I'm trying to prove the following:

Let $G$ be a group with finite symmetric generating set $S$ and let $\Gamma(G,S)$ be the corresponding Cayley graph. Let $X_1, X_2,\cdots$ be a simple random walk starting from a fixed vertex $x\in\Gamma(G,S)$. For each $n$ let $\mu_n$ denote the law of $X_n$. Fix an integer $k>0$. Suppose that for some integer $m$, the support of $\mu_m$ and the support of $\mu_{m+k}$ are non-disjoint. Construct a coupling to show that $X_n$ and $X_{n+k}$ meet in finite time almost surely.

So, I think $X_n$ and $X_{n+k}$ meet in finite time a.s if and only if there exists a word of length exactly $k$ with letters from $S$ representing the identity element $e$? Because on the one hand, if no such word exists then $X_n$ and $X_{n+k}$ will never meet. On the other hand, since there are only finitely many words of length $k$, and $k$ steps of the random walk correspond to a uniform choice of one of them, eventually the random walk will come across the sequence of length $k$ corresponding to $e$ with probability limiting to 1 (so for example, for all even $k$, $X_n$ and $X_{n+k}$ will trivially meet in finite time a.s).

When it says that the supports of $\mu_m$ and $\mu_{m+k}$ are non-disjoint, does this mean that there is a word lying in the set of values that $X_m$ can take and also in the set of values $X_{m+k}$ can take? i.e there is some $y\in G$ so that $x^{-1}y$ can be expressed using exactly $m$ letters from $S$ and also so that it can be expressed using exactly $m+k$ letters from $S$? If so is there a purely group theoretic way to do the problem? The problem says to construct a coupling of $X_n$ and $X_{n+k}$ but I can't think of one that would work.