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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
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1 Answer

up vote 2 down vote accepted

The question has been answered in the comments, so I'm adding an actual answer to keep this question from staying "unanswered".

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.

Also, in the comments darij grinberg linked an interesting article which is about the problem in my question (link here). I was surprised to find out that this had a real-world "application"! In the article there is also a construction which gives a better solution (that is, a shorter word) to the problem than the iterated commutator.

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