# Question related to commutator collection process and free groups

Let $F$ be a free group, generated by $x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n$. Assume that $n > 1$. For every $i$, let $\varphi_i: F \rightarrow F$ be the endomorphism of $F$ defined by $\varphi_i(x_i) = \varphi_i(y_i) = 1$ and $\varphi_i(x_j) = x_j$, $\varphi_i(y_j) = y_j$ for $j \neq i$.

Is it true that $K = \bigcap_{i=1}^n \operatorname{Ker}(\varphi_i)$ is trivial? It is not when $n = 1$, because then $[x_1, y_1] \in K$. What about $n > 1$? I'm not sure where to start with this one. If a word $w \in K$ is nontrivial, then $w$ must be a product where each $x_i$ and $y_j$ occurs at least once. Furthermore, for each $i$ we can make $w$ trivial by removing every occurence of $x_i$ (or $y_i$) in $w$.

What about if we consider $F$ generated by $x_1, \ldots x_n$ ($n > 2$) and $\varphi_i: F \rightarrow F$ defined by $\varphi_i(x_i) = 1$ and $\varphi_i(x_j) = x_j$ for $j \neq i$?

The reason for the title of the question is that the $\varphi_i$ are used in Suzuki's Group Theory II to prove Philip Hall's formula for $(xy)^n$.

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$K$ is not trivial when $n>1$ either. For $n=2$, you get a counterexample from $\left[x_1,x_2\right]$. See arxiv.org/pdf/1203.3602v1.pdf . – darij grinberg Jan 26 '13 at 13:32
Aha, I took a look at the article and it seems that $[x_1, x_2, x_3, \ldots, x_n]$ works? Any $\varphi_i$ makes $x_i$ in the commutator the identity, so the commutator maps to identity! I should have thought about this a bit more.. – spin Jan 26 '13 at 13:46
In your definition of $\,\phi_i\,$ I think you meant $\,\phi_i(x_j)=x_j\,\,,\,i\neq j\,$ , and likewise for the $\,y_j'$s... – DonAntonio Jan 26 '13 at 22:28
Yes, spin, the iterated commutator does the trick. – darij grinberg Jan 26 '13 at 23:32
@DonAntonio: Right, thanks. – spin Jan 27 '13 at 11:31

The kernel is not trivial. In the two generator case $[x_1, x_2]$ works (as I noted in the question), and more generally $[x_1, x_2, \ldots, x_n]$ works.