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I'm reading a proof of the proposition bellow:

If $\Gamma=G_1*\cdots*G_n$ where each $G_i$ is residually finite, then the profinite topology of each $G_i$ is induced by the profinite topology of $\Gamma$.

and have some questions:

Firstly, is $\Gamma$ still residually finite?

Secondly, the author want to prove that for every $i$ and every subgroup $H$ of finite index in $G_i$, there exists a finite index subgroup $H_i$ of $\Gamma$, s.t. $H_i\cap G_i$ is contained in $H$. However, when trying to prove this, the author claims that we can assume that $H$ is normal in $G_i$, but I don't know why.

Thirdly, before the proof, the author provide a proof that each $G_i$ is a closed subset in the profinite topology of $\Gamma$, but I don't know the necessity of this.

Fourthly, even in the proof that each $G_i$ is a closed subset in the profinite topology of $\Gamma$, there may be some problem: the author wants to prove that the intersection of all finite index subgroups in $\Gamma$ containing $G_i$ is precisely $G_i$. Below is the proof:

let $G$ denote one of the $G_i$, and let $g\in \Gamma \backslash G$. Since $g\notin G$, the normal form for $g$ contains at least one element $a_k\in G_k\neq G$. Since $G_k$ is residually finite, there is a finite quotient $A$ of $G_k$ for which the image of $a_k$ is non-trivial. Using the projection homomorphism $G_1*\cdots *G_n\rightarrow G_k\rightarrow A$ defines a homomorphism for which the image of $G$ is trivial but the image of g is not.

Why is $g$ not trivial? What if there's some ${a_k}^{-1}$ in the normal form of $g$?

Finally, below is the original proof:

enter image description here

xtwxtw
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    The first question is a duplicate of https://math.stackexchange.com/questions/3145363/is-the-free-product-of-residually-finite-groups-always-residually-finite. The second follows from the fact that every finite index subgroup contains a normal finite index subgroup. – Moishe Kohan Jan 16 '25 at 04:28
  • @MoisheKohan Fine! Thinks! – xtwxtw Jan 16 '25 at 06:02
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    The element $g$ is non-trivial as it is in $\Gamma\setminus G$, and $1\in G$. I think your fourth point is good though. and the proof you highlighted is incorrect. For example, if $g=a_kba_k^{-1}b$ for $b\in G\setminus{1}$ then the image of $g$ is trivial although $g$ itself is not. The linked argument to show that $\Gamma$ is residually finite is much more careful than this, ensuring that every syllable in the normal form is not killed in a quotient to a residually finite group; in particular, $G$ is not necessarily killed. Are you willing to share where you found this proof? – user1729 Jan 16 '25 at 12:14
  • @user1729 A survey titled profinite properties of discrete groups written by Reid, page 9. Thanks! – xtwxtw Jan 17 '25 at 02:23
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    @xtwxtw Ah, cool. Alan Reid is a quality mathematician, which makes me doubt myself! I guess if noone gives a decent answer here, you could write to him? – user1729 Jan 17 '25 at 09:57
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    @user1729: You are right and Reid's argument is wrong. This happens. – Moishe Kohan Jan 18 '25 at 18:14

1 Answers1

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This was mostly answered in the comments, so I'll combine these into a proper answer now.

  1. Yes, $\Gamma$ is still residually finite. The idea is for a finite word $w=x_1x_2\cdots x_k$, find a quotient $K_i$ of each $G_i$ such that for each $x_j\in G_i$ which are in $w$, the image of $x_i$ is non-trivial. This induces a homomorphism $\Gamma\to K_1*K_2*\dots*K_n$, which is residually finite as it is the free product of finite groups. The result follows. Details are here.

  2. Every subgroup $K$ of finite index in $G$ contains a normal subgroup $H$, so we have $H\leq K\leq G_i$ with $H\lhd K$. The result follows from this. There is a proof here.

  3. The profinite topology on $\Gamma$ is efficient if it induces the full profinite topology on $G_v$ and $G_e$ for all vertex and edge groups, and $G_v$ and $G_e$ are closed in the profinite topology on $\Gamma$. So the author proves that the vertex groups $G_i$ are closed because...it is part of the definition... The first sentence of the proof - that the trivial group is residually finite - establishes that the edge groups are residually finite.

  4. The proof given is incorrect. Note that the element $g$ is non-trivial as it is in $\Gamma\setminus G$, and $1\in G$. To see the issue, if $g=a_kba^{-1}_kb$ for $b\in G\setminus\{1\}$ then the image of $g$ is trivial although $g$ itself is not. Argument in (1), above, to show that $\Gamma$ is residually finite is much more careful than this, ensuring that each $x_i$ in the normal form is not killed in a quotient to a residually finite group; in particular, $G$ is not necessarily killed. I'm not sure what the author was aiming for here.

user1729
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