LIFTING SUBGROUPS OF SYMPLECTIC GROUPS OVER Z/ℓZ

arXiv:1607.04698v2 [] 25 Mar 2017

AARON LANDESMAN, ASHVIN A. SWAMINATHAN, JAMES TAO, AND YUJIE XU
Abstract. For a positive integer g, let Sp2g (R) denote the group of 2g × 2g symplectic
matrices over a ring R. Assume g ≥ 2. For a prime number ℓ, we show that any closed
subgroup of Sp2g (Zℓ ) that surjects onto Sp2g (Z/ℓZ) must in fact equal all of Sp2g (Zℓ ).
Our result is motivated by group theoretic considerations that arise in the study of Galois
representations associated to abelian varieties.

1. Introduction
Let g be a positive integer, and for a ring R, denote by Sp2g (R) the group of 2g × 2g
symplectic matrices over R. Let Zℓ denote the ring of ℓ-adic integers, and consider the
natural projection map Sp2g (Zℓ ) → Sp2g (Z/ℓZ). In this paper, we show that when g > 1,
there are no proper closed subgroups of Sp2g (Zℓ ) that surject via this projection map onto
all of Sp2g (Z/ℓZ).
The case g = 1, in which Sp2 = SL2 , is well-understood. Indeed, as proven in [Ser98,
Lemma 3, Section IV.3.4], if ℓ ≥ 5 and Hℓ ⊂ SL2 (Zℓ ) is a closed subgroup that surjects onto
SL2 (Z/ℓZ), then Hℓ = SL2 (Zℓ ). The corresponding result for ℓ ∈ {2, 3} simply does not
hold: in each of these cases, there are nontrivial subgroups of Sp2g (Z/ℓ2 Z) that surject onto
SL2 (Z/ℓZ). See [Ser98, Section IV.3.4, Exercises 1 – 3] for exercises outlining a proof, and
also see for more comprehensive descriptions [DD12] for the case ℓ = 2 and [Elk06] for the
case ℓ = 3.
The objective of the present article is to generalize [Ser98, Lemma 3, Section IV.3.4], to
hold for all g ≥ 2. Our main theorem is stated as follows:
Theorem 1. Let g ≥ 2, let ℓ be a prime number, and let Hℓ ⊂ Sp2g (Zℓ ) be a closed
subgroup. If the mod-ℓ reduction of Hℓ equals all of Sp2g (Z/ℓZ), then Hℓ = Sp2g (Zℓ ), and
in particular, the mod-ℓk reduction of Hℓ equals all of Sp2g (Z/ℓk Z) for each positive integer k.
Remark 1. A more general version of Theorem 1 is proven for a large class of semisimple
Lie groups G in [Wei96, Theorem B] (except for the case that g = 3 and ℓ = 2) and also
in [Vas03, Theorem 1.3]. In the present article, we provide an elementary and self-contained
proof for the special case G = Sp2g . In particular, our inductive method circumvents the use
of Lie theory, and is therefore suitable for somewhat more general groups (e.g. certain finiteindex subgroups of matrix groups over Zℓ ) which arise in the study of Galois representations
associated to abelian varieties, cf. [LSTX16b].
Remark 2. To give a typical application, one can directly use Theorem 1 to reduce the
problem of checking that the ℓ-adic Galois representation associated to an abelian variety
has maximal image to the simpler problem of checking that the mod-ℓ reduction has maximal
image. Indeed, the conclusion of Theorem 1 has been applied many times in the study of
Galois representations, such as in [Zyw15, Proof of Lemma 2.4], [HL16, Proof of Theorem
Date: March 28, 2017.
1

