Simplified Version: $N \unlhd F$, $a,b \in N$. $[F:N]=p$, prime. $a \sim b$ in $F$ (conjugates). Are $a \sim b$ in $N$? (more inside) I'm trying to understand the proof for Lemma 4.8 in Lyndon & Schupp Combinatorial Group Theory, page 26, proposition 4.8.
In the very end of this proof, we have: $N \unlhd F$ where $F$ is a finitely generated free group. $[F:N]=p$, $p$ prime. $K \unlhd N$, $K \unlhd F $. $u,v \in N$ such that their images in $N/K$ are not conjugate.
It is then claimed that the images of $u,v$ in $F/K$ are not conjugate as well. Why is that? 
 A: I don't have the book handy, but in your setup, a,b could easily be non-conjugate in N; similarly u,v could easily be conjugate in F/K.  Take F/K to be nonabelian of order 6, N/K to be index 2 and order 3.  None of its distinct elements are conjugate in N/K, but its non-identity elements are all conjugate in F/K.
In other words, something you haven't mentioned so far is critical for their proof.
A: As you have stated it, I think the result is not true. 
Here is a counterexample: for odd prime $p$, take the nonabelian group of order $p^3$ and exponent $p$, 
$$G = \langle a,b\,\bigm|\, a^p = b^p = [a,b]^p = [a,b,a]=[a,b,b]=1\rangle.$$
Let $F(x,y,z)$ be the free group on three generators. Consider $\varphi\colon F\to G$ given by $x\mapsto a$, $y\mapsto b$, and $z\mapsto [a,b]$. Let $K$ be the kernel of $\varphi$, and let $N$ be the preimage under $\varphi$ of the subgroup $\langle b,[a,b]\rangle$. Then $N\triangleleft F$, $K\triangleleft N$, $K\triangleleft F$, and $[F:N] = [G:\langle b,[a,b]\rangle] = p$. Note that since $N/K$ is abelian, if $u,v\in N$ are conjugate in $N/K$, then they are equivalent modulo $K$. Now consider $y$ and $yz$ in $N$; viewing $N/K$ as being $\langle b,[a,b]\rangle$, you have the elements $b$ and $b[a,b]$, which are not equal and hence not conjugate in $N/K$. However, $y^x = y[x,y]$. But $y[x,y] \equiv yz \pmod{K}$, since $z$ maps to $[a,b]$, and so does $[x,y]$. So $y$ and $yz$ 
are conjugate in $F/K$, even though they are not conjugate in $N/K$. 
You can cook up a similar example with the quaternion group or the dihedral group with $p=2$. 
So either the lemma is incorrect, or else your simplification has eradicated some crucial bit of information.
