Subgroups of finitary symmetric groups 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 $m$?
 A: 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).  
A: 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.
A: The notes at my site have moved to here: Finitary Permutation Groups
