Consider a finitely generated group. Assume that the first Betti number of the ball of radius n in the Cayley graph is at most polynomial in n. This property is satisfied by free groups and groups of polynomial growth. Are there any others?

  • 2
    $\begingroup$ The property seems to depend on the generating set. Consider the free group $\langle a,b\rangle$ with the generating set $\\{a,b,ab\\}$. The Cayley graph is a tree-graded space where pieces are triangles. A ball of radius $k$ has exponentially many triangles, each contributing 1 to the Betti number. So it seems that the growth is exponential if you use this generating set. Perhaps you should correct the question. $\endgroup$
    – user6976
    May 4, 2013 at 1:11
  • $\begingroup$ @Anton: My comment is equivalent to your answer. $\endgroup$
    – user6976
    May 4, 2013 at 1:18
  • $\begingroup$ @Mark: well, almost equivalent, almost simultaneous :) $\endgroup$ May 4, 2013 at 1:48

1 Answer 1


Assume the group is not free. Then its Cayley graph has a nontrivial loop. Applying shifts of the group you get a loop which starts at any element.

If the group grows faster then polynomially, then the number of disjoint cycles and therefore first Betti number grows faster than polynomial.

So, the answer is "yes" --- only free groups and groups of polynomial growth.


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