Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $F_n$ be a free group of finite rank $n$ and let $\gamma_m(F)$ denote the $m$-th term of the lower central series of $F$ where $m \ge 1$ is a natural number. Suppose that $x,y$ are primitive elements of $F$ that are equivalent modulo $\gamma_m(F)$: $$ x \equiv y\, (\mathrm{mod}\, \gamma_m(F)). $$ Can then both $\{x\}$ and $\{y\}$ be extended to bases $\{x,x_2,\ldots,x_n\}$ and $\{y,y_2,\ldots,y_n\}$ of $F_n$ so that $$ x_k \equiv y_k\, (\mathrm{mod}\, \gamma_m(F)) $$ for all $k=2,\ldots,n?$ Apparently, the statement is true for $n=2,$ but what about greater $n?$

share|improve this question
In $F(a, b)$, if $A=X\text{ mod }\gamma_m(F)$ ($m$ arbitrary) with $A$, $X$ primitive then (up to changing notation) $A=a^v$ and $X=a^w$ for some words $v, w$. The same does not hold for free groups of higher rank, and I think will be a sticking point. If you think about what I've said hard enough you should be able to prove it using some well-known result (the fact that $Out(F_2)\cong Out(\mathbb{Z}\times \mathbb{Z})$ looks promising). However, there is a paper called "What Does a Basis of $F(a, b)$ Look Like" by Cohen, Metzler and Zimmerman which will give you what I said almost immediately! –  user1729 Mar 1 '12 at 10:45
(what I am saying is that there is only one basis modulo $G^{\prime}$ in $F_2$, so the result is immediate.) –  user1729 Mar 1 '12 at 10:57
I said above that the case $n=2$ is pretty clear for me. Thanks for the comment, anyway. I could only add that if you think hard enough, you'll see that bases of $F_2$ are not conjugate modulo $F_2',$ but rather commutators $[a,b]$ where $\{a,b\}$ is any basis of $F_2.$ –  user23386 Mar 1 '12 at 12:02
According to Andy Putman, see this discussion at mathoverflow, the answer is affirmative for $m=2.$ –  user23386 Mar 4 '12 at 8:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.