# A Normal subgroup of a free group generated via a fixed integer

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 $$v_i \pmod{n}=0$$.

• As Lee Mosher pointed out in his answer, the fact that $F$ is a free group is irrelevant and a distraction. This is true for any group $F$. Prove first that $g^{-1}R_ng = R_n$ for all $g \in F$. You then have $g^{-1}\langle R_n\rangle g = \langle g^{-1}R_ng \rangle = \langle R_n \rangle$. – Derek Holt Nov 24 '19 at 8:09

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.