How to prove this group is free? Let $H:=\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}=\langle p,q| q^n=1\rangle.$ I consider the normal closure of $p$ denoted $N:=\langle q^kpq^{−k}|k\in\{0,\dots n−1\}\rangle.$I would like to prove that $N$ is free of rank $n$ on the set $\{p,p^q,p^{q^2},\ldots,p^{q^{n-1}}\}$.
I don't know if it is a good method but here is what I did. 
Let $\forall k\in\{0,\dots,n-1\},p_k=q^kpq^{-k}=q^kpq^{n-k}.$ I consider a word in $N$ $w=p_0^{k_0}\cdots p_{n-1}^{k_{n-1}}$ with $k_0,\dots,k_{n-1}\in\mathbb{Z}.$ I can rewrite it as $w=p_0^{l_0}\cdots p_{n-1}^{l_{n-1}}$ with $l_0,\dots,l_{n-1}\in\{0,\dots n-1\}.$ We have $w=p^{l_0}qp^{l_{1}}q\cdots p^{l_{n-1}}q.$ I thought that to prove $N$ is free, I should write $w$ as a word in $\{p,p^q,p^{q^2},\ldots,p^{q^{n-1}}\}$. But I don't know how to do because I don't see any link between $w$ and $p^{q^i}$ for $i\in\{0,\dots,n-1\}.$
By advance thank you
 A: The geometric proof proposed by Lee Mosher has the advantage that the method apply much more generally.
But it is not hard to solve this particular problem by completely elementary calculations. You have to prove that a general nonempty word of the form $p_{i_1}^{k_1}p_{i_2}^{k_2}\cdots p_{i_m}^{k_m}$, in which each $k_j \in {\mathbb Z} \setminus \{0\}$ and $i_j \ne i_{j+1}$ for $1 \le j < m$ in not equal to the identity in $G$.
This word reduces to
$$q^{i_1}p^{k_1}q^{i_2-i_1}p^{k_2} \cdots p^{k_j}q^{i_{j+1}-i_j}p^{k_{j+1}} \cdots p^{k_{m-1}}q^{k_m-k_{m-1}}p^{k_m}q^{-k_m},$$
and since each $k_j \ne 0$ and $i_{j+1} \ne i_j$, this is a reduced word in the free product, and it is manifestly not equal to the identity.
A: I would suggest instead a geometric proof using Bass-Serre theory. There's a nice picture that goes with this, although I'll instead describe the (less than) a thousand words that the picture is worth.
First I'll describe a Bass-Serre tree $T$ using the "graph of spaces" approach developed by Scott and Wall. Then I'll apply $T$ to show that $N$ is free.
To construct $T$, start with your favorite cell complex $L$ having fundamental group $\mathbb{Z}/n\mathbb{Z}$. Let $X$ be the cell complex constructed by attaching an edge $E$ with endpoints $x_0,x_1$ to the disjoint union of $S^1$ and $L$, with $x_0$ attached to a point of $S^1$ and $x_1$ to a point of $L$. By Van Kampen's theorem, $X$ has fundamental group $\mathbb{Z} * \mathbb{Z} / n \mathbb{Z}$.
Consider the universal covering map $f : \widetilde X \to X$. Form $T$ as the quotient space of $\widetilde X$ where each component of $f^{-1}(L)$ is collapsed to a point. The deck transformation action of $H$ on $\widetilde X$ descends to an action of $H$ on $T$. The fact that $T$ is a tree is a consequence of the Scott-Wall presentation of Bass-Serre theory. More formally, $T$ is the Bass-Serre tree of the graph of groups obtained from $X$ by collapsing $L$ to a point labelled by the group $\mathbb{Z}/n\mathbb{Z}$, this graph being the disjoint union of a circle and that labelled point with an edge connecting the circle and the point. 
To use this setup to prove that $N$ is free one uses the restriction to $N$ of the action of $H$ on $T$. There's a few ways to proceed, each of which is really about showing that $N$ acts freely on $T$, which implies that $N$ is free. Here's one very concrete way to do this.
It suffices to exhibit a finite subtree $\tau \subset T$ such that the intersection of $\tau$ with its translates under each of $p,p^q,p^{q^2},\ldots,p^{q^{n-1}}$ and their inverses is a collection of $2n$ distinct points. The reason this suffices is that one shows by induction that for any reduced word $w$ in those generators, the successive translates of $\tau$ under the sequence of terminal subwords of $w$ are pairwise disjoint except for single points of intersection for adjacent subwords in the sequence; hence $\tau \ne w\tau$ and so $w$ is nontrivial.
To construct $\tau$, start with the point $A \in T$ stabilized by the $\mathbb{Z}/n\mathbb{Z}$ subgroup of $H$. This point $A$ has valence $n$, connected by edges $E_0,\ldots,E_{n-1}$ to vertices $B_0,\ldots,B_{n-1}$, such that $B_i$ is contained in the line $L_i$ which is the axis of $p^{q^i}$, and such that $L_i \cap E_i = B_i$. Let $J_i \subset L_i$ be an interval with endpoint $B_i$ that is a fundamental domain for the action of the cyclic group $\langle p^{q^i} \rangle$ on $L_i$; let $C_i$ be the opposite endpoint of $J_i$.
On the line $L_i$ let $J_i$ be a fundamental domain for the action of the cylic group $\langle p^{q^i} \rangle$. Then we take
$$\tau = (E_0 \cup \cdots \cup E_{n-1}) \bigcup (J_0 \cup \cdots \cup J_{n-1})
$$
The intersection of $\tau$ with its translate under $p^{q^i}$ is $C_i$, and the intersection with its translate under the inverse of $p^{q^i}$ is $B_i$, and the $2n$ points $B_0,C_0,\ldots,B_{n-1},C_{n-1}$ are all distinct, as required to complete the construction.
