Does there exist a group $G$ such that
- for any finite $K$ there is a monomorphism $K \to G$
- for any $H$ with property 1 there is a monomorphism $G \to H$
If yes, is it the only one?
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asked Oct 24, 2022 at 0:34
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3$\begingroup$ Remark 1. Related question mathoverflow.net/questions/28945/… 2. It is obvious that if such $G$ is necessarily Cohopfian, then it is unique. $\endgroup$– Arshak AivazianCommented Oct 24, 2022 at 0:37
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2$\begingroup$ This wouldn't answer the question as stated, but: let $G_1 \cong \bigsqcup_n S_n$ be the coproduct of the symmetric groups, and let $G_2 \cong \bigoplus_n S_n$ be the direct sum (meaning the subgroup of the direct product with finite support). Both satisfy the first condition. Does anyone see an embedding of either of these groups into each other? $\endgroup$– Qiaochu YuanCommented Oct 24, 2022 at 3:59
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1$\begingroup$ Hall's Universal Group is a countable locally finite group that contains all countable locally finite groups (in particular all finite groups) as subgroups (so in 2. the arrows are the wrong way round - for $H$ countable). $\endgroup$– j.p.Commented Oct 24, 2022 at 6:59
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2$\begingroup$ The question has a negative answer (I wrote an answer) but it is natural to think about the locally finite case. I think I can show that the group $A=\bigoplus_{n\ge 5}\mathrm{Alt}_n$ has the property that every subgroup $H$ of $A$ that contains isomorphic copies of all finite groups, actually contains an isomorphic copy of $A$. Hence, if there exists a group $G$ that satisfies the required property restricted to locally finite groups, then $A$ itself satisfies this property. But I don't know if it's the case, i.e. if $A$ embeds into every locally finite group containing all finite groups. $\endgroup$– YCorCommented Oct 24, 2022 at 8:01
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2$\begingroup$ As a follow-up to this question I asked the question suggested in my previous comment separately. $\endgroup$– YCorCommented Oct 28, 2022 at 12:39
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2 Answers
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No. To show that it doesn't exist it is enough to produce two groups $G,H$ which contain isomorphic copies of all finite groups, but such that no group containing isomorphic copies of all finite groups embeds into both $G$ and $H$.
Let $(G_n)$ be an enumeration of all finite groups. Let $G=\bigoplus G_n$ be the restricted direct sum and $H={\Large\ast}_nG_n$ the free product.
If $K$ is a subgroup of $G$ then $K$ is locally finite, hence freely indecomposable. Hence if $K$ is also isomorphic to a subgroup of $H$, then by Kurosh's subgroup theorem, $K$ is finite. In particular, $K$ doesn't contain isomorphic copies of all finite groups.
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7$\begingroup$ "To show that it doesn't exist it is enough to produce two groups 𝐺,𝐻 which contain isomorphic copies of all finite groups, but which don't embed into each other." Don't you need to show that there is no third group that contains all finite subgroups that embeds into both $G$ and $H$? $\endgroup$ Commented Oct 24, 2022 at 5:12
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7$\begingroup$ @JimConant indeed. This is fixed now. $\endgroup$– YCorCommented Oct 24, 2022 at 5:25
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The answer is yes if:
- you furthermore impose that every finitely generated subgroup of $G$ is finite, and
- you replace your requirement that the group be a smallest group containing all finite groups by the requirement that the group be a largest group containing all finite groups (within the class of group with the property that every finitely generated subgroup is finite).
This group is known as the Fraïssé limit of the category of finite groups.
See:
https://en.wikipedia.org/wiki/Fra%C3%AFss%C3%A9_limit
and
https://math.stackexchange.com/questions/88169/fra%C3%AFss%C3%A9-limits-and-groups
answered Oct 27, 2022 at 21:44
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1$\begingroup$ Thank you very much! In fact, I was trying to find an inductive limit (or something like it) of all finite structures of some type, and then, thinking about the existence of such objects in general, I asked my question. $\endgroup$ Commented Oct 28, 2022 at 21:04