1. ## 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 $N_x$ in G such that $x \notin N_x$.

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

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

I don't know whether I followed the correct path, but i'm stuck here.
I want to prove that $x \notin N'$. Please give me some hints.
Thank you.

2. Originally Posted by deniselim17

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

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