I study profinite group theory, and I need to show the following property:
Let $a_{n}(G)$ the number of subgroups of index $n$ in $G$ and $a_{n}(\hat{G})$ the number of open subgroups of index $n$ in $\hat{G}$;
Prove that $a_{n}(G)=a_{n}(\hat{G})$.
I think that it's enough to prove that start from the injection $j:G\longrightarrow \hat{G}$
(I suppose $G$ residually finite) we get a bijection between the set of subgroups $H$ of index $n$ in $G$ and the set of open subgroups $K$ of index $n$ in $\hat{G}$ given by $H \longrightarrow \bar{H}$ and $K \longrightarrow K\cap G$ and that the index in both case is preserved.
But how I can prove this?
Another little question: Does this implies that $\hat{G}/\bar{H} \cong G/H$, if $H,\bar{H}$ are normal subgroups ?
I prove first that a map as above is actually a bijection:
in fact, let $K\leq_{o} \hat{G}$, then since $G$ is dense in $\hat{G}$ (in $G$ i take the profinite topology), $K\cap G$ is dense in $K$ so $\overline{K\cap G}=K$.
Conversely, let $H\leq G$, then $H\subset \bar{H}\cap G$ (obvious), but $\bar{H}=\underleftarrow{\lim}_{N\trianglelefteq G}HN/N$, where the limit is over all the normal subgroup of $G$ with finite index. Now if $(gN)_{N}\in \bar{H}\cap G$ (with the identification $G\hookrightarrow \hat{G},$ $g\rightarrow (gN)_{N}$) then $gN\in HN$, $\forall N$, so $g\in HN$; but if I take $H=H_{G}$ then $g\in HH_{G}=H$ and we get $\bar{H}\cap G\subset H$.
Now I prove that this bijection preserve the index:
Let $H\leq_{o} G$ and let $n=[\hat{G}:\bar{H}]$; Since $G$ is dense in $\hat{G}$, $G\bar{H}=\hat{G}$ and so $\hat{G}=\bigcup_{i=1,..,n} \bar{H}t_{i}$; but we can take the transversal $t_{1},...,t_{n}$ in $G$ and we have $G=\bigcup_{i=1,..,n} Ht_{i}$, so $[G:H]=n$.
Finally I prove that if $H\trianglelefteq G$ then $\bar{H}\trianglelefteq \hat{G}$ and we have: $G/H\cong \hat{G}/\bar{H}$:
in fact, if $H\trianglelefteq G$ then $HN/N\trianglelefteq G/N$, $\forall N$, so $\bar{H}=\underleftarrow{\lim}_{N\trianglelefteq G}HN/N\trianglelefteq \hat{G}$;
Now if $H\trianglelefteq G$ the natural morphism $G\rightarrow \hat{G}/\bar{H}$ have kernel $G\cap \bar{H}=H$; since $[G:H]=[\hat{G}:\bar{H}]$ we get the conclusion.