subgroup that has the congruence subgroup property.

Proof: We are given now a perfect $ d $-generator group $ P = \langle X_1, \ldots, X_d \rangle $ and for each $ n \geq 1 $ an epimorphism $ \pi_n: P \to S_n $. For $ S_n $ we take the generating set $ \{ \alpha(i)_n = \pi_n(X_i): i = 1, \ldots, d \} $, and form the corresponding group $ A $ as in Lemma 1. If $ w(X_1, \ldots, X_d) = 1 $ is a relation in $ P $ then for each $ n $ we have $ w(b(1)_n, \ldots, b(d)_n) = 1 $ in $ A $, so $ X_i \mapsto b(i)_n $ ($ i = 1, \ldots, d $) defines a homomorphism $ \pi(n): P \to A $; it follows that the subgroup $ \pi(n)(P) = \langle b(1)_n, \ldots, b(d)_n \rangle $ is perfect.

Now repeated applications of Lemma 3 show that

\[
\operatorname{top}_A(v)|_{T_v} = \langle S_n, b(1)_n, \ldots, b(d)_n \rangle = \langle S_n, \pi(n)(P) \rangle
\]

for each vertex $ v $ of level $ n $, and hence that $ \operatorname{rist}_A(n) $ is a perfect group. It follows as in the proof of theorem 2 (using Lemma 3 in place of Lemma 4) that $ A $ is dense in $ W $ and that every non-identity normal subgroup of $ A $ contains $ \operatorname{St}_A(n) = A \cap K_{n-1} $ for some $ n $; and $ A = \langle S_0, b(1), \ldots, b(d) \rangle $ can be generated by $ d + r $ elements if $ S_0 $ is generated by $ r $ elements. $\blacksquare$

Thus every profinite group $ W $ of this form is the profinite completion of a finitely generated just-infinite group. For example, we can take $ S_n = \operatorname{PSL}(2, p_n) $, where $(p_n)$ is any sequence of primes greater than or equal to 5. It turns out that each $ S_n $ is an image of the perfect 2-generator group $ \operatorname{SL}(2, \mathbb{Z}[1/6]) $. This gives the following.

Corollary 7: Let $(p_n)$ be any sequence of primes exceeding 3. Then there exists a 4-generator group $ G $ such that

\[
\hat{G} \cong W = \varprojlim_{n} W_n
\]

where

\[
W_n = \operatorname{PSL}(2, p_n) \wr \operatorname{PSL}(2, p_{n-1}) \wr \cdots \wr \operatorname{PSL}(2, p_0),
\]

the permutational wreath product using the natural permutation representation of each $ \operatorname{PSL}(2, p) $ on the points of the projective line over $ \mathbb{F}_p $.

To deduce Theorem 1 from Theorem 6, Segal directly constructs a 61-generator perfect group that has every non-abelian finite simple group as a quotient.

Kassabov--Nikolov

The second result concerns the following stronger question: what does it mean for a family of finite groups $\mathcal{X}$ to be precisely the set $\mathcal{F}(\Gamma)$ of (isomorphism types of) all finite quotients of some finitely generated group $\Gamma$? Equivalently, what does it mean for a profinite group $G$ to be the profinite completion of a finitely generated (abstract) group? This holds if and only if $G$ contains a dense finitely generated subgroup $\Gamma$ that has the congruence subgroup property; so the question may be seen as finding necessary and/or sufficient conditions on a profinite group $G$, expressed in terms of the family $\mathcal{F}(G)$, for the existence of such a subgroup (when $G$ is a profinite group, $\mathcal{F}(G)$ denotes the set of continuous finite quotient groups of $G$).

