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Let $G$ be a group. In the literature, I have encountered the following two definitions of residually finite group $G$:

Definition 1: $G$ is called residually finite if for all $x\ne 1$, there exists a normal subgroup $N\lhd G$ of finite index such that $x\notin N$.

This is e.g. the definition on Wikipedia, where equivalent characterisations are given.

On the other hand, I have also encountered the following definition in e.g. Brown-Ozawa's book "$C^*$-algebras and finite-dimensional approximations" or other $C^*$-literature:

Definition 2: $G$ is called residually finite if there exists a descreasing sequence of normal subgroups $$G \supseteq G_1 \supseteq G_2 \supseteq G_3 \supseteq \dots$$ such that $G_i$ is of finite index in $G$ for all $i\ge 1$ and such that $\bigcap_{i=1}^\infty G_i = \{1\}$.

My question: Are these definitions equivalent? It is clear to me that Definition 2 implies Definition 1, so concretely, I want to know why Definition 1 implies Definition 2.

I don't see how to construct the desired decreasing sequence of normal subgroups. Maybe, we need to assume that the group $G$ is countable?

Any help will be highly appreciated!

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    $\begingroup$ It's easy to see that Defn 1 implies Defn 2 if $G$ is countable, because you can just choose $G_n$ to exclude elements $g_1,\ldots,g_n$ of $G$, but I don't see it when $G$ is uncountable. $\endgroup$
    – Derek Holt
    Sep 6, 2023 at 7:59
  • $\begingroup$ @DerekHolt Thanks. The groups I'm interested in are countable, so this is great already. But for the sake of curiosity I will wait if someone can say something more in the general case. $\endgroup$
    – Andromeda
    Sep 6, 2023 at 8:25
  • $\begingroup$ Just a thought: a subgroup of a residually finite group is again residually finite. If $x \neq 1$, and $N \lhd G$, with $x \notin N$, $|G:N| \lt \infty$, then apply induction: $N$ is residually finite and the members of a sequence $\{N_i\}$ in $N$ still have finite index in $G$. $\endgroup$ Sep 6, 2023 at 8:40
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    $\begingroup$ If $G$ is residually finite in the sense of Defn 2 then the natural map from $G$ to the direct product $\prod G/G_i$ is injective, so $|G| \le 2^{\aleph_0}$. $\endgroup$ Sep 6, 2023 at 8:43
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    $\begingroup$ @SeanEberhard This shows, that in general, both definitions are not equivalent, right? Simply take a direct product of finite groups where the index set is large enough $(> 2^{\aleph_0})$). This is residually finite in the sense of definition 1, but not in the sense of definition 2. $\endgroup$
    – Andromeda
    Sep 6, 2023 at 8:55

1 Answer 1

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If $G$ is residually finite in the sense of Definition 2 then the natural map from $G$ to $\prod_{i=1}^n G/G_i$ is injective, so $|G| \le |\prod_{i=1}^n G/G_i| = 2^{\aleph_0}$. Therefore any residually finite (in the usual sense, Definition 1) group of cardinality $>2^{\aleph_0}$ is a counterexample. For example let $G$ be the direct product of $2^{\aleph_0}$ copies of $C_2$.

There are some uncountable groups obeying Definition 2, e.g., the direct product of $\aleph_0$ copies of $C_2$.

Here is a smaller counterexample. Let $G$ be the direct sum of uncountably many copies of $A_5$, i.e., $G = \bigoplus_{i \in I} H_i$ where each $H_i \cong A_5$ and $|I|$ is uncountable (e.g., $|I| = \aleph_1$). Then $|G| = |I|$. Any normal subgroup $G_i \trianglelefteq G$ is the direct sum of a subcollection, i.e., $G_i = \bigoplus_{j \in I_i} H_j$ for some subset $I_i \subset I$, and $G_i$ has finite index if and only if $I_i$ is cofinite. The intersection of a countable collection of such subgroups has the form $\bigoplus_{j \in J} H_j$ where $J \subset I$ is cocountable, so the intersection is nontrivial.

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  • $\begingroup$ Thanks for your answer! This completely solves my question. $\endgroup$
    – Andromeda
    Sep 6, 2023 at 9:17
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    $\begingroup$ Nice example Sean, +1 from me! And @Andromeda, great question! $\endgroup$ Sep 6, 2023 at 14:05

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