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Introduction of terminology: Let $G$ be an infinite group and let $S$ be a finite generating subset of $G$ that is symmetric, i.e. $x\in S$ implies $x^{-1}\in S$. Then the relation $g\sim g':\Leftrightarrow g^{-1}g'\in S$ is an equivalence relation on $G$. Consider the (undirected) graph which has $G$ as its set of vertices and an edge between $g$ and $g'$ iff $g\sim g'$. In this graph there exists a path from $g$ to $g'$ iff there exist $s_1,\ldots,s_n\in S$ such that $g'=gs_1\cdots s_n$. A subset of $G$ shall be called $S$-connected if the corresponding subgraph is connected. Since $S$ generates $G$, $G$ is $S$-connected.

For any finite subset $T$ of $G$, define $B_T^\infty$ to be the set of infinite $S$-connected components of $G-T$. Suppose now we have another symmetric finite generating subset $S'$ of $G$ such that $S\subset S'$. Define $B_T'^ \infty$ just like $B_T^\infty$ with $S'$ in place of $S$. Obviously, $S$-connected sets are also $S'$-connected.

My question: I want to prove that, if there is $T$ such that $|B_T^\infty|\geq 2$, there is $U$ (maybe a superset of $T$?) such that $|B_U'^\infty|\geq 2$ (and the same with $3$ instead of $2$).

What I have tried: Let $A,B$ be two distinct elements of $B_T^\infty$. The problem is that, though $A\cup B$ is not $S$-connected, it may well be $S'$ -connected. My strategy would be to successively enlarge $T$ in a way that all possible $S'$-paths from $A$ into $B$ are "cut off". My problem with this is that $A$ and $B$ are infinite, and it is not clear at all to me that all possible paths from $A$ into $B$ have to pass through a certain finite set. Even if we remove points from $A$ and $B$, these sets have to stay infinite and the problem persists.

Background information: This comes from an exercise in Bourbaki's Algebra which concerns itself with the notion of the number of ends of a group. For $T\subset T'$ a mapping $f_{TT'}$ is defined of $B_{T'}^\infty$ into $B_{T}^\infty$ mapping $Z'$ to the unique component of $G-T$ that contains $Z'$. The set of ends is defined as the inverse (projective) limit of the family $(B_T^\infty)_{T\in\mathfrak{T}}$ with respect to the mappings $f_{TT'}$, where $\mathfrak{T}$ is the set of finite subsets of $G$. This set of ends depends on $S$, but its cardinality (the number of ends) does not. That's exactly the part where I am stuck. I've already shown that the cardinality can only be $0$ (iff $G$ is finite), $1$, $2$ or $2^{\aleph_0}$. This can be translated into conditions on the sets $B_T^\infty$. For example, the set of ends has exactly two elements iff all of the $B_T^\infty$ have at least one element, some have two, and none have three.

I've seen the term number of ends in the literature, but never defined in this particular way (at least I couldn't detect any connection). So I've tried to make the question self-contained without this background information.

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up vote 3 down vote accepted

Here is an idea that I think works, but I'll leave the details to you. Express every generator of $S'$ as a word in $S$ and let $n$ be the maximum length of all such words. Now define $U'$ to be the set of vertices at distance at most $n$ from a vertex in $U$.

The idea is that if two vertices are in different components with respect to $S$ but in the same component with respect to $S'$, then there must be two such vertices connected by a single $S'$-edge. If you wrote this edge out as a word in $S$, then at least one of the vertices of this word (in the $S$-Cayley graph) would lie in $U$. By extending $U$ to $U'$ you would eliminate this edge and thereby prevent the two components getting connected.

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That looks promising. I'll comment again, when I have worked out the details. – Stefan Apr 22 '12 at 19:37
Got it. Thank you! – Stefan Apr 30 '12 at 19:39

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