Two obvious necessary conditions for such a family $\mathcal{X}$ are

that $\mathcal{X}$ is quotient-closed, and
that all the groups in $\mathcal{X}$ can be generated by some bounded number of elements;

but it seems very difficult to find further, less obvious ones. Applying a deep `strong approximation' theorem due to Nori and Weisfeiler, Lubotzky established the following important result:

Theorem 8: (`Lubotzky alternative'): Let $\Gamma$ be a finitely generated linear group over a field of characteristic zero. Then one of the following holds:

$\Gamma$ is virtually soluble;
there exist a connected, simply connected simple algebraic group $\mathfrak{G}$ over $\mathbb{Q}$, a finite set of primes $S$ such that $\mathfrak{G}(\mathbb{Z}_S)$ is infinite, and a subgroup $\Gamma_1$ of finite index in $\Gamma$ such that the profinite group $\mathfrak{G}(\mathbb{Z}_S)$ is an image of $\bar{\Gamma}_1$.

(Here $\mathbb{Z}_S = \mathbb{Z}[\frac{1}{p}; p \in S]$, and $\mathfrak{G}(\mathbb{Z}_S)$ is isomorphic to the product $\prod_{p \notin S} \mathfrak{G}(\mathbb{Z}_p)$.)

Suppose for example that $\mathcal{X}$ contains a subgroup $X_i$ of $\operatorname{GL}_d(F_i)$ for $i = 1, 2, \ldots$ where $F_i$ is a finite field of characteristic $p_i$ and $p_1, p_2, \ldots$ is an infinite sequence of distinct primes. Then $\Gamma$ has a quotient $\bar{\Gamma}$ which satisfies the hypotheses of Theorem 21, so $\bar{\Gamma}$ is a finitely generated characteristic-zero linear group; if we assume also that the groups $X_i$ are simple and of unbounded orders (or some suitable weaker condition), we find that $\bar{\Gamma}$ is not virtually soluble. Applying the Lubotzky alternative to the group $\bar{\Gamma}$, we may deduce that the set $\mathcal{X}$ must contain many other groups in addition to the $X_i$: for each prime $ p \notin S $ and each $ n $, a group $ Q_p $ containing $ \mathfrak{G}(\mathbb{Z}/p^n\mathbb{Z}) $ as a subgroup, the indices $ |Q_p : \mathfrak{G}(\mathbb{Z}/p^n\mathbb{Z})| $ being bounded above by a constant.

Thus if $ P $ is an infinite set of primes, a set of groups like

\[
\left\{ \prod_{p \in T} \operatorname{PSL}_d(\mathbb{F}_p) \mid T \text{ a finite subset of } P \right\}
\]

cannot be the whole of $ \mathcal{F}(\Gamma) $ for a finitely generated group $ \Gamma $, while of course it is equal to $ \mathcal{F}(G) $ where $ G = \prod_{p \in P} \operatorname{PSL}_d(\mathbb{F}_p) $. Thus the 2-generator profinite group $ G $ cannot be the profinite completion of a finitely generated group.

The problem with this group $ G $ is that the finite simple factor groups have bounded ranks. In an amazing feat of ingenuity, Kassabov and Nikolov have shown that this is essentially the only obstacle, when it comes to products of finite simple groups. For a group $ S $ they write $ l(S) $ to denote the largest integer $ k $ such that $ S $ contains a copy of the alternating group $ \operatorname{Alt}(k) $, and they prove

Theorem 9: Let $(S_n)$ be a sequence of finite simple groups such that $ l(S_n) \to \infty $ as $ n \to \infty $, and let

\[
G = \prod_{n=1}^\infty S_n.
\]

If $ G $ is finitely generated (as a profinite group), then $ G $ is the profinite completion of a finitely generated group.

I mentioned earlier that one hopes the discrete group will 'remember' the ambient group. For the proof of theorem 9, Kassabov--Nikolov turn this on its head and try to find an ambient group which looks like it is built out of the discrete group we want. Let $ G = \prod_{n=1}^\infty S_n $ be a Cartesian product of finite groups.

Definition: A finitely generated subgroup $ G < \mathcal{G} $ is a \emph{frame} for $ \mathcal{G} $ if the following hold:

(a): $ G $ contains $ \bigoplus_{n=1}^\infty S_n $.
(b): The natural surjection $ \hat{G} \to \mathcal{G} $ is an isomorphism.

