I am stuck at an exercise concerning a special subgroup of given free group.
Let $F$ be free group, $n$ a fixed integer and $N = \langle R_n\rangle$ the subgroup generated by the set $R_n:=\{x^n: x \in F\} $. Show that $N$ is a normal subgroup.
I know that $N$ is normal if and only if the left cosets $G/N$ form a group under the induced binary operation. So I considered the projection $\pi: F \to G/N, \ g \mapsto \ gN$. If I can show that $gNhN=ghN$ for all $g,h \in F$ I get a group structure on $G/N$.
I know that there is a set $X$ such that $F$ is the free group over X.. Hence $N$ can be written as $N=\langle\{(x_1^{k_1} \dots x_m^{k_m})^n: x_1,\dots ,x_m \in X, k_i \in\Bbb Z\}\rangle$.
Moreover, since $F$ is a free group, I can find $g_1,\dots ,g_a,h_1,\dots ,h_b \in X$ and $u_1,\dots ,u_a, v_1,\dots ,v_b \in Z$ such that $g=g_1^{u_1} \dots g_a^{u_a}$ and $h=h_1^{v_1} \dots h_b^{v_b}$.
So I get $gNhN=(g_1^{u_1} \dots g_a^{u_a})\ N \ (h_1^{v_1} \dots h_b^{v_b}) \ N$. But now I do not see how to proceed.
I think I have to rewrite $h_1^{v_1} \dots h_b^{v_b}$ in such a manner that $h_1^{v_1} \dots h_b^{v_b} \in N$. But I do not see how to do this, since I do not know whether $v_i \pmod{n}=0$.
Best Answer
Here's another useful way to think of a normal subgroup: $N$ is normal if for every $g \in N$ and every $h \in F$ we have $hgh^{-1} \in N$.
Let's first show that the generating set $R_n$ has the desired property: given $x^n \in R_n$ and $h \in F$ we have $$h x^n h^{-1} = \underbrace{(hxh^{-1}) \cdot (hxh^{-1}) \cdot \ldots \cdot (hxh^{-1})}_{\text{$n$ times}} = (hxh^{-1})^n \in R_n $$ And now let's show that everything in the subgoup $\langle R_n \rangle$ has this property: given $x_1^n x_2^n \ldots x_k^n \in \langle R_n \rangle$ and $h \in F$ we have \begin{align*} h(x_1^n \cdot x_2^n \cdot \ldots \cdot x_k^n)h^{-1} &= (h x_1^n h^{-1}) \cdot (h x_2^n h^{-1}) \cdot \ldots \cdot (h x_k^n h^{-1})\\ & = (hx_1h^{-1})^n \cdot (hx_2h^{-1})^n \cdot \ldots \cdot (h x_k h^{-1})^n \\ &\in R_n \end{align*} By the way, notice that this proof does not use that $F$ is a free group. It follows that the subgroup generated by the $n$th powers is normal in any group.