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AARON LANDESMAN, ASHVIN A. SWAMINATHAN, JAMES TAO, AND YUJIE XU

8], [AP07, Proof of Corollary 3.5], [AP08, Proof of Lemma 5.1], [Wal14, p. 467], as well in
the authors’ own papers [LSTX16a] and [LSTX16b].
The rest of this paper is organized as follows. In Section 2.1, we introduce the basic
definitions and properties of the symplectic group. Next, in Section 2.2, we compute the
commutator subgroups of Sp2g (Zℓ ) and Sp2g (Z/ℓk Z) for every prime number ℓ and positive
integer k. Finally, in Section 3, we prove Theorem 1.
2. Background on Symplectic Groups
In this section, we first detail the basic definitions and properties of symplectic groups,
and then proceed to prove a number of lemmas that are used in our proof of Theorem 1.
2.1. Symplectic Groups. Fix a commutative ring R, let Mat2g×2g (R) denote the space of
2g × 2g matrices with entries in R, and let Ω2g ∈ Mat2g×2g (R) be defined by
"
#
0
idg
Ω2g ··=
,
− idg 0
where idg denotes the g × g identity matrix. We define the symplectic group Sp2g (R) as the
set of M ∈ SL2g (R) so that M T Ω2g M = Ω2g .
In the proof of Theorem 1, we will make heavy use of the “Lie algebra” sp2g (R), which is
defined by
sp2g (R) ··= {M ∈ Mat2g×2g (R) : M T Ω2g + Ω2g M = 0}.
It is easy to see that M T Ω2g + Ω2g M = 0 is equivalent to M being a block matrix with g × g
blocks of the form
"
#
A B
M=
,
C −AT
where B and C are symmetric.
In what follows, we specialize to studying symplectic groups over R = Z, R = Zℓ , or
R = Z/ℓk Z for ℓ a prime number and k a positive integer. We will adhere to the following
notational conventions:
• Let Hℓ ⊂ Sp2g (Zℓ ) be a closed subgroup.
• Let H(ℓk ) ⊂ Sp2g (Z/ℓk Z) be the mod-ℓk reduction of Hℓ .
• Notice that the map S 7→ id2g +ℓk S gives an isomorphism of groups
sp2g (Z/ℓZ) ≃ ker(Sp2g (Z/ℓk+1 Z) → Sp2g (Z/ℓk Z))
for every k ≥ 1. We will use the Lie algebra notation sp2g (Z/ℓZ) to denote the above
kernel when we want to think of its elements additively, and we will use the kernel
notation when we want to view its elements multiplicatively.
• For any group G, let [G, G] be its commutator subgroup, and let Gab = G/[G, G] be
its abelianization.

LIFTING SUBGROUPS OF SYMPLECTIC GROUPS OVER Z/ℓZ

3

2.2. Commutators. We shall now compute the abelianizations of Sp2g (Zℓ ) and Sp2g (Z/ℓk Z)
for every integer g ≥ 2, prime number ℓ, and positive integer k. It will first be convenient
for us to compute the abelianization Sp2g (Z)ab .
Lemma 2. The group Sp2g (Z)ab is trivial when g ≥ 3 and is isomorphic to Z/2Z when g = 2.
Proof. The case g ≥ 3 follows from [BMS67, Remark, p. 123], so it only remains to deal with
the case g = 2. By [Ben80, Satz], Sp4 (Z) has two generators K and L that satisfy several
relations, three of which are given as follows:
K 2 = id2g ,
L12 = id2g ,
(K · L5 )5 = (L6 · K · L5 · K · L7 · K)2 .
By the universal property of the abelianization, we have that Sp4 (Z)ab is a quotient of the
rank-2 free abelian group KZ ⊕ LZ generated by K and L. Thus, from the aforementioned
multiplicative relations between K and L in Sp4 (Z), we obtain the following additive relations
in the abelianization
2K = 0,
12L = 0,
5(K + 5L) = 2 · (6L + K + 5L + K + 7L + K).
Substituting the first two relations above into the third relation, we find L = K, which
implies that Sp4 (Z)ab is a quotient of (KZ ⊕ LZ)/(2K, K − L) ≃ Z/2Z.
It remains to show that Sp4 (Z) maps surjectively onto Z/2Z. Postcomposing the surjection
Sp4 (Z) ։ Sp4 (Z/2Z) with the isomorphism Sp4 (Z/2Z) ≃ S6 by [O’M78, 3.1.5] and then
applying the sign map S6 ։ Z/2Z yields the desired result.

