I am currently studying combinatorial group theory and am trying to prove the ping pong lemma. Many of the proofs I have come across seem to use a result of the following flavour.
Let $G$ be a group and $H_1,H_2 \leq G$ if every reduced word in $H_1 \backslash\{e\}\sqcup H_2 \backslash \{e\}$ is non-trivial, then $\left<H_1,H_2\right>=H_1 \ast H_2$ the free product. How would one go about proving this? I am currently attempting to use the universal property of free products.
Let $K$ be a group and $\varphi_1:H_1 \to K$ and $\varphi_2: H_2 \to K$ be homomorphisms then I have recursively defined the map $\varphi: \left<H_1,H_2\right> \to K$ by $$ g_0\ldots g_n \mapsto \begin{cases} \varphi_1(g_0) &\text{if $n=0$ and $g_0 \in H_1$}\\ \varphi_2(g_0) &\text{if $n=0$ and $g_0 \in H_2$}\\ \varphi_1(g_0)\varphi(g_1\ldots g_n) &\text{if $g_0 \in H_1$}\\ \varphi_2(g_0)\varphi(g_1\ldots g_n) &\text{if $g_0 \in H_2$} \end{cases} $$ Showing that this map preserves structure seems simple enough but how do I prove that it is well defined. My gut is telling me that it is here that the assumption that no reduced word is trivial comes into play but I do not know how to proceed. My intention is to continue by showing that $\varphi_1 = \varphi\circ\iota_{1}$ where $\iota:H_1 \to \left<H_1,H_2\right>$ is the inclusion map etc. hence $\varphi$ satisfies the universal property and $\left< H_1,H_2\right> = H_1 \ast H_2$.
I have a tendency of becoming entangled in the details and missing obvious results, especially when it comes to showing maps are well defined so I apologise if that is the case and appreciate any help. Thanks