2

Question: Let $G$ be a group. The finite residual of $G$ is the intersection of all finite index subgroups of $G$. We denote this with $\text{fr}(G)$. Note that $\text{fr}(-)$ is functorial on the category of groups and group homomorphisms. My question is: how does $\text{fr}(-)$ and free products relate?

Motivation: A group $G$ is called residually finite if $\text{fr}(G)$ vanishes, and residually finite groups have been studied a lot. It is a classical result that a residually finite and finitely generated group is Hopfian (Malcev 1940). That is, it is not (abstractly) isomorphic to any of its proper quotients. In a similar vein, one can use finite residuals to check whether certain quotients of groups $G/H$ are (abstractly) isomorphic to $G$. Indeed, finite residuals behave nicely with quotients and is functorial hence an invariant.

Elaboration: It is not too difficult to show that $\text{fr}(G)*\text{fr}(H)\subset \text{fr}(G*H)$. A proof is given below. I am interrested in when the reverse inclusion holds and examples of when it does not. For example, it is well-known that the free product of residually finite groups are residually finite. See e.g. here. Therefore, the reverse inclusion holds when $\text{fr}(G)=\text{fr}(H)=0$. A particular case i am interrested in is when either $G$ or $H$ is residually finite.

Proof that $\text{fr}(G)*\text{fr}(H)\subset \text{fr}(G*H)$: An alternative description of the finite residual is $$ \text{fr}(G)=\{ g\in G \mid \text{for all } \varphi:G\to K\text{ with $K$ finite}, \varphi(g)=1_K \}. $$ Consider a word $g_1h_1\cdots g_nh_n\in \text{fr}(G)*\text{fr}(H)$. Then since $g_i\in \text{fr}(G)$, we get that any homomorphism $\varphi:G\to K$ with $K$ finite maps $g_i$ to the identity. Similar for $h_i$. Since the free product is the categorical coproduct, we get by the universal property that any homomorphism $G*H\to K$ with $K$ finite is of the form $\varphi_G*\varphi_H$ where $\varphi_G:G\to K$ and $\varphi_H:H\to K$. These map all $g_i$'s and all $h_i$'s to the identity respectively hence $$ \varphi_G*\varphi_H(g_1h_1\cdots g_nh_n)=1_K. $$ This shows that $g_1h_1\cdots g_nh_n\in \text{fr}(G*H)$.

Frederik
  • 1,199
  • 2
    If $G$ is residually finite and $H$ is infinite simple then $\operatorname{fr}(G\ast H)$ is $\langle\langle H\rangle\rangle$, i.e. the smallest normal subgroup of $G\ast H$ containing $H$. This is strictly larger than the embedded copy of $\operatorname{fr}(G)\ast \operatorname{fr}(H)=H$. – user1729 May 03 '21 at 15:21
  • 2
    ${\rm fr}(GH)$ is a normal subgroup of $GH$, whereas ${\rm fr}(G)*{\rm fr}(H)$ is usually not, so you would not expect them to be equal. – Derek Holt May 03 '21 at 15:24
  • From stability of residual finiteness into free products it follows that $\mathrm{fr}(G\ast H)=\langle!\langle \mathrm{fr}(G)\ast\mathrm{fr}(H)\rangle!\rangle$ in general. – YCor May 04 '21 at 23:16
  • @YCor Could you elaborate on how you got this equality? I don't see how it follows that $\langle \langle \text{fr}(G)\text{fr}(H)\rangle \rangle\supset \text{fr}(GH)$. – Frederik May 05 '21 at 10:39

0 Answers0