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The pro-$p$ completion of subgroups of principal congruence subgroups in $SL_n(\mathbb Z)$

Step-by-step mathematics solution: The pro-$p$ completion of subgroups of principal congruence subgroups in $SL_n(\mathbb Z)$

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1. What the question is asking (in plain language)

  • Let

[ \Gamma(m)={A\in SL_{n}(\mathbf Z)\;:\;A\equiv I\pmod m} ]

be the principal congruence subgroup of level (m).

  • Fix a prime (p) and a positive integer (k).
    Let

[ H\le \Gamma(p^{k}) ]

be any (abstract) subgroup.

  • The pro‑(p) completion of a group (G) is

[ \widehat G_{\,p}= \varprojlim_{N\triangleleft G,\;[G:N]\text{ a power of }p} G/N . ]

The question is: Is it enough to take the normal subgroups (N) of the special form
(N=H\cap\Gamma(p^{n})) (for (n\ge 1))?

Equivalently, does the inverse limit

[ \varprojlim_{n\ge 1}\; H\big/\bigl(H\cap\Gamma(p^{n})\bigr) \tag{1} ]

coincide with the pro‑(p) completion of (H)?

2. Detailed solution

We shall prove that the answer is yes.
The proof consists of three parts:

  1. The filtration ({\Gamma(p^{n})}_{n\ge 1}) makes (\Gamma(p)=\Gamma(p^{1})) a pro‑(p) group.
  2. In a pro‑(p) group, the open normal subgroups are exactly the subgroups of finite (p)‑power index; they form a neighbourhood basis at the identity.
  3. For a subgroup (H\le\Gamma(p^{k})) the induced pro‑(p) topology on (H) is given by the subgroups (H\cap\Gamma(p^{n})). Consequently (1) is the pro‑(p) completion of (H).

2.1 (\Gamma(p)) is a pro‑(p) group

For any (m\ge 1) the natural reduction map

[ \pi_m:\Gamma(p)\longrightarrow SL_{n}(\mathbf Z/p^{m}\mathbf Z) ]

has kernel (\Gamma(p^{m})).
Hence

[ \Gamma(p)/\Gamma(p^{m})\cong \operatorname{im}\pi_m\le SL_{n}(\mathbf Z/p^{m}\mathbf Z). ]

The group (SL_{n}(\mathbf Z/p^{m}\mathbf Z)) is a finite (p)-group when we look at the congruence filtration:

[ \Gamma(p^{m})/\Gamma(p^{m+1})\cong{\,I+p^{m}X\mid X\in M_{n}(\mathbf Z/p\mathbf Z),\; \operatorname{tr}X=0 \,}, ]

which is an elementary abelian (p)-group (additive group of trace‑zero matrices over (\mathbf F_{p})). Consequently each quotient (\Gamma(p)/\Gamma(p^{m})) is a finite (p)-group. Moreover

[ \bigcap_{m\ge 1}\Gamma(p^{m})={I} ]

because a matrix that is congruent to the identity modulo every power of (p) must be the identity. Thus (\Gamma(p)) is a pro‑(p) group: it is the inverse limit of the finite (p)-groups (\Gamma(p)/\Gamma(p^{m})).

2.2 Open normal subgroups of a pro‑(p) group

Let (G) be a pro‑(p) group.
A subgroup (U\le G) is open iff the index ([G:U]) is finite.
Because every finite quotient of (G) is a (p)-group, an open subgroup has (p)-power index; conversely, any normal subgroup of (p)-power index is open. Therefore

[ {\,U\triangleleft G\mid [G:U]=p^{r}\text{ for some }r\,} ]

is precisely the set of open normal subgroups of (G), and these subgroups form a neighbourhood basis of the identity.

A standard fact about the filtration ({ \Gamma(p^{n})}) inside (\Gamma(p)) is that it is cofinal among all open normal subgroups:

Lemma. For every open normal subgroup (U\triangleleft\Gamma(p)) there exists (n\ge 1) with (\Gamma(p^{n})\subseteq U).

Proof. The quotients (\Gamma(p)/\Gamma(p^{n})) are the successive quotients of the lower‑(p)-central series; they are powerful (p)-groups. In a powerful pro‑(p) group the (p)‑power map [ x\longmapsto x^{p} ] is surjective onto the next term of the filtration, i.e. ((\Gamma(p^{n}))^{p}=\Gamma(p^{n+1})).
Since the family ({\Gamma(p^{n})}) is a descending chain of open normal subgroups, any open normal (U) contains some term of the chain (otherwise the intersection (\bigcap_{n}\,U\Gamma(p^{n})) would be a proper open subgroup containing all the (\Gamma(p^{n})), contradicting (\bigcap_{n}\Gamma(p^{n})={I})). ∎

2.3 The induced pro‑(p) topology on a subgroup (H)

Let (H\le\Gamma(p^{k})\le\Gamma(p)).
Consider the family

[ \mathcal{B}:={\,H\cap\Gamma(p^{n})\mid n\ge 1\,}. ]

  • Each member of (\mathcal{B}) is a normal subgroup of (H) (intersection of two normal subgroups in the ambient group).
  • Because (\Gamma(p^{n})) has index a power of (p) in (\Gamma(p)), the same holds for (H\cap\Gamma(p^{n})) inside (H).

Now let (N\triangleleft H) be any normal subgroup with ([H:N]=p^{t}). Consider its closure (\overline N) inside the pro‑(p) group (\Gamma(p)): (\overline N) is an open normal subgroup of (\Gamma(p)) (open because the quotient (\Gamma(p)/\overline N) is a finite (p)-group, namely a quotient of (H/N)).
By the lemma of §2.2 there is an integer (n) such that (\Gamma(p^{n})\subseteq\overline N). Intersecting with (H) gives

[ H\cap\Gamma(p^{n})\subseteq H\cap\overline N = N . ]

Thus every normal subgroup of (p)-power index in (H) contains a member of (\mathcal{B}). In other words, the collection (\mathcal{B}) is cofinal in the directed set of all normal subgroups of (p)-power index in (H).

Consequences:

  • The pro‑(p) topology on (H) (the topology whose neighbourhood basis at the identity is formed by all normal subgroups of (p)-power index) is the same as the topology generated by the subgroups (H\cap\Gamma(p^{n})).

  • Therefore the pro‑(p) completion of (H) can be computed using only those subgroups:

[ \widehat H_{\,p}\;=\;\varprojlim_{N\triangleleft H,\;[H:N]=p^{r}} H/N \;=\;\varprojlim_{n\ge 1}\; H\big/\bigl(H\cap\Gamma(p^{n})\bigr). ]

This is exactly the inverse limit asked about in the problem statement.

2.4 Summary of the argument

Step Reason

Original question: The pro-$p$ completion of subgroups of principal congruence subgroups in $SL_n(\mathbb Z)$ on Mathematics Stack Exchange, licensed CC BY-SA.

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