Let $G=\pi_1(\mathbb{C}-\{z_1,z_2\})$ be the fundamental group of the plane punctured twice. There exist two homotopy classes of loops $\mathfrak{g}_1=[\gamma_1]$ and $\mathfrak{g}_2=[\gamma_2]$ such that $G$ is the free group generated by $\mathfrak{g}_1,\mathfrak{g}_2$. Define $\mathfrak{g}_3=\mathfrak{g}_1\mathfrak{g}_2$. Let $e_1,e_2,e_3\in\mathbb{N}/{\{1\}}$ and define $H=\left\langle\mathfrak{g}_1^{e_1},\mathfrak{g}_2^{e_2},\mathfrak{g}_3^{e_3}\right\rangle\leq G$. Denote by $H^G=\left\langle\{ghg^{-1}\mid g\in G,h\in H\}\right\rangle$ the normal closure of $H$ in $G$.
I am trying to argue that for $i=1,2,3$ the following holds:$\quad \mathfrak{g}_i^m\in H^G\;\Rightarrow\;m\in e_i\mathbb{Z}$.
I know that $$\mathfrak{g}_i^m\in H^G\;\Rightarrow\;\mathfrak{g}_i^m=\prod_{k=1}^lg_k\mathfrak{g}_{i_k}^{\pm e_{i_k}}g_k^{-1}.$$ From here on, it seems super intuitive to me that the claim holds (at least I'd be very surprised if it does not). I'd like to just write that it follows immediately from $G$ being free that $g_1,...,g_l=1$. The rest would then be trivial. However, I don't feel confident just stating it like that (or should I?). I tried to get rid of the conjugates using the abelianization $\varphi: G\rightarrow G/{[G, G]}$, but then the images of $\mathfrak{g}_1,\mathfrak{g}_2,\mathfrak{g}_3$ have the relation $(\varphi(\mathfrak{g}_3))^n=(\varphi(\mathfrak{g}_1^n))(\varphi(\mathfrak{g}_2^n))$ for all $n\in\mathbb{N}$ (unlike as in $G$), which ruined the approach.
Thanks in advance for any suggestions on how to write this down nicely.
Using the theorem mentioned by Arturo Magidin, the claim is easily proven:
Let $K$ be a group with $a,b\in K$ such that $\operatorname{ord}(a)=e_1$, $\operatorname{ord}(b)=e_2$ and $\operatorname{ord}(ab)=e_3$, which exist due to the theorem. Define $\varphi:\{\mathfrak{g}_1,\mathfrak{g}_2\}\rightarrow K$ as the map that sends $\mathfrak{g}_1$ to $a$ and $\mathfrak{g}_2$ to $b$. By the universal property of $G$, there is a group homomorphism $\Phi:G\rightarrow K$ that extends $\Phi$, i.e. $\Phi(\mathfrak{g}_1)=a$ and $\Phi(\mathfrak{g}_2)=b$. Let $N=\operatorname{Ker}(\Phi)$, then $N$ is a normal subgroup of $G$ that contains $H$. Since $H^G$ is the intersection of all normal subgroup of $G$ that contain $H$, it follows that $H^G\leq N$. This proves the claim because $\left\langle\mathfrak{g}_i\right\rangle\cap N=\left\langle\mathfrak{g}_i^{e_i}\right\rangle$ for $i=1,2,3$. $\quad\square$