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Question 1: Does there exist an intrinsic characterization of groups $G$ isomorphic to some subgroup of some finitary symmetric group (i.e. all the permutations of a given set that fix all but finitely many elements)?

Clearly every such $G$ enjoys local finiteness, but I see where (for a fixed $p$) the multiplicative group $\{e^{2\pi i k/p^n}\}$ shows that this does not suffice (because all non-identity elements have $m$-th roots for every $m$).

Question 2: Do there exist locally finite groups not isomorphic to any subgroup of any finitary symmetric group such that no non-identity element has $m$-th roots for every $m$?

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If nobody here is able to answer this question, you could try to contact Ulrich Meierfrankenfeld. I'd be surprised, if he can't help you with this questions. – j.p. May 17 '12 at 7:33
up vote 6 down vote accepted

The answer to Question 2 is "yes". Take the relatively free group with law $x^4=1$ and infinite set of generators. That group is locally finite (Sanov, I. N. Solution of Burnside's problem for exponent 4. Leningrad State Univ. Annals [Uchenye Zapiski] Math. Ser. 10, (1940). 166–170.). It is not inside the group of finitary permutations (every permutation of order 4 is a product of 4-cycles and 2-cycles) and no non-identity element has roots of order 4.

Update It is even easier to consider the relatively free group $G$ of exponent $3$ instead. That group is solvable (M. Hall, The Theory of Groups, pages 320-324), every non-zero element does not have a root of degree 3, and is not inside $S_\infty$, the group of finitary permutations of an infinite set. Indeed, suppose $G$ is inside $S_\infty$. We can assume that it is transitive (exercise). But this contradicts Corollary 4.7 in the notes cited by Igor Rivin (Wiegold's theorem).

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Could you clarify the link between "It is not inside the group of finitary permutations" and "every permutation of order 4 is a product of 4-cycles and 2-cycles"? – YCor May 17 '12 at 16:12
@Yves: I added an update. For exponent 4, the proof is not much more difficult and in fact also can be deduced from some results in the notes. – Mark Sapir May 17 '12 at 16:33
OK. Here I quote the results you use, in case the notes are removed: Thm 4.6 (P.M.Neumann): an infinite transitive permutation group $G$ has the property that every normal transitive subgroup contains the derived subgroup, and the derived subgroup is perfect. Corollary 4.7 (Wiegold): $G$ can't be solvable. Added note: if $G$ is your initial example (free exponent 4 group on countable set), it's not solvable (Razmyslov) but its derived subgroup is not perfect because $G$ admits quotients of nilpotency length exactly 2. – YCor May 17 '12 at 16:45
On the other hand, I'm not able to solve your exercise, which amounts to show that if $G$ is the relatively free group of exponent 3, then it does not admit a faithful finitary action with finite orbits, i.e. does not embed into the infinite direct sum $\bigoplus Sym(n)$. However this statement is clear for any non-residually finite group. An example is the restricted wreath product $F\wr L$ of a finite nonabelian group $F$ with any infinite locally finite group $L$. – YCor May 17 '12 at 16:53
Hint: a direct product of finite groups is an FC-group. – Mark Sapir May 17 '12 at 17:22

Well, for the first question, there is the amazing result that the only simple infinite such group is the finitary alternating group, see Chris Pinnock's notes (the Mihles-Tyskevic theorem). That tells you that the theory is not too much richer than the theory of finite groups.

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Reading those notes prompted my question! – David Feldman May 17 '12 at 0:36

The notes at my site have moved to here: Finitary Permutation Groups

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Welcome to MathOverflow! – Igor Rivin May 17 at 10:28

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