Is the free product of residually finite groups always residually finite?

Solution 1:

Yes. This reduces to the case of a free product of finite groups, which is free-by-finite, and therefore residually finite.

To see this, take your alternating product $a = g_1 h_1 \cdots g_n h_n\neq 1$ in $G\ast H$, and choose normal subgroups $N$ and $K$, of finite index in $G$ and $H$, respectively, such that $a_1 , a_2 ,\ldots , a_n\not\in N$ and $b_1, b_2, \ldots, b_n \not\in K$. (This can be done since, by definition, there are normal subgroups of $G$ excluding each of the $a_i$, and then their intersection excludes all of them and still has finite index in $G$. Likewise for the $b_i$ in $H$.) Then the natural homomorphisms $G\to \overline{G} = G/N$ and $H\to\overline{H}=H/K$ extend to a homomorphism $\phi : G\ast H \to \overline{G}\ast\overline{H}$ for which $\overline{a} = \phi(a)\neq 1$. (Indeed, $\overline{a} = \overline{g_1}\overline{h_1}\cdots\overline{g_n}\overline{h_n}$ is a reduced alternating product in $\overline{G}\ast\overline{H}$ as all of the $\overline{a_i}$ and $\overline{b_j}$ are non-trivial in their respective finite factors.) Since $\overline{G}\ast\overline{H}$ is a free product of the finite groups $\overline{G}$ and $\overline{H}$, it is free-by-finite, so it is residually finite. Hence, there is a homomorphism $\psi : \overline{G}\ast\overline{H} \to Q$, where $Q$ is finite, such that $\psi(\overline{a})\neq 1$. Then $\psi(\phi(a))\neq 1$, and $\psi\circ\phi$ is a homomorphism from $G\ast H$ to the finite group $Q$ for which $(\psi\circ\phi)(a)\neq 1$.

(There is a tiny corner case to tidy up, where $a\in G\cup H$ but, as $G$ and $H$ are each homomorphic images of $G\ast H$, this is easy to handle.)