One can think of condition (a) as saying that $ G $ is a good approximation of $ \mathcal{G} $ from `within' while condition (b) says that $ G $ approximates very well $ \mathcal{G} $ from `above'. More precisely, for a finite subset $ V \subset \mathbb{N} $ of integers define the $ V $-principal congruence subgroup $ G_V $ to be the kernel of the projection of $ G $ onto $ \prod_{n \in V} S_n $. Let $ G(V) $ be the projection of $ G $ onto $ \mathcal{S}(V) := \prod_{n \notin V} S_n $. The $ m $-th principal congruence subgroup $ G_m $ is just $ G_{\{1,\ldots,m\}} $ and $ G(m) := G(\{1,\ldots,m\}) $.

Part (a) of the above definition is now equivalent to

\[
G = \left( \prod_{n \in V} S_n \right) \times G_V, \quad G_V = G \cap \mathcal{S}(V).
\]

Therefore the congruence subgroup $ G_V $ can be identified with the projection $ G(V) $.

On the other hand, part (b) says that the profinite topology of $ G $ is the same as its congruence topology: Every subgroup of finite index in $ G $ contains a congruence subgroup $ G_m $ for some $ m \in \mathbb{N} $. It isn't obvious how to find a frame, but Kassabov--Nikolov show that one frame begets many:

Lemma 10: Let $ A_n, M_n, B_n $, ($n = 1, 2, \ldots$) be finite groups such that $ M_n = A_n \ltimes B_n $. For each $ n $ let $ b_{n,1}, \ldots, b_{n,k} $ be elements in $ B_n $, such that $ M_n $ is generated by $ A_n $ and $[A_n, b_{n,s}]$ ($1 \leq s \leq k$).

Suppose that $ X = \langle x_1, \ldots, x_m \rangle $ is a frame subgroup of the product $\prod_{n=1}^\infty A_n$. Then $ X $ can be considered as a subgroup of $ \mathfrak{M} := \prod_{n=1}^\infty M_n $ in the natural way. Define $ b_s = (b_{n,s})_n $ for $ s = 1, \ldots, k $. Then the group

\[
Z = \langle X, [x_j, b_s] \mid 1 \leq j \leq m, \, 1 \leq s \leq k \rangle < \mathfrak{M}
\]

is a frame in $ \mathfrak{M} $.

The question of how to find a frame at all remains to be answered. Kassabov--Nikolov manage to get the ball rolling by proving

Theorem 11: For every odd prime $ p $, there exists a 10-generated group $ G_1 $ which is a frame for the Cartesian product

\[
\prod_{n=3}^\infty \operatorname{Alt}(u_{n,p}),
\]

where $ u_{n,p} = (p^{3n} - 1)(p - 1)^{-1} $.

I think this is the most interesting subresult they prove by far. It is a very clever proof involving calculations in rings of matrices that I confess I do not understand at all despite following it line by line. I would say this is where a lot of the ingenuity lies, but unfortunately I will have to leave the interested reader to consult the original paper for the proof. We will move on with the proof of their main result by quoting some

Facts from the theory of finite simple groups

Theorem 12: For every $ m \geq 5 $ there is an integer $ r = r(m) $ such that if $ S $ is a finite simple group with $ l(S) > r $ then $ S $ is generated by two subgroups isomorphic to $ \operatorname{Alt}(m) $.

Proposition 13: For any finite simple group $ S $ and any integers $ c, m \in \mathbb{N} $ such that $ m \leq |\operatorname{Aut}(S)|^c $ there exist $ c+1 $ embeddings

\[
f_i : S \to S^m \quad (i = 1, \dots, c+1)
\]

such that $ S^m = \langle f_1(S), \dots, f_{c+1}(S) \rangle $.

Proposition 14: For every $ c \in \mathbb{N} $, every finite simple group $ S $ and $ m > |S|^c $, the direct product $ D = S^m $ is not generated by $ c $ elements.

