Generating set for abelianization of "mod $p$ commutator subgroup" of a free group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:44:59Z http://mathoverflow.net/feeds/question/87509 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87509/generating-set-for-abelianization-of-mod-p-commutator-subgroup-of-a-free-grou Generating set for abelianization of "mod $p$ commutator subgroup" of a free group Caleb 2012-02-04T05:58:36Z 2012-02-04T06:17:21Z <p>Let $F_n$ be a free group on $n$ letters, and fix some prime $p \geq 2$. Define</p> <p>$$K_{n,p}=\text{ker}(F_n \rightarrow H_1(F_n;\mathbb{Z}/p))$$</p> <p>and</p> <p>$$V_{n,p} = H_1(K_{n,p};\mathbb{Q}).$$</p> <p>For $x \in K_{n,p}$, let <code>$[x]_{n,p} \in V_{n,p}$</code> be the associated element. An element $x \in F_n$ is <strong>primitive</strong> if there is a free basis for $F_n$ containing $x$. Observe that for $x \in F_n$, we have $x^p \in K_{n,p}$. Define</p> <p><code>$$S_{n,p} = \{\text{[x^p]_{n,p} | x \in F_n is primitive}\}.$$</code></p> <p>Question : Does $S_{n,p}$ span $V_{n,p}$? My guess is that the answer is no, but I cannot figure out how to prove it.</p> <p>It is clear that $S_{n,p}$ would span if the primitivity condition were dropped. Of course, every element of $F_n$ is a product of primitive elements; however, these observations are not as helpful as they might appear since</p> <p><code>$$[(xy)^p]_{n,p} \neq [x^p]_{n,p} + [y^p]_{n,p}.$$</code></p> <p>The correct formula is actually</p> <p><code>$$[(xy)^p]_{n,p} = [x^p]_{n,p} + [y^p]_{n,p} + [y^{-p} x^{-p} (xy)^p]_{n,p},$$</code></p> <p>where of course we have $x^{-p} x^{-p} (xy)^p \in [F_n,F_n] \subset K_{n,p}$. Alas, however, in general we will not have $[x^{-p} x^{-p} (xy)^p]_{n,p} = 0$.</p>