b denote the profinite completion of Z. It follows immediately from Lemma 2,
Remark 3. Let Z
d
d
ab
ab
c1 ≃ G
c2 then G
together with the fact if G1 and G2 are groups with G
1 ≃ G2 , that the group
b ab is trivial for g ≥ 3 and is isomorphic to Z/2Z when g = 2.
Sp2g (Z)
Using Lemma 2, we can now compute the abelianizations of all aforementioned groups.

Proposition 3. We have the following results:
(a) The group Sp2g (Zℓ )ab is trivial except when g = ℓ = 2, in which case it is isomorphic
to Z/2Z.
(b) Let k ≥ 1. The group Sp2g (Z/ℓk Z)ab is trivial except when g = ℓ = 2, in which case
it is isomorphic to Z/2Z.
b ab is trivial, and since we have
Proof. We first verify Statement (a). For g ≥ 3, since Sp2g (Z)
b ։ Sp2g (Zℓ ), the result follows immediately. Now take g = 2. First,
a surjection Sp2g (Z)
b ։ Sp4 (Z2 ) ։ Sp4 (Z/2Z) ∼
we have surjections Sp4 (Z)
= S6 . Since the former and the latter
have abelianizations isomorphic to Z/2Z, using Lemma 2, it follows that Sp4 (Z2 )ab ∼
= Z/2Z.
Then, since we have
Y
Y
∼ Sp (Z2 )ab ×
b ab =
Sp (Zℓ )ab ,
Sp (Zℓ )ab ∼
Z/2Z ∼
= Z/2Z ×
= Sp (Z)
4

ℓ6=2

it follows that Sp4 (Zℓ )ab is trivial for ℓ 6= 2.

4

4

4

ℓ6=2

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AARON LANDESMAN, ASHVIN A. SWAMINATHAN, JAMES TAO, AND YUJIE XU

Now, observe that Statement (b) follows from Statement (a): indeed, notice that the
surjection Sp2g (Zℓ ) → Sp2g (Z/ℓk Z) induces a surjection Sp2g (Zℓ )ab → Sp2g (Z/ℓk Z)ab and in
the case of ℓ = g = 2, Sp4 (Z/2k Z)ab is nontrivial as it surjects onto Sp4 (Z/2Z)ab ∼
= Z/2Z. 
3. Proof of Theorem 1
In this section, we provide a complete proof of the main theorem of this paper, namely
Theorem 1. The basic strategy has two steps: lift from ℓ2 to ℓ∞ (see Section 3.1), and lift
from ℓ to ℓ2 (see Section 3.2). Considerable care must be taken in dealing with the cases
where ℓ = 2, 3, so we treat these situations separately (see Sections 3.3 and 3.4). We execute
this strategy as follows:
3.1. Lifting from ℓ2 to ℓ∞ for ℓ ≥ 3, and from 8 to 2∞ .
Lemma 4. If ℓ ≥ 3, then H(ℓ2 ) = Sp2g (Z/ℓ2 Z) implies Hℓ = Sp2g (Zℓ ). If ℓ = 2, then
H(8) = Sp2g (Z/8Z) implies H2 = Sp2g (Z2 ).
Proof. This is done in the g = 1 case in in [Ser98, Lemma 3, Section IV.3.4], which readily
generalizes to the case of g ≥ 2.

