# Equivalent definitions of residual finite groups

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!

• 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. Commented Sep 6, 2023 at 7:59
• @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. Commented Sep 6, 2023 at 8:25
• 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 $G$. Commented Sep 6, 2023 at 8:40
• 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 $|G| \le 2^{\aleph_0}$. Commented Sep 6, 2023 at 8:43
• @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. Commented Sep 6, 2023 at 8:55

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.