@MaxHorn, Now i get it thank you so much for your time . . . is it true that if we have $H$ a subgroup of finite index $n$ in $G$, we can say that for $g \in G$, $g^n \in H$ ?? can we say that $G/H \cong \mathbb{Z}_n$ ??? because in our example we use the fact that $G/H \cong \mathbb{Z}_2$.

@MaxHorn, Thanks for your time . . . For the second part, we know that $<g^2>$ is a subgroup in $<g>$ and $g^2 \in F$, then $<F,g^2>$ has the same rank as $<F>$ now how I can prove that $<F,g>$ and $F$ have the same rank ?? $ is <g^2>$ and $<g>$ have different generator??

@MaxHorn $N$ has index $2$ in $G$ means that there exist $g_1$, $g_2$ in $G$ such that {$g_1N$,$g_2$} form a partition of $G$ and i defined $G/N$ to be the set of all elemants of the form $gN$ where $g \in G$. The group $\mathbb{Z}^{n-1}\times {0}$ is normal in $\mathbb{Z}^n$ but it is not isomorphic to $\mathbb{Z}^n$.

@MaxHorn thanks in advance for your time .... Yes, I know that if $N$ has index $2$ then it is normal in $G$ but can I prove that the quotient is of order 2?? For the second comment, you just use what I am trying to prove as a result to conclude that $<F,g^2>$ has the same rank as $<F,g>$

@MaxHorn How did you conclude or prove that $G/N \cong C_2$? and if we prove that $g^2 \in F$ we can only conclude that $<F,g^2>$ has the same rank as $F$ because $<g^2>$ is a subgroup in $<g>$ .... but we need to prove that $<F,g>$ has the same rank as $F$

@user43326 can you please tell me why this is true: Since $Sl(n,Z)$ has index $2$, its square is in $Sl(n,Z)$,

Yes, I know that $SL(n,\mathbb{Z})$ has a finite index in $GL(n,\mathbb{Z})$, my question is why the rank of their maximal free abelian subgroups are the same?

@YCor I really appreciate your help, it was extremely helpful.

I know that we have equality in the case of nilpotent Lie group ... to be specific I am looking for the case where G is semi-simple.

what about $G=\mathbb{Z}^n$ instead of $\mathbb{R}^n$?

I think the Lemma is not true, (2) and (3) are not necessarily equal, could you please tell me how did prove that?

So by using you notation, is $d(X)$ equal to the maximum integer such that $\mathbb{R}^{d(X)}$ act properly on X?