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Thread: residually finite problem

  1. #1
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    residually finite problem

    i need to show that subgroups of residually finite groups are residually finite.
    i have 2 definitions for residually finite.

    definition 1: A group G is called residually finite if for all nontrivial element in G, there exists a normal subgroup with finite index $\displaystyle N_x$ in G such that $\displaystyle x \notin N_x$.

    definition 2: A group G is called residually finite if for all nontrivial element x in G, there is a homomorphism $\displaystyle f$ from G to a finite group such that $\displaystyle f(x)\ne1$.

    NonCommAlg proved the above by using definition 2.
    now i try to prove by using definition 1. im stuck somewhere and quite confused where to continue. please give me some hints.
    here is my proof:

    Let $\displaystyle G$ be a residually finite group.
    Let $\displaystyle H$ be a subgroup of $\displaystyle G$ and $\displaystyle 1 \ne x \in H$.
    By def. 1, there exists a normal subgroup with finite index $\displaystyle N$ in $\displaystyle G$ such that $\displaystyle x \notin N$.
    Let $\displaystyle N'$ be a normal subgroup with finite index in $\displaystyle H$.
    Since intersection of subgroups is also a subgroup, i have $\displaystyle N'$ as a subgroup of $\displaystyle G$.
    So, now i have $\displaystyle N \vartriangleleft G$ and $\displaystyle N'<G$.
    I apply 2nd isomorphism theorem here and obtained $\displaystyle N \cap N' ={\{1\}} \triangleleft N'$ and $\displaystyle N'/(N \cap N') \cong NN'/N$.

    I don't know whether I followed the correct path, but i'm stuck here.
    I want to prove that $\displaystyle x \notin N'$. Please give me some hints.
    Thank you.
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  2. #2
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    Quote Originally Posted by deniselim17 View Post

    Let $\displaystyle G$ be a residually finite group. Let $\displaystyle H$ be a subgroup of $\displaystyle G$ and $\displaystyle 1 \ne x \in H$. By def. 1, there exists a normal subgroup with finite index $\displaystyle N$ in $\displaystyle G$ such that $\displaystyle x \notin N$.
    so $\displaystyle N'=N \cap H$ is a normal subgroup of $\displaystyle H$ and $\displaystyle x \notin N'.$ the only thing we need to prove now is that $\displaystyle [H:N']< \infty$: define $\displaystyle f: \frac{H}{N'} \longrightarrow \frac{G}{N}$ by $\displaystyle f(hN')=hN.$ obviously $\displaystyle f$ is a one-to-one group

    homomorphism. thus $\displaystyle \frac{H}{N'}$ is isomorphic to a subgroup of $\displaystyle \frac{G}{N},$ which is a finite group because $\displaystyle N$ is finite index. hence $\displaystyle \frac{H}{N'}$ has to be finite too, i.e. $\displaystyle [H:N']< \infty.$
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