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Given a finite subset $S$ not containing the identity element in a residually finite group $G$, does there always exist a normal subgroup of $G$ which has finite index (in $G$) and which avoids $S$? (If $S$ is a singleton, this is of course the definition of a residually finite group.)

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Yes. Simply take the intersection of the finitely many normal subgroups, each avoiding one of the elements of $S$. –  Keivan Karai Oct 28 '10 at 16:57
Thanks. (I was just trying to put something analogous as a comment myself, since this idea ocured to me a moment ago.) –  Roland Bacher Oct 28 '10 at 17:02

2 Answers 2

up vote 6 down vote accepted

Yes, take the intersection of the normal subgroups $N_1, N_2, ..., N_k$ of finite index avoiding elements $x_1,x_2,...,x_k$ of your set. It is normal and of finite index (at most the product of indices of $N_i$).

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Thank you (this was an easy question put on MO a little bit quickly as I discovered myself after submission.) –  Roland Bacher Oct 28 '10 at 17:04
Yes, it was easy. In terminology of the book by Hanna Neumann "Varieties of groups", a residually finite group is discriminated by finite groups (meaning exactly the property you are after). If you replace "finite" by some other property $A$, the property "residually $A$" and "disciminated by A" may be different. For example "residually free" and "discriminated by free" are two different concepts. –  Mark Sapir Oct 28 '10 at 17:34

The answer is yes. Since $G$ is residually finite, given $g\in S$ there is a finite group $F_g$ and a homomorphism $\phi_g: G\to F_g$ whose kernel does not contain $g$. Now $S$ is disjoint from the kernel of the product of these homomorphisms $G\to \prod_{g\in S}F_g$, and the target is a finite group.

More generally if $C$ is a class of groups closed under finite products, and $G$ is residually $C$, then $G$ is fully residually $C$.

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