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Aug 22, 2021 at 13:08 comment added markvs Try replacing $\mathbb Z$ by a residually finite hyperbolic linear group with property (T), say $SO(n,1)$.
Aug 22, 2021 at 12:35 comment added Doryan Temmerman I agree that for the example $\mathbb{Z} \ast C_2$ the set $\mathcal{F}$ is profinitely dense, and so that it can happen, thanks! The reason why I did not think of this example is that actually the groups $G$ I am considering also satisfy a stronger property (Kazhdan's Property (T)) which would imply that it can not be a free product itself. Note that for such groups the set $\mathcal{F}$ can be non-empty. Do you think that there is still an example among such groups?
Aug 22, 2021 at 0:20 history edited markvs CC BY-SA 4.0
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Aug 21, 2021 at 22:47 history answered markvs CC BY-SA 4.0