most recent 30 from http://mathoverflow.net 2013-05-21T08:31:55Z http://mathoverflow.net/feeds/question/58462 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58462/is-the-collection-process-for-commutators-potentially-infinite Anvita 2011-03-14T19:04:57Z 2011-03-14T21:43:22Z

<p>Let $F$ be a free group of finite rank $r>1$ and let $c_1,c_2,...$ be a full set of basic commutators for $F$. Upon completing the $k$-th step of the collection process for an arbitrary $g\in F$, one obtains a representation </p> <p>$g=c_1^{n_1}c_2^{n_2}\ldots c_{t}^{n_t}g_k$,</p> <p>where </p> <p>$c_1,\ldots,c_t$ are all basic commutators of weight less than $k$ ($t$ depending on $k$),</p> <p>$n_1,\ldots,n_t$ are integers,</p> <p>$g_k\in \gamma_k$, the $k$-th term of the lower central series for $F$.</p> <p><strong>Question.</strong> Is there always $g\in G$ such that the series $g_1,g_2,\ldots$ never terminates (i.e. has nonidentity terms $g_i$ for arbitrarily large $i$)?</p>

http://mathoverflow.net/questions/58462/is-the-collection-process-for-commutators-potentially-infinite/58482#58482 Jack Schmidt 2011-03-14T21:43:22Z 2011-03-14T21:43:22Z

<p>Yes. I believe (ab)^3 or so, at least (ab)^n for high enough n, never terminates. As far as I recall, only (ab)^1 and (ab)^2 terminate. If you start to collect them, I think you'll see the proof.</p>