Projective Limits and Ultraproducts of Nonabelian Finite Groups

Since $P_{n}$ has strictly increasing commutator width, each $P_{n}$ must have a commutator width of at least $n$. Pick $g_{n}\in P_{n}$ such that it cannot be written as the product of less than $n$ commutators.

If $\lim_{n\to\omega}g_{n}\in P$ can be written as the product of $d$ commutators, then we must have $\{n\in\mathbb{N}:g_{n}\text{ is the product of $d$ commutators}\}\in\omega$. However, this set is finite, while $\omega$ is a non-principal ultrafilter, a contradiction. Hence $\lim_{n\to\omega}g_{n}$ is not in the commutator subgroup of $P$. In particular, $P$ is not perfect. ∎

Suppose $G\in\mathcal{P}(\mathcal{C})$, and it is the internal direct product $G=\prod_{i\in I}G_{i}$ where $G_{i}\in\mathcal{C}$ for all $i\in I$. It is enough to show that the only $\mathcal{C}$-normal subgroups are kernels of the projection maps onto some factor group.

We perform induction on the number of factor groups, i.e., on the cardinality of the index set $I$. If $I$ has only one element, then $G$ is a group in $\mathcal{C}$. If $N$ is a $\mathcal{C}$-normal subgroup of $G$, then the quotient map $q$ from $G$ to $G/N$ is a surjective homomorphism between groups in $\mathcal{C}$. In particular, we must have $q(G)\mathrm{C}_{G/N}(q(G))=G/N$, so $q$ is either trivial or an isomorphism. Since the trivial group (which is abelian) is not in $\mathcal{C}$, therefore the quotient map must be an isomorphism. Hence the only $\mathcal{C}$-normal subgroup of $G$ is the trivial subgroup, and the statement is true.

Now suppose $I$ has more than one element. Let $N$ be any $\mathcal{C}$-normal subgroup. Consider the quotient map $q:G\to G/N$. For any $i\in I$, since $G_{i}$ is a direct factor of $G$, any element of $q(G_{i})$ must commute with any element of $q(G_{j})$ for $j\neq i$. In particular, we have $q(G_{i})\mathrm{C}_{G/N}(q(G_{i}))=G/N$. So $q$ restricted to $G_{i}$ must either be trivial or an isomorphism.

If $q$ is trivial on all $G_{i}$, then $G/N$ is trivial and not in $\mathcal{C}$, which is impossible. Hence $q$ restricts to an isomorphism to some $G_{i}$. Furthermore, if $q$ restricted to $G_{i}$ and $q$ restricted to $G_{j}$ are both isomorphisms for some $i\neq j$, then all elements in $q(G_{i})=G/N$ and in $q(G_{j})=G/N$ must commute, so $G/N$ is an abelian group and not in $\mathcal{C}$, also impossible. Hence, there is a unique $i\in I$ such that $q$ restricted to $G_{i}$ is an isomorphism, and $q$ kills all other $G_{j}$ where $j\neq i$.

In particular, $q$ is a projection map onto a factor group $G_{i}$. So we are done. ∎

Let $G$ be a pro-$\mathcal{P}(\mathcal{C})$ group. Then it is the projective limits of $G/U\in\mathcal{P}(\mathcal{C})$ for all open $\mathcal{P}(\mathcal{C})$-normal subgroups $U$.

But for each open $\mathcal{P}(\mathcal{C})$-normal subgroup $U$, by Lemma 2.2, we have an isomorphism $G/U\simeq\prod_{N\in\mathcal{M}_{U}}G/N$ where $\mathcal{M}_{U}$ is the set of open $\mathcal{C}$-normal subgroups containing $U$. By taking inverse limits of these isomorphisms, we have an isomorphism $G\simeq\prod_{N\in\mathcal{M}}G/N$ where $\mathcal{M}_{U}$ is the set of open $\mathcal{C}$-normal subgroups of $G$. ∎

Note that if $G$ is an arbitrary product of non-abelian finite simple groups, then each factor group of $G$ has commutator width $1$ by the classification of non-abelian finite simple groups. So $G$ has commutator width $1$, and thus it is perfect. So $G/M$ is also a perfect group.

Since $M$ is maximal, $G/M$ is perfect and simple. So by Corollary 2.9 below, if $G=\prod_{i\in I}S_{i}$ for a family of non-abelian finite simple groups $S_{i}$, then the quotient map $G\to G/M$ must factor through some ultraproduct $\prod_{i\to\mathcal{U}}S_{i}$ for some ultrafilter $\mathcal{U}$ on $I$. So we are done. ∎

By Proposition 2.3, we know that we can find an ultrafilter $\mathcal{U}$ such that $N_{\mathcal{U}}\subseteq M$. But by [15], $G/N_{\mathcal{U}}$ is either isomorphic to $S_{i}$ for some $i$, or it cannot have any non-trivial finite dimensional unitary representation. But $G/M$ is a finite quotient of $G/N_{\mathcal{U}}$, which must have a non-trivial finite dimensional unitary representation. Hence $G/N_{\mathcal{U}}$ is either isomorphic to $S_{i}$ for some $i$, and thus $M=N_{\mathcal{U}}$ and $G/M$ is isomorphic to $S_{i}$ for some $i$. ∎

Using Corollary 2.4, let these maximal normal subgroups of finite indices be $\{N_{\mathcal{U}_{k}}\}_{k\in K}$ for some index set $K$ and ultrafilters $\mathcal{U}_{k}$ on $I$. For each $g=(g_{i})_{i\in I}\in G$, let $\mathrm{supp}(g)=\{i\in I:g_{i}\neq e\}$. Then $g\in N_{\mathcal{U}_{k}}$ if and only if $\mathrm{supp}(g)\notin\mathcal{U}_{k}$. So $g\in\bigcap_{k\in K}N_{\mathcal{U}_{k}}$ if and only if $\mathrm{supp}(g)\notin\bigcup_{k\in K}\mathcal{U}_{k}$.

In particular, $\bigcap_{k\in K}N_{\mathcal{U}_{k}}$ is generated by $G_{J}$ where $J\notin\bigcup_{k\in K}\mathcal{U}_{k}$. Since each $G_{J}$ is also a semisimple profinite group, hence a perfect group, therefore the subgroup they generate must also be perfect. So we are done. ∎

Say $G=\prod_{i\in I}S_{i}$ for a family of non-abelian finite simple groups $S_{i}$. Suppose for contradiction that our statement is not true, and let $B$ be a minimal counterexample, i.e., $B$ is a finite quotient of $G$, but it is not semisimple. Let $q:G\to B$ be the quotient map.

If $B$ has a non-trivial center $Z$, then by minimality, $B/Z$ is a finite semisimple group. Therefore, $Z$ is an intersection of maximal normal subgroups of $B$ with finite indices. Pulling back to $G$, we see that $q^{-1}(Z)$ is an intersection of maximal normal subgroups of $B$ with finite indices. Therefore according to Corollary 2.5, it is a perfect group. Hence its image $Z$ must also be a perfect group. But $Z$ is also abelian, hence it must be trivial, a contradiction.

So $B$ must be centerless. Furthermore, if $B$ is the direct product of two non-trivial groups, then by minimality of $B$, the two factor groups must both be semisimple groups. Hence $B$ itself is a semisimple group, a contradiction.

So $B$ must be a centerless group, and it is not the direct product of two non-trivial groups. By Corollary 2.8 below, this means that $q:G\to B$ must factor through an ultraproduct. But by [15] again, this means that $B$ is either infinite, or $B$ is isomorphic to $S_{i}$ for some $i$. Neither case is possible.

To sum up, such a counterexample is impossible. ∎

First let us show that the negation of $(i)$, $(ii)$, and $(iii)$ are equivalent.

