Null sequences subgroup of Cauchy sequences Proof

Question

First some definitions:

Let $G$ be a group. Let $F$ be the family of all subgroups of finite index. Let $\{x_n\}$ be a sequence in $G$. We define this sequence to be Cauchy if given $H\in F$ there exists $n_0$ such that for $m,n\geqslant n_0$ we have $x_n x_m^{-1} \in H$. We define multiplication of two sequences componentwise.

Now, we define $\{x_n\}$ to be a null sequence if given $H\in F$ there exists $n_0$ such that for $n\geqslant n_0$ we have $x_n\in H$.

The question asks to prove that the null sequences $N$ form a normal subgroup of the Cauchy sequences $C$. I am able myself to prove that C is a group and that $N$ is a subgroup of $C$, but I am struggling on the fact that $N$ is normal. Does anyone have any hints/answers or help they could give? Thank you

Answer 1

Let $\mathcal C$ be the group of Cauchy sequences and $\mathcal N$ the group of null sequences of elements $\in G$. Let $F^\lhd$ be the set of all normal subgroups of finite index in $G$. Clearly $F^\lhd\subseteq F$. Every $H\in F$ gives rise to a an action of $F$ on the cosets, i.e., a homomorphism $F\to \operatorname{Sym}(F/H)$. The group on the right is finite, hence the kernel $N$ of this action is $\in F$, and clearly $N\le H$. That is, for every $H\in F$, we find $N\in F^\lhd$ with $N\le H$. Thus

$\{x_n\}$ is Cauchy if and only if for every $N\in F^\lhd$, there exists $n_0$ such that $x_nx_m^{-1}\in N$ for all $n,m\ge n_0$.

and

$\{x_n\}$ is null if and only if for every $N\in F^\lhd$, there exists $n_0$ such that $x_n\in N$ for all $n,m\ge n_0$.

For $N\in F^\lhd$, we have a homomorphism $\phi_N\colon \mathcal C\to G/N$ as follows: Given $\{x_n\}\in\mathcal C$, pick $n_0$ such that $x_nx_m^{-1}\in N$ for all $n,m\ge n_0$ and set $\phi_N(\{x_n\})=x_{n_0}N$. This is well-defined. By combining all $\phi_N$, we obtain a homomorphism $$\Phi\colon \mathcal C\to \prod_{N\in F^\lhd}G/N. $$ We see that $\ker\Phi=\mathcal N$. Therefore, $\mathcal N\lhd\mathcal C$.