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.
Best Answer
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 $G$.