Suppose $(i)$ is false. Then let $f:A_{1}\times A_{2}\to B$ be a counterexample. Let $N_{1}=f(A_{1}\times\{e\})$ and $N_{2}=f(\{e\}\times A_{2})$. Then $N_{1}$ and $N_{2}$ are non-trivial. Since $f$ is surjective, $N_{1}$ and $N_{2}$ are normal subgroups of $B$, and $B=N_{1}N_{2}$. Finally,

So $(ii)$ is false.

Suppose $(ii)$ is false. Let $N_{1},N_{2}$ be a pair of normal subgroups of $B$ and a counterexample to $(ii)$. If $N_{1}\cap N_{2}=\{e\}$, then $B=N_{1}\times N_{2}$, and $(iii)$ is false. Suppose $N_{1}\cap N_{2}$ is non-trivial. For any element $a\in N_{1}\cap N_{2}$, since $a\in N_{2}$ and $[N_{1},N_{2}]=\{e\}$, $a$ must commute with every element of $N_{1}$. Similarly, since $a\in N_{1}$, it must commute with every element of $N_{2}$. And since $B=N_{1}N_{2}$, we must have $a\in Z(B)$. So $Z(B)\supseteq N_{1}\cap N_{2}$ is non-trivial. So $(iii)$ is false.

Now suppose $(iii)$ is false. If $B$ is the direct product of two non-trivial groups, then obviously $(i)$ is false. Suppose $Z(B)$ is non-trivial. Then let $f:Z(B)\times B\to B$ be the homomorphism induced by the inclusion homomorphism of $Z(B)$ into $B$ and the identity homomorphism of $B$. This provides a counterexample of $(i)$.

Now we shall show that $(i)$ and $(iv)$ are equivalent.

$(i)$ is a special case of $(iv)$ when $I=\{1,2\}$. So we only need to show that $(i)$ implies $(iv)$. But this follows easily from Lemma 2.7. ∎

Since $B$ is perfect and simple, it cannot have a non-trivial center, and it cannot be the direct product of two non-trivial subgroups. So the statement is true by Corollary 2.8. ∎

Suppose $G$ is a perfect central extension of a finite semisimple group $S$, which is the internal direct product of non-abelian finite simple subgroups $S_{1},\dots,S_{k}$. Let the quotient map be $q:G\to S$, let $M_{i}=q^{-1}(S_{i})$, i.e., the pull-back in the following diagram.

Let $Q_{i}$ be the unique maximal perfect normal subgroup of $M_{i}$. Now each $Q_{i}$ is perfect by construction, and since $M_{i}$ is finite, $M_{i}/Q_{i}$ is solvable. Therefore $q(Q_{i})=S_{i}$. So, we have the following diagram.

Finally, since the kernel of $q$ is the center of $G$, we see that $Q_{i}$ must be a perfect central extension of $S_{i}$, i.e., it is quasisimple. I now claim that $G$ is the central product of all these $Q_{i}$.

Let $Z$ be the center of $G$. Since $q(Q_{i})=S_{i}$, we see that $q$ restricted to $Q_{1}\dots Q_{k}$ is already surjective onto $S\simeq G/Z$. Hence we must have $G=Q_{1}\dots Q_{k}Z$.

If $Q_{1}\dots Q_{k}$ is a proper normal subgroup of $G$, then $G/(Q_{1}\dots Q_{k})$ is non-trivial and abelian, which cannot happen since $G$ is perfect. So $G=Q_{1}\dots Q_{k}$. It remains to be shown that $[Q_{i},Q_{j}]$ is trivial for all $i\neq j$.

Now $Q_{j}$ acts on $Q_{i}$ via conjugation, but $[Q_{i},Q_{j}]\subseteq Q_{i}\cap Q_{j}$ is in the kernel of $q$, hence inside of the center $Z$. So $Q_{j}$ acts trivially on $Q_{i}/(Q_{i}\cap Z)\simeq S_{i}$. But any automorphism of a finite quasisimple group must be trivial if it acts trivially on the corresponding simple group. So $[Q_{i},Q_{j}]$ is in fact trivial.

