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 ?



  • 2
    $\begingroup$ If $G$ is a multiplicative abelian group, and $H$ is a subgroup of $G$, then $$H^2 = \{ xy : x, y \in H \}$$ is a subgroup of $G$ which is equal to $H$. Is this what you mean by $(1+p^r\mathbb{Z}_p)^2$? $\endgroup$ – D_S Jul 18 '16 at 13:57
  • $\begingroup$ Or do you mean $\{ x^2 : x \in H\}$? $\endgroup$ – D_S Jul 18 '16 at 14:00
  • $\begingroup$ In agreement with @D_S, I say that I might perhaps be able to help you with this question once I knew what $(1+p^r)^2$ was, $\endgroup$ – Lubin Jul 19 '16 at 17:32
  • $\begingroup$ @D_S Oh sorry, I meant the set of squares ! $\endgroup$ – Wolker Jul 20 '16 at 10:05

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) .

  • $\begingroup$ This is a perfect answer, and deserves more up-votes than the one I gave it. $\endgroup$ – Lubin Jul 28 '16 at 16:23
  • 1
    $\begingroup$ @nguyen quang do (1) I think what you mean by $U_1 \cong \mathbb Z_p$ being cyclic is that it topologically cyclic. Please confirm. (2) As far as I know all subgroups of $\mathbb Z_p$ do not have index equal to a power of $p$, only closed subgroups do. For example $\mathbb Z$ has infinite index in $\mathbb Z_p$. (3) I know the isomorphism $U_1 \cong (\mathbf Z_p,+)$ using $p$-adic exponentiation, do you know any proof that is shorter? (4) The index of $U_r$ in $U_1$ is $p^{r-1}$, not $p^r$? $\endgroup$ – Sameer Kulkarni Aug 1 '16 at 6:33
  • $\begingroup$ (1) Yes; (2) By a theorem of Nikolov and Segal, every finite index subgroup of a finitely generated profinite group is actually open, and hence closed. So anyway, using this cute hammer, these are all the subgroups by (1); $\endgroup$ – Shoutre Aug 12 '17 at 0:31

@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


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