# Universal group such that every finite group is a quotient

We say that a permutation $$\varphi:\mathbb{N}\to\mathbb{N}$$ is finitary if there is $$k\in\mathbb{N}$$ such that $$\varphi(i) = i$$ for all $$i\in\mathbb{N}$$ with $$i\geq k$$. Let $$I_\mathbb{N}$$ denote the group of finitary permutations of $$\mathbb{N}$$, with composition as group operation. Every finite group can be embedded into $$I_\mathbb{N}$$.

Turning arrows around, is there a group $$S_\mathbb{N}$$ with the following strong properties?

1. For every finite group $$F$$ there is a surjective group homomorphism $$\pi:S_\mathbb{N}\to F$$, and
2. If $$G^*$$ is a group such that for every finite group $$F$$ there is a surjective group homomorphism $$\pi:G^*\to F$$, then there is a surjective group homomorphism $$s:G^*\to S_\mathbb{N}$$.

(Note that in the embedding case, there is no such group, see the comment section; in particular $$I_\mathbb{N}$$ is not "universal" in the above sense.)

• I think there is a Typo and in 2 it should be $\pi:G\to F$. Nov 15 at 8:51
• I don't think $I_{\mathbb N}$ (which is often denoted something like $\operatorname{FSym}(\mathbb N)$) has the universal property you say. For example every finite group embeds into $\prod_{n=1}^\infty S_n$, but there is no injective homomorphism from $\operatorname{FSym}(\mathbb N)$ to $\prod_{n=1}^\infty S_n$. Nov 15 at 10:17
• What you call "is almost-identical" is widely known as "has finite support" or "is finitely supported", or "is finitary".
– YCor
Nov 15 at 11:40
• Concerning the subgroup case, as I already said somewhere else on this site (I don't remember where), there is no group that contains all finite groups and embeds into every group containing every finite group. Indeed, the groups $\bigoplus_n S_n$ and $\ast_n S_n$ both contain every finite group but no infinite group embeds into both.
– YCor
Nov 15 at 12:02
• For your new version of the question with no universality requirement, just take the direct sum (rather than product) of the finite groups. Nov 15 at 17:49

Not a complete answer: Let $$G = \prod_F F$$ be the direct product of all finite groups and let $$F_\omega$$ be the free group on a countably infinite collection of generators. Obviously $$G$$ and $$F_\omega$$ both cover every finite group. To show that there is no group $$S$$ as sought in the question it suffices to show that $$G$$ has no countable quotient that still covers every finite group. Does it?

• I agree with @YCor that my answer should not be accepted in its current form. Although I think it almost certain that such a group $S$ does not exist, it's not obvious, and I would be interested to see the demonstration. Also, I am actually not at all sure that $G = \prod_F F$ has no countable quotient covering every finite group, it was just a thought. Nov 20 at 10:50

Again not a complete answer: If we consider the analog situation, where we replace the finite groups by simple finite groups, then there exists no such universal group $$S$$.

Indeed, assume such a hypothetical universal group $$S$$ exists. Then it must be countable by an analogous argument to the answer of Sean Eberhard and of course $$S$$ cannot be finite. Analogously, as above, we consider $$G = \prod_{F \in \mathscr{F}} F$$, where the product runs over a system of representatives $$\mathscr{F}$$ of all finite simple groups up to isomorphism. We call $$S \subseteq G$$ a product subgroup, if $$S = \prod_{F \in \mathscr{F}} S_F$$ where $$S_F$$ is a subgroup of $$F$$. A product subgroup $$S$$ of $$G$$ is normal if and only if $$S_F$$ is trivial or equal to $$F$$ for all $$F \in \mathscr{F}$$. In the next paragraph we show that every normal subgroup of $$G$$ is a product subgroup of $$G$$. This will prove that every quotient of $$G$$ is either uncountable or finite, and hence, there exists no surjective homomorphism $$G \to S$$, contradiction.

