I would like to ask the following question:

Does there exist a group $G$ such that

  • every finite group can be embedded in $G$, and
  • every proper subgroup of $G$ is finite?

The closest example to this I have seen is the group $S_{\omega}$, i.e. bijections of $\mathbb{N}$ fixing all but finitely many elements. However, this group contains isomorphic copies of itself as its proper subgroups (even continuum many - for every $S \subseteq \mathbb{N}$ that is not cofinite, bijections from $S_{\omega}$ fixing all points in $S$ form a group isomorphic to $S_{\omega}$).

So obviously such a group needs to contain some subgroup isomorphic to $S_n$ for every $n$, but not infinite ascending chain of these. I am not sure how to even look for something like that (to be honest, I am inclined to believe that such a group cannot exist).

I would also appreciate pointing me towards some infinite subgroups of $S_{\omega}$ not isomorphic to $S_{\omega}$ (the more bizarre, the better. : ) ).

Thanks in advance for any help.

Edit: OK I see that even the $S_\omega$ is far from having these properties - one can, for example, take an arbitrary countable family of finite groups $G_n$ and then find $\bigoplus_{n<\omega}G_n$ as a subgroup by partitioning the set $\mathbb{N}=\bigcup_{n < \omega}M_n$ and realizing $G_n$ as a subgroup of $S(M_n). $ So there are in fact even uncountably many up to isomorphism infinite subgroups of $S_{\omega}$.

Edit2: The only idea I came up with is the following: Obviously such a group, assuming it exists, is a union of all its proper subgroups, hence it is a direct limit (i.e. directed colimit) of some diagram consisting of a set of finite groups s.t. every finite group is isomorphic to some of theose groups. Finding the "right diagram" could resolve the problem (by showing the colimit has desired properties, or showing that if some group does, it is necessarily that one, and finding a proper infinite subgroup there). However, it is not clear to me which morphisms should one use. My first idea was to use all possible injections (something like "whenever some injection exists, fix one"), but that obviosly doesn't work - the resulting limit would contain $S_{\omega}$ as a subgroup. So maybe one needs to use less injections (which I suspect would make the limit even bigger) or more general morphisms, than injections.

  • 3
    $\begingroup$ Relevant discussion on MathOverflow here: mathoverflow.net/q/28945, mathoverflow.net/q/28999 $\endgroup$
    – curious
    Oct 29, 2014 at 22:43
  • 4
    $\begingroup$ I do not know about the "all subgroups are finite" stipulation, but if you enumerate all finite groups and take their direct dum you obtain a recursively presentable group which can then be embedded into a finitely presented group. So there exists a finitely presented group which contains every finite group. The resulting group does contain infinite proper subgroups, but all the ones I can think of are infinitely generated (and they come from the direct sum construction). $\endgroup$
    – user1729
    Oct 29, 2014 at 23:05
  • 1
    $\begingroup$ @DerekHolt Can you, please, elaborate on that? I do not see why something like that should be true. $\endgroup$ Oct 30, 2014 at 8:23
  • 2
    $\begingroup$ Sorry, I am not sure if I can prove that $G$ is $2$-generated, but it must be finitely generated. $G$ must have a finite $2$-subgroup $S$ that is maximal among finite $2$-subgroups, since otherwise we could form an infinite ascending chain and deduce that the whole group was a $2$-group, which is clearly false. Now let $T$ be a finite $2$-subgroupof $G$ with $|T|>|S|$. Then we must have $\langle S,T \rangle=G$, proving the claim, since otherwise, by Sylow's Theorem, $S$ would not be a maximal $2$-subgroup of $\langle S,T \rangle$, contradiction. $\endgroup$
    – Derek Holt
    Oct 30, 2014 at 8:40
  • 3
    $\begingroup$ Nice question. For what it's worth, any example must be quasisimple, and its simple quotient must also be an example. So if there is such a group then there's also one that's simple. $\endgroup$ Oct 31, 2014 at 8:38

1 Answer 1


No. Jeremy already commented that we may assume the center to be trivial.

Now look at the involutions (= elements of order 2). Their centralizers are finite, so there are countably many conjugacy classes of involutions. Two involutions generate a dihedral group. If both involutions are not conjugate (in $G$ and hence within the dihedral group), the dihedral group has a non-trivial center (contained in the centralizer of either involution). As there are infinitely many involutions not conjugated to a fixed involution $\tau$, there has to be a non-trivial element $\sigma$ of the (finite) centralizer of $\tau$ that lies in the centralizers of infinitely many of those involutions. Hence the centralizer of $\sigma$ is a proper infinite subgroup of $G$.

  • $\begingroup$ Thank you. Could you meybe comment on how one can see that $G$ is wlog simple (or at least centerless)? I actually do not see why that should be true. $\endgroup$ Dec 2, 2014 at 15:49
  • 1
    $\begingroup$ @PavelC: The center $Z$ is finite and intersects subgroups of $G$ isomorphic to $S_n$ (n>2) trivially. If you factor out the center, therefore $G/Z$ still has your property. If it has a non-trivial center, repeat. This has to stop after finitely many steps, as otherwise you would get an infinite subgroup not containing any simple subgroup of $G$. $\endgroup$
    – j.p.
    Dec 2, 2014 at 16:15
  • $\begingroup$ Thank you. This is truly a wonderfully elementary argument. $\endgroup$ Dec 2, 2014 at 16:20
  • 1
    $\begingroup$ I don't understand the beginning of the argument: assuming that each involutions has a finite centralizer, we immediately deduce that the conjugacy classes are infinite, so the remainder of the argument works as soon as we show there are at least 2 non-conjugated involutions. But what prevents us from the possibility that there is a single conjugacy class of involutions? An additional argument is needed: $\endgroup$
    – YCor
    Aug 21, 2016 at 10:33
  • 1
    $\begingroup$ If there were finitely many conjugacy classes of involutions (I still assume the center is trivial), they would have centralizers of bounded order. But since the group contains copies of, say, $C_2^k$ for all $k$, there exist involutions with arbitrary large centralizer, whence a contradiction. $\endgroup$
    – YCor
    Aug 21, 2016 at 10:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .