Index for p-adic subgroups I always got some problems in computing precisely and understanding indexes for congruence-like subgroups. My problem seems quite simple: what is the index of $(1 + p^r)^2$ in $\mathbf{Z}_p^\times$ (I write $1 + p^r$ for $1 + p^r \mathbf{Z}_p$) ?
Since $1 + p^{2r} \subseteq (1+p^r)^2$, I could dominate it roughly, but it is not sufficient to me and it seems a quite bad estimate. Following Serre, it would give:
$$[\mathbf{Z}_p^\times:(1+p^r)^2] \leqslant [\mathbf{Z}_p^\times:1+p^{2r}] = \phi(p^{2r}) $$
Am I right ? Is it a better explicit understanding or computation of those squares ?
And a further question : is there any change if I replace $\mathbf{Z}_p$ by $F_p$ for $F$ a more general number field ?
Regards,
Wolker
 A: Although your notation ($1 + p^r$) is awful, I interpret it as being the subgroup $U_r = 1 + p^r \mathbf Z_p$ of the group of $p$-adic units $\mathbf Z_p^{*}$. Obviously $U_1 \cong (\mathbf Z_p,+)$, hence is a cyclic group such that all its subgroups have index equal to a power of $p$. In particular, $U_r$ is the unique subgroup of $U_1$ having idex $p^r$. Now, squaring elements in $U_1$ is equivalent to doubling elements in $\mathbf Z_p$, hence if $p$ is odd, the subgroup of squares of elements of $U_r$ is equal to $U_r$. Its index in $\mathbf Z_p^{*}\cong (\mathbf Z/(p-1) , +)$ x $(\mathbf Z_p , +)$ is then $(p-1)p^r$ . The case $p=2$ requires to take care of extra factors 2 which I leave to you (this is purely computational) . 
A: @Sameer Kulkami
(1) and (2): you are right of course. But since the beginning, the question was about multiples in $\mathbf Z_p$, which explains my sloppiness
(4): right again. Just a misprint
(3): you can consider the filtration of  the $U_r$'s and notice that $U_1$ is the inverse limit of the quotients $U_1 / U_r$ ; then determine the successive quotients $U_r / U_{r+1}$ , as is done e.g. in the first chapter of Cassels-Fröhlich                                                                              
