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Let $G=A*B$. And let $N\unlhd A$ be a normal subgroup of A. Let $H\leq G$ be the kernel of the following map: $$\Psi:A*B\to A/N*1.$$ With Kurosh's Theorem there exists a splitting $H=(H\cap A)*(H\cap B)*F$, where $F\leq G$ is a free subgroup of $G$. Why can I choose $F\neq 1$? Or is $F$ always nontrivial? And if so, why?

Thanks for help.

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"Splitting" is a bad name for this, since in the context of Group Theory, a "splitting" generally refers to a retract of a projection (that is, expressing the group/subgroup as a semidirect product). "Factorization" is better for this. –  Arturo Magidin May 29 '12 at 16:44
    
@Arturo: The term "splitting" is pretty well established now for any graph of groups description of a group $G$, in particular for a free factorization of $G$, or a free factorization with amalgamated subgroup, or an HNN amalgamation, etc. –  Lee Mosher Jun 2 '12 at 18:39
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I got it by myself. For example elements of the form $a_1*b_1*a_2^{-1}*b_2$ aren't in $(H\cap A)*(H\cap B)$, if $a_1N=a_2N$ and $a_1,a_2\not\in N$, since $(H\cap A)=N$. But these elements have to lie in $\mathrm{Kern}\Psi$. So since the splitting exists because of Kurosh's Theorem, $F$ has to be nonitrivial, right?

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