So $G$ is indeed the central product of $Q_{1},\dots,Q_{k}$. ∎

Say $G=\prod_{i\in I}Q_{i}$ for a family of finite quasisimple groups $Q_{i}$. Let $Z_{i}$ be the center of $Q_{i}$, and set $Z=\prod_{i\in I}Z_{i}$. Then $Z$ is the center of $G$, and $G/Z$ is a semisimple profinite group.

Suppose we have a surjective homomorphism $q:G\to B$. Then $B/q(Z)$ is a finite quotient of the semisimple profinite group $G/Z$. Hence $B/q(Z)$ is semisimple. Furthermore, since $Z$ is the center of $G$ and $q$ is surjective, $q(Z)$ is inside the center of $B$. So $B$ is a central extension of the semisimple group $B/q(Z)$.

Finally, by classification of finite simple groups, any finite quasisimple group has commutator width at most $2$. So $G$ has commutator width at most $2$. So $G$ is perfect, and its quotient $B$ is also perfect.

So $B$ is a finite perfect central extension of a finite semisimple group. ∎

Suppose $G,H$ are finite superperfect quasisimple groups, and $f:G\to H$ is any group homomorphism such that $f(G)\mathrm{C}_{H}(f(G))=H$. Since the subgroups $f(G)$ and $\mathrm{C}_{H}(f(G))$ both normalize $f(G)$, therefore $f(G)\mathrm{C}_{H}(f(G))=H$ implies that $f(G)$ is normal in $H$.

If $f(G)\neq H$, then note that all proper normal subgroups of a quasisimple group are central, so $f(G)$ is abelian. But since $G$ is perfect, $f(G)$ must be trivial.

If $f(G)=H$, then $G$ is a central perfect extension of $H$. But since $H$ is super perfect, it is already a maximal central perfect extension, hence $f$ can only be an isomorphism. ∎

Let $\mathcal{C}$ be the class of superperfect quasisimple groups. Then for each finite central product $Q$ of finite quasisimple groups $\{Q_{i}\}_{i\in I}$, then $Q$ is a quotient of the direct products of these $Q_{i}$ by a central subgroup. Let $\widetilde{Q}_{i}$ be the universal perfect central extension of $Q_{i}$, then $Q$ is a central quotient of $\widetilde{Q}=\prod_{i\in I}\widetilde{Q}_{i}\in\mathcal{P}(\mathcal{C})$.

Now let $G$ be the projective limit of a directed family of finite central products of finite quasisimple groups $\{G_{i}\}_{i\in I}$. Let $\widetilde{G}_{i}$ be the universal perfect central extension of $G_{i}$, so that $\widetilde{G}_{i}\in\mathcal{P}(\mathcal{C})$. Let $Z_{i}$ be the kernel of the quotient map from $\widetilde{G}_{i}$ to $G_{i}$.

Note that any surjective homomorphism $\phi_{ij}:G_{i}\to G_{j}$ must induce a surjective homomorphism $\widetilde{\phi}_{ij}:\widetilde{G}_{i}\to\widetilde{G}_{j}$, and $\widetilde{\phi}_{ij}$ must also send the center into the center. So the directed family $\{G_{i}\}_{i\in I}$ must induce directed families $\{\widetilde{G}_{i}\}_{i\in I}$ and $\{Z_{i}\}_{i\in I}$, and thus we have corresponding induced projective limits $\widetilde{G}$ and $\widetilde{Z}$.

Note that the projective limit on inverse systems of finite groups is an exact functor. (See, e.g., Proposition 2.2.4 of [12].) Therefore, $G$ is a quotient of $\widetilde{G}$ by the closed subgroup $Z$. Since $\widetilde{G}$ is a pro-$\mathcal{P}(\mathcal{C})$ group and $\mathcal{C}$ is a central-simple class, $\widetilde{G}$ must be a direct product of superperfect quasisimple groups, and thus any finite quotient of $\widetilde{G}$ must be a central product of finite quasisimple groups.

Since $G$ is a quotient of $\widetilde{G}$, the result holds for $G$ as well. ∎

Let us call the three classes $\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}$.

