Probability that a random element of a group is trivial Let $G$ be an infinite group with a finite generating set $S$.  For $n \geq 1$, let $p_n$ be the probability that a random word in $S \cup S^{-1}$ of length at most $n$ represents the identity.  Is it possible for $p_n$ to not go to $0$ as $n$ goes to $\infty$?
 A: The answer is "no", and it has nothing to do with free groups, cogrowth, or Schreier graphs. We are talking here about the return probabilities of the simple random walk on $G$ (i.e., the one whose step distribution is equidistributed on the set $S\cup S^{-1}$). The reason is the following simple property differentiating finite and infinite groups: the random walk on a countable group $G$ determined by a step distribution $\mu$ has a finite stationary measure if and only the group is finite (under, for simplicity, the assumption that the random walk is non-degenerate, i.e., the support of the step distribution generates the whole group as a semigroup), and this distribution is then uniform.
The argument is very simple: if a stationary distribution is finite, then it has a maximal weight atom, and then by the maximum principle the weights of all other atoms have to be same. Or, instead of stationary measures one can argue in terms of harmonic functions, then the claim is that the finite groups are the only ones that admit summable harmonic functions (which are constants).
Thus, one can not have a positively recurrent random walk on an infinite group, which by general Markov theory (e.g., see Theorem 1.8.5 in Norris' textbook) implies the answer.
EDIT This argument only uses that the transition matrix of a random walk on a group is bi-stochastic, i.e., that the counting measure on the state space is stationary. The underlying general fact is then: if a Markov chain has an infinite stationary measure, then it can not be positively recurrent.