Proposition 15: Given any sequence $\{S_n\}$ of distinct finite simple groups of essentially infinite rank, there is a 22-generated group $G_2$ which is frame for the Cartesian product

\[
\mathcal{G} = \prod_{n=1}^\infty S_n.
\]

Since there should be at least one proof somewhere in this section, here's a proof of proposition 15:

Proof: Without loss of generality we may assume that $S_n$ are numbered so that $l(S_1) \leq l(S_2) \leq \ldots$.

Assume first that $l(S_1) \geq r(3^9)$, where $r(m)$ is the number from Theorem 12.

Define a function $h : \mathbb{N} \to \mathbb{N}$ inductively by

\[
h(1) = 1,
\]
$ h(k) $ is the smallest $ n > h(k-1) $ such that $ l(S_n) \geq r(u_k) $, where $ u_k := (3^{3(k+2)} - 1)/2 = u_{k+2,3} $.

The existence of such $ h(k) $ follows from the fact that $ l(S_i) \to \infty $.
Set
\[
B_k = \prod_{h(k) \leq n < h(k+1)} S_n.
\]

Then $ \mathfrak{G} = \prod_{k=1}^{\infty} B_k $.
For every $ n \in [h(k), h(k+1)) $ we have that $ l(S_n) \geq r(u_k) $ and therefore by Theorem 12 the group $ S_n $ is generated by two copies of $ \operatorname{Alt}(u_k) $. In other words we have two embeddings

\[
f_{n,j} : \operatorname{Alt}(u_k) \to S_n, \quad j = 1, 2,
\]

such that $ S_n = \langle f_{n,1}(\operatorname{Alt}(u_k)), f_{n,2}(\operatorname{Alt}(u_k)) \rangle $.
Now $ B_k $ is generated by two copies of $ \operatorname{Alt}(u_k) $ as follows:
For $ j = 1, 2 $ define $ P_{k,j} $ to be the image of

\[
\operatorname{Alt}(u_k) \ni a \mapsto (f_{n,j}(a))_n \in B_k, \quad (h(k) \leq n < h(k+1)).
\]

Then $ \{P_{k,1}, P_{k,2}\} $ is the whole of $ B_k $ because it is a subdirect product of distinct simple groups.

Now by Theorem 11 there are two embeddings, $ t_j $ ($ j = 1, 2 $) of $ G_1 $ into $ \prod_{k=1}^{\infty} P_{k,j} $ such that the images $ t_j(G_1) $ are frame subgroups. Lemma 10 gives that these two copies of $ G_1 $ embedded in $ \prod_{k=1}^{\infty} B_k = \mathfrak{G} $ via the $ t_j $ generate a frame subgroup $ G_2 $ for $ \mathfrak{G} $.

In general we can write $ \mathfrak{G} $ as $ \mathfrak{G} = K \times \mathfrak{G}' $ where $ K $ is a finite product of distinct simple groups and $ \mathfrak{G}' $ is a Cartesian product of simple groups $ S $ with $ l(S) > r(3^9) $. Taking a 20-generated frame $ G' $ in $ \mathfrak{G}' $ together with 2 generators $ a, b \in K $ gives the frame $ G = \langle a, b, G' \rangle $ in $ \mathfrak{G} $. \hfill $\square$

Proof of Kassabov--Nikolov: Suppose $ \mathfrak{G} = \prod_n S_n^{f(n)} $ is topologically finitely generated.
By Proposition 15 there is a 22-generated group $ G_2 $ which is a frame in
$ \prod_n S_n $. Since $ \mathfrak{G} $ is assumed to be finitely generated, by Proposition 14 there is
$ c \in \mathbb{N} $ such that

\[
f(n) < |S_n|^c \leq |\operatorname{Aut}(S_n)|^c, \quad \text{for all } n.
\]

Now by Proposition 13, for each $ n = 1, 2, \ldots $ we find $ (c+1) $ copies of $ S_n $ inside
the product $ S_n^{f(n)} $, which together generate it. By an application of Lemma 8 we deduce that there exists a frame subgroup $ \Gamma < \mathfrak{G} $ generated by $ (c+1) $ copies of $ G_2 $, and Theorem 4 is proved. $\blacksquare$

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