Since central products are quotients of the direct products, $\mathcal{C}_{1}$ is closed under central products. Also note that any cyclic group $Z$ can be realized as a center for some quasisimple group over $\mathrm{PSL}_{n}(q)$ for some $n,q$. Let $Q$ be such a quasisimple group over a non-abelian finite simple group $S$ with center $Z$. Then $Q^{2}$ has a quotient $S^{2}$, and the pullback of the diagonal subgroup of $S^{2}$ in $Q^{2}$ is a sub-direct product of two copies of $Q$, and it is isomorphic to $Q\times Z$. Hence $\mathcal{C}_{1}$ must also contain all finite abelian groups. So $\mathcal{C}_{3}\subseteq\mathcal{C}_{1}$.

Suppose $G\in\mathcal{C}_{2}$. Then we have a quotient map $q:G\to S$ such that $S$ is semisimple, and the kernel of $q$ is the center $Z$ of $G$. Since $S$ is perfect, the unique largest perfect normal subgroup $N$ in $G$ must have $q(N)=S$, and the restriction of $q$ to $N$ must have kernel $N\cap Z$, which is in the center of $N$. So $N$ is a perfect central extension of $S$. In particular, by Lemma 2.10, $N$ is a finite central product of finite quasisimple groups.

But since $q(N)=S$ and $Z$ is the kernel of $q$, we have $G=NZ$. And since $Z$ is the center, $[N,Z]$ must be trivial. So $G$ is the central product of $N$ and the abelian group $Z$. So $\mathcal{C}_{2}\subseteq\mathcal{C}_{3}$.

It remains to show that $\mathcal{C}_{1}\subseteq\mathcal{C}_{2}$. Since $\mathcal{C}_{2}$ contains all finite quasisimple groups, we only need to show that it is closed under sub-direct products and quotients.

The class $\mathcal{C}_{2}$ is obviously closed under quotients. We now prove that sub-direct products of groups in $\mathcal{C}_{2}$ are still in $\mathcal{C}_{2}$.

Suppose $G$ has two normal subgroups $N_{1},N_{2}$ such that $N_{1}\cap N_{2}$ is trivial, and $G_{i}=G/N_{i}\in\mathcal{C}_{2}$ for $i=1,2$. Then $G$ is a finite group. Let $M$ be the intersection of all maximal normal subgroups of $G$ whose corresponding quotients are non-abelian finite simple groups. Then $G/M$ is a finite semisimple group. We need to show that $M$ is in the center of $G$.

Consider the quotient map $q_{i}:G\to G_{i}$. Then $q_{i}(M)$ is in the intersection of all maximal normal subgroups of $G_{i}$ whose corresponding quotients are non-abelian finite simple groups. Since $G_{i}\in\mathcal{C}_{2}$, $q_{i}(M)$ is the center of $G_{i}$. In particular, for any $g\in G$ and $x\in M$, we have $xgN_{i}=gxN_{i}$. So, since $N_{1}\cap N_{2}$ is trivial, we have

So $x$ is in the center of $G$. So $G$ is a central extension of $G/M$, and $G\in\mathcal{C}_{2}$. So we conclude that $\mathcal{C}_{1}\subseteq\mathcal{C}_{2}$. ∎

Now if $G$ is a projective limit of a family of groups $G_{i}$, each a finite central extension of a finite semisimple group $S_{i}$. Let $Z_{i}$ be the center of $G_{i}$. Then we have a corresponding projective limit $Z$ of the abelian groups $Z_{i}$, and a projective limit $S$ of the semisimple groups $S_{i}$. Note that the projective limit on inverse systems of finite groups is an exact functor. (See, e.g., Proposition 2.2.4 of [12].) So $Z$ is a closed central subgroup of $G$ and $G/Z\simeq S$, a semisimple profinite group.