Let $$N$$ be a normal subgroup of $$G$$. Choose a maximal normal product subgroup $$S$$ of $$G$$ such that $$S \cap N$$ is trivial. Consider the quotient map $$\pi \colon G \to G/S$$. Then $$\pi$$ restricts to an injection $$N \to G/S$$ which is also surjective by the maximality of $$S$$ (in fact, $$G/S = \prod_{F \in \mathscr{F}_S} F$$ for some subset $$\mathscr{F}_S$$ of $$\mathscr{F}$$. If $$N \to G/S$$ is not surjective, then there exists $$F_0 \in \mathscr{F}_S$$ such that $$T = \prod_{F \in \mathscr{F}_S} R_F$$ intersects $$\pi(N)$$ trivially, where $$R_F$$ is trivial for all $$F \in \mathscr{F}_S \setminus \{F_0\}$$ and $$R_{F_0} = F_0$$. But in this case $$\pi^{-1}(T)$$ is a normal product subgroup of $$G$$ that intersects $$N$$ trivially but contains $$S$$ properly, contradiction). Let $$\rho \colon G \to \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F$$ be the natural projection. Then the homomorphism $$\prod_{F \in \mathscr{F}_S} F \simeq N \xrightarrow{\rho|_N} \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F$$ is trivial, since every homomorphism $$F \to F'$$ is trivial for distinct $$F, F' \in \mathscr{F}$$. This shows that $$N$$ lies in the kernel of $$\rho$$. Since, $$N$$ and $$\ker(\rho)$$ are both sections of $$\pi \colon G \to G/S$$, we get $$N = \ker(\rho)$$. Hence, $$N$$ is a normal product subgroup of $$G$$.

EDIT: The argument doesn't work, as the homomorphism $$\rho |_N \colon N \to \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F$$ is not necessarily trivial!

• This proof sounds incorrect to me. Indeed if $\eta$ is a nonprincipal ultrafilter, then the set of sequences $(g_i)$, $g_i\in F_i$ such that $\lim_{i\to\eta}g_i=1$ (that is, such that $\{i:g_i=1\}\notin \eta$) is a normal subgroup (that's the kernel of the canonicap map onto the ultrafilter) and is not a subproduct.
– YCor
Nov 20 at 8:53
• This is in addition false, because you allow abelian finite simple groups. However this might become true if you allow only non-abelian ones. Related results can be found in: Simon Thomas, Infinite products of finite simple groups II, J. Group Theory 2, 401-434) proved the following a sequence of finite nonabelian simple groups $F_n$ satisfies that $\prod F_n$ has no non-open subgroup of countable index iff the rank of $F_n$ tends to infinity. This however doesn't apply to the sequence of all ones taken once — but the non-open subgroups this produces are not normal.
– YCor
Nov 20 at 9:02
• So I believe you definitely need some nontrivial fact relying on the classification of finite simple groups. One would like an intermediate lemma, of the spirit: for every normal subgroup of $\prod F_i$, there exists an ideal of subsets such that it consists of sequences whose support belongs to the ideal... but even this is too optimistic. For instance, start from the product of all alternating groups $A_n$, $n\ge 5$. Let $g$ be an element consisting of one 3-cycle in each $A_n$. Then the normal subgroup generated by $g$ is not everything, although $g$ has full support in $\prod A_n$.
– YCor
Nov 20 at 9:12
• Well, here's a proof (using the Thomas 1999 paper [T] above as well as another paper of Saxl-Wilson). Namely I prove: if $G=\prod F_n$ (product of distinct finite simple groups ) surjects onto a countable group $H$, then there $H$ surjects onto the alternating group $A_n$ for only finitely many $n$. Proof: write $A=\prod_{n\ge 5}A_n<G$; let $K$ be the kernel of $G\to H$, and $L=K\cap A$. Then $L$ has countable index in $A$. By [T], $L$ is open in $A$. So $L$, and hence $K$, contains $A_n$ for say $n\ge n_0\ge 5$. (...)
– YCor
Nov 20 at 10:48
• (...) Then Saxl-Wilson [A note on powers in simple groups, M.P. Cambridge Ph. Soc. 122 (1997), 91–94] proved that nontrivial homomorphisms $G\to A_n$ are continuous (their result apply to any product of nonabelian finite simple group with order tending to infinity), hence nontrivial on $A_n$. Hence for $n\ge n_0$ they don't factor through $H$. Thus $A_n$, for $n\ge n_0$, is not a quotient of $H$. Note that these SW and T papers rely on the classification.
– YCor
Nov 20 at 10:52