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There are infinite graphs which contain all finite graphs as induced subgraphs, e.g. the Rado graph or the coprimeness graph on the naturals.

Are there infinite groups which contain all finite groups as subgroups?

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What's wrong with just taking the direct sum of all finite groups? – Jason DeVito Jun 21 '10 at 12:51
There are only countably many finite groups, up to isomorphism, so just take a direct sum of them all. – Joel David Hamkins Jun 21 '10 at 12:52
This can also be achieved with a straightforward compactness argument. – Carl Mummert Jun 21 '10 at 12:54
How about finitely generated? – Victor Protsak Jun 21 '10 at 19:27
@Carl: Could you be a bit more specific, please? – Hans Stricker Jun 22 '10 at 7:28
up vote 51 down vote accepted

Yes, plenty. The group only has to contain all finite permutation groups. Perhaps the most straightforward example would be the permutations of a countable set. That is bijections which fix all but a finite set.

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There is a finitely presented group which contains all finitely presented groups, as proved by Higman.

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Do you know an "easy" example if we only require that it contains all finite subgroups? More ambitiously, is there $F_\infty$ such group? – Victor Protsak Jun 22 '10 at 6:18
Higman's construction is fairly explicit (see Rotman's introductin to group theory text) for details, so one can keep track of some things, e.g. there is a finitely presented torsion free group that contains any finitely presented torsion free group. I forgot which finiteness condition is denoted $F_\infty$ and off hand I am not aware of the example you want. – Igor Belegradek Jun 22 '10 at 11:17
Well, finite groups aren't usually torsion free:) I read the construction of Higman's group in Manin's "Computable and not computable" and it didn't leave the impression of being easy. Pick your favorite finiteness condition -- for example, that $K(G,1)$ may be realized as a CW complex with finitely many cells of every dimension (I think that this is $F_\infty$), or some cohomological weakening of it. – Victor Protsak Jun 22 '10 at 11:36
Higman's embedding involves taking various free products and HNN extensions which behave reasonably under some cohomological conditions (via Mayer-Vietoris), so maybe there is a chance here. – Igor Belegradek Jun 22 '10 at 12:30

There many examples for instance the direct product of all finite groups. A more interesting example of a finitely generated (topologically) profinite groups is $\Pi_{n \geq 5} A_n$. Of course you can also construct many examples using direct limits of finite groups.

Another interesting example is the Nottingham group which is for $p>3$ a finitely presented pro-$p$ group (proved by Mikhail Ershov) which contains every countably based pro-$p$ group (proved by Rachel Camina and aslo by Ivan Fesenko) and in particuar every finite $p$-group. Of course it does not contain every finite group, but still in the spirit of your question. The Nottingham group has many other remarkable properties and is a worth knowing example of a pro-$p$ group.

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Here's a finitely generated group that contains all finite groups: permutations of the integers that are almost shifts; that is, each permutation differs form the shift-by-$n$ permutation by a permutation with finite support. This maps to the integers with kernel Bruce's group. Is it finitely presented?

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What are the generators? – HJRW Jun 21 '10 at 21:25
The shift and an adjacent transposition generate; conjugating the transposition by powers of the shift gives all adjacent transpositions, which generate all finitely supported permutations, which is everything up to shifts. – Ben Wieland Jun 21 '10 at 21:39
@Ben: A presentation for this group has to make disjoint transpositions commute; in particular, it has to make transpositions at distance $n$ commute for all $n$ (except possibly some small numbers, of course). This should imply that $H_2(G)$ has infinite rank and so the group is not finitely presented. For a simpler example along the same lines, consider the lamplighter group $\mathbb{Z}\wr \mathbb{Z}=\langle t,s_i| ts_it^{-1}=s_{i+1}, [s_i,s_j]=0\rangle$. Hopf's formula shows that $H_2(\mathbb{Z}\wr \mathbb{Z})=\mathbb{Z}^\infty$, with basis e.g. $\{[s_7,s_{7+k}]|k\in\mathbb{Z}\}$. – Tom Church Jul 22 '10 at 0:11
The homology is finitely generated in all degrees (also true without the $\mathbb Z$ action), so that won't do it. Your list is only finitely many relations away from a presentation. I think that the theory of Coxeter groups says that if you omit any of those relations, you get a different group, so it is not finitely presented. – Ben Wieland Jul 22 '10 at 3:55
Although not the most natural way to realize the "shift and permutation" group, it shows up in the list of Houghton's groups in Ken Brown's paper as the n=2 case. Again, not the most natural way to get this result, but in Brown's paper it is shown to be f.g., and not f.p. – Matt Brin Jan 9 '12 at 23:46

My favorite is Thompson's group $V$.

My favorite picture of $V$ is to take a set $X$ which is a disjoint union of subsets $L$ and $R$ having fixed bijections $l:X\rightarrow L$ and $r:X\rightarrow R$. Finite words in $l$ and $r$ map $X$ to "fragments" of $X$. Two "fragments" $W$ and $U$ that are the images, respectively, of words $w$ and $u$ in $l$ and $r$ are connected by the bijection $uw^{-1}:W\rightarrow U$. The two "fragments" will be disjoint iff neither of $w$ nor $u$ is a prefix of the other. If this condition holds, let $(w,u)$ represent the permutation on $X$ that is $uw^{-1}$ on $W$, is $wu^{-1}$ on $U$, and is the identity elsewhere. The group $V$ is generated by all such $(w,u)$. It is finitely presented and contains all finite groups.

Other f.p. groups containing all finite groups are known as Houghton groups. See the section on Houghton groups K. S. Brown "Finiteness properties of groups" in Journal of Pure and Applied Algebra, 44 (1987), 45-75. I forget how they are indexed, but from about n=3 on up, they are all f.p.

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If you want your group to be countable, you may consider the direct sum of all S_n's.

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All the groups mentioned in answers here (so far) are countable. – Joel David Hamkins Jun 22 '10 at 12:07
Can one likewise build the direct sum of all finite graphs which would contain all finite graphs? Can the resulting infinite graph be characterized otherwise? – Hans Stricker Jun 22 '10 at 13:04
Hans, yes, you can just take the disjoint union of all finite graphs---place them side by side and think of them as one graph. This graph is different from the random graph, since every node is in a finite component. – Joel David Hamkins Jun 22 '10 at 13:12

Another interesting example is Hall's universal group

Hall's universal group is a countable locally finite group, say U, which is uniquely characterized by the following properties.

* Every finite group G admits a monomorphism to U.

* All such monomorphisms are conjugate by inner automorphisms of U.
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Hall's universal group has even the stronger property to contain all countable locally finite groups (see – Someone Jul 22 '10 at 7:30

In the same spirit, there do exist infinite groups, all of whose proper subgroups are finite: the Tarski Monster. Are there any other "easier" examples?

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A standard example is Q/{a/b:b is odd}, the Prüfer group Z(2∞). It is infinite, but all of its proper subgroups are cyclic groups of order some power of 2. – Jack Schmidt Dec 16 '10 at 23:50
Jack, thanks, Nicky – Nicky Hekster Dec 20 '10 at 10:09

I can't resist mentioning that the group of unitary elements in the hyperfinite $II_{1}$ factor (The group von Neumann algebra of Bruce Westbury's "locally finite" permutation group above) in fact contains every countable discrete amenable group as a subgroup.

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