Suppose we have a surjective homomorphism $q:G\to B$ where $B$ is finite. Then $q(Z)$ is a central subgroup of $B$, and $B/q(Z)$ is a finite quotient of the semisimple profinite group $G/Z$. By Proposition 2.6, $B/q(Z)$ is a finite semisimple group. So $B$ is a finite central extension of a finite semisimple group. ∎

Let $S$ be the CR-radical of $G$. Then $G$ acts on $S$ by conjugation, which induces a homomorphism from $G$ to $\mathrm{Aut}(S)$. Let $N$ be the kernel of this homomorphism. Assume for contradiction that it is not trivial.

Since $N$ is a normal subgroup of $G$ and $\mathrm{CR}(N)$ is characteristic in $N$, therefore $\mathrm{CR}(N)$ is a semisimple normal subgroup of $G$. Therefore $\mathrm{CR}(N)\subseteq\mathrm{CR}(G)=S$ and by construction, every element of $\mathrm{CR}(N)$ commutes with every element of $S$. Hence $\mathrm{CR}(N)$ is in the center of $S$. But $S$ is a non-abelian finite simple group. So $\mathrm{CR}(N)$ is trivial.

Therefore, any minimal normal subgroup of $N$ must be a direct product of cyclic subgroups of prime order. In particular, the solvable radical $M$ of $N$ (the unique largest solvable normal subgroup of $N$) is non-trivial. Since $M$ is characteristic in $N$, it is a non-trivial solvable normal subgroup of $G$, a contradiction.

So we must conclude that $N$ is trivial. Hence $G$ is isomorphic to a subgroup of $\mathrm{Aut}(S)$. And since $G$ contains $S$, the image of $G$ in $\mathrm{Aut}(S)$ would contain $\mathrm{Inn}(S)$. ∎

Suppose $G,H$ are finite almost simple groups, and $f:G\to H$ is any group homomorphism such that $f(G)\mathrm{C}_{H}(f(G))=H$. Then $f(G)$ is a normal subgroup of $H$.

Suppose it is not trivial. Let $S$ be the unique minimal normal subgroup of $H$. Then it is a finite simple group, and any non-trivial normal subgroup of $H$ must contain $S=\mathrm{CR}(H)$. So $f(G)$ contains $S$. However, since $H$ can be seen as a subgroup of $\mathrm{Aut}(S)$, the centralizer of $S$ in $H$ is trivial. Hence $\mathrm{C}_{H}(f(G))$ is trivial. So $f(G)\mathrm{C}_{H}(f(G))=H$ implies that $f(G)=H$.

Finally, note that proper quotients of an almost simple group must be solvable. But $f(G)=H$ is not solvable. So $f$ has a trivial kernel, and thus it is an isomorphism. ∎

Note that $\mathcal{C}$ is obviously closed under direct products. So we only need to show that it is closed under central quotients.

Suppose $G\in\mathcal{C}$, say $G$ is the internal direct product of almost simple subgroups $G_{1},\dots,G_{k}$ and a solvable subgroup $S$. Now $G$ acts on its normal subgroup $\mathrm{CR}(G_{1})\dots\mathrm{CR}(G_{k})$ by conjugation. Since $G_{1},\dots,G_{k}$ are all almost simple, the kernel of the induced homomorphism $G\to\mathrm{Aut}(\mathrm{CR}(G_{1})\dots\mathrm{CR}(G_{k}))$ is exactly $S$. So the center $Z$ of $G$ is contained inside of $S$. Hence $G/Z\simeq G_{1}\times\dots\times G_{k}\times(S/Z)\in\mathcal{C}$. ∎

Consider a potential counterexample. Suppose for contradiction that we have a pro-$\mathcal{C}$ group $G=\prod_{i\in I}G_{i}$, each $G_{i}$ is either finite solvable or finite almost simple, and $\phi:G\to B$ is a surjective homomorphism onto a finite group $B$ NOT in $\mathcal{C}$. Since $B$ is finite, if such counterexample exists, then we can in fact let $B$ be a minimal counterexample.

Then by Corollary 2.30 below, $B$ cannot exist. Hence we are done. ∎