**Original answer**

This isn't an answer, but since there has been little activity on this (very interesting) question I guess I might as well say it.

What if we consider a set of generators of $G_1$ and keep extending it by adding some elements so that it generates $G_n$ at each step. Then, looking at the Cayley graphs we can put some distance on each group, making this a sequence of finite metric spaces. The idea would be to do this in such a way so that the spaces are converging in the Gromov-Hausdorff sense to some metric space $(X,d)$ and the uniform probabilities $\mu_n$ on each $(G_n,d_n)$ are converging to some probability measure $\mu$ on $X$.

**Added later...**

**Compactification and weak limit of probabilities in the case of finite Abelian cyclic groups**

Consider the case when all the groups are finite cyclic so that $G_k = \mathbb{Z}_{n_k}$ (i.e. the finite cyclic group of $n_k$ elements) for some non-decreasing sequence of natural numbers each of which divides the next $n_1 | n_2 | \cdots$.

Let $S = \mathbb{R}/\mathbb{Z}$, we can identify each $G_k$ with the subgroup $(\frac{1}{n_k}\mathbb{Z})/\mathbb{Z} \subset S$. The uniform probability measures on these finite subgroups either converge weakly to Lebesgue measure on $S$ or (if the sequence of numbers $n_k$ is eventually constant) are eventually equal to a constant measure supported on a finite subgroup.

**Compactification and weak limit of probabilities in the case of general Abelian finite groups**

Consider now the case in which all groups $G_k$ are abelian. By the clasification of finite abelian groups, one can decompose each $G_k$ into a direct sum of cyclic groups with orders that are powers of primes. This implies that one can obtain a group isomorphic to $G_{k+1}$ from $G_k$ by replacing some of these powers of primes by higher powers, and by forming the direct product with another cyclic group of power of prime order.

Let $G = S^{\mathbb{N}}$ be the cartesian product of countably many copies of $S$. This is a compact and metrizable group. One can identify each group $G_k$ with a subgroup of $G$ generated by a finite number of elements with power of prime order and only one non-null corrdinate. The extension from $G_k$ to $G_{k+1}$ is obtained by replacing one of these generators by another with the same non-null coordinate but whose order is a higher power of the same prime number, or by adding a new generator with power of prime order whose only non-null coordinate is distinct from that of all the other generators.

This procedure gives rise to an increasing family of subgroups of $G$. The sequence of uniform measures $\mu_n$ on these groups can be seen to have a weak limit since each projection to a coordinate does (it is either eventualy a constant uniform measure on a finite subgroup of prime power order, or converges to Lebesgue measure on the circle).

**Further directions**

The answer to the question posed here implies that there are countable groups that are not a subgroup of a compact group. Hence it might be possible to construct a counter-example using one of these countable groups as the union of the $G_n$. The simplest possible candidate seems to be obtained by taking $G_k = \text{SL}(2,\mathbb{F}_{2^k})$ where $\mathbb{F}_p$ is the field with $p$ elements.

Also, here several countable groups which contain all finite subgroups are defined. This might serve to reduce the discussion to one concrete chain such as $G_1 = S_3, G_{n+1} = S_{G_n}$ where $S_G$ denotes the group of permutations of the elements of $G$ (which contains $G$ a subgroup since each element of $G$ acts on $G$ as a permutation).

satisfyin the sentence: "a k-tuple of group elements satisfies the word w"? $\endgroup$ – Pietro Majer Sep 25 '10 at 21:251more comment