3.2. Lifting from ℓ to ℓ2 for ℓ ≥ 5.
Lemma 5. Fix g ≥ 2. If ℓ ≥ 5, then H(ℓ) = Sp2g (Z/ℓZ) implies H(ℓ2 ) ⊂ Sp2g (Z/ℓ2 Z).
Proof. It suffices to show that H(ℓ2 ) contains all of
ker(Sp2g (Z/ℓ2 Z) ։ Sp2g (Z/ℓZ)) = id2g +ℓ · sp2g (Z/ℓZ).
We prove this by taking ℓ-th powers of specific matrices. We want to pick M ∈ sp2g (Z/ℓZ)
such that id2g +M lies in Sp2g (Z/ℓZ) and such that M 2 = 0. As it happens, these two
conditions are equivalent: indeed, since M T Ω2g + Ω2g M = 0, we have
(id2g +M)T Ω2g (id2g +M) = Ω2g + M T Ω2g + Ω2g M + M T Ω2g M
= Ω2g + M T Ω2g M
= Ω2g − Ω2g M 2 ,
so the condition that id2g +M is symplectic is equivalent to the condition that M 2 = 0.
Moreover, by expanding in terms of matrices, we see that M 2 = 0 if and only if
"
#
A2 + BC
AB − BAT
= 0.
CA − AT C CB + (AT )2
So, if A2 = 0 and two of A, B, and C are zero, the matrix id2g +M will be symplectic.
For M ∈ sp2g (Z/ℓZ), choose an arbitrary lift of M in Mat2g×2g (Z/ℓ2 Z), and by abuse
of notation, also denote it M. By assumption, H(ℓ2 ) contains an element of the form
id2g +M + ℓV for some V ∈ Mat2g×2g (Z/ℓZ). This means that H(ℓ2 ) contains


 
ℓ
ℓ
2
ℓ
(M + ℓV )ℓ−1 + (M + ℓV )ℓ
(M + ℓV ) + · · · +
(id2g +M + ℓV ) ≡ id2g +ℓ(M + ℓV ) +
ℓ−1
2
≡ id2g +ℓM (mod ℓ2 ),
where the last step above relies crucially upon the assumption that ℓ ≥ 5. We conclude that
H(ℓ2 ) contains id2g +ℓM for every M satisfying the above conditions.

LIFTING SUBGROUPS OF SYMPLECTIC GROUPS OVER Z/ℓZ

5

Taking A = C = 0, we see that H(ℓ2 ) contains
"
#
0 B
id2g +ℓ ·
0 0
for any symmetric matrix B. Similarly, taking A = B = 0 shows that H(ℓ2) contains
#
"
0 0
id2g +ℓ ·
C 0
for any symmetric matrix C. Taking B = C = 0 shows that H(ℓ2 ) contains
"
#
A
0
id2g +ℓ ·
0 −AT
for any matrix A with A2 = 0. It is a standard fact that the span of such matrices A is the
space of trace zero matrices. Observe that tr A = 0 is a single linear condition on sp2g (Z/ℓZ)
that singles out a codimension-one linear subspace W ⊂ sp2g (Z/ℓZ), so that
(3.1)

id2g +ℓW ⊂ H(ℓ2 ) ∩ ker(Sp2g (Z/ℓ2 Z) ։ Sp2g (Z/ℓZ)).

Observe that if the inclusion in (3.1) were strict, then the right-hand side would be all of
id2g +ℓ · sp2g (Z/ℓZ), and the desired result follows. Therefore, suppose the inclusion in (3.1)
is an equality. Since W has index ℓ in sp2g (Z/ℓZ), it follows that H(ℓ2 ) has index ℓ in
Sp2g (Z/ℓ2 Z). We obtain a surjection
Sp2g (Z/ℓ2 Z) ։ Sp2g (Z/ℓ2 Z)/H(ℓ2 ) ∼
= Z/ℓZ,
which contradicts Proposition 3.



3.3. Lifting from 4 to 8.
Lemma 6. Fix g ≥ 2. Then H(4) = Sp2g (Z/4Z) implies H(8) = Sp2g (Z/8Z).
Proof. We modify the proof of Lemma 5. As in that proof, we consider a matrix
"
#
A B
M=
∈ sp2g (Z/2Z)
C −AT
with the property that id2g +2M lies in Sp2g (Z/4Z) and M 2 = 0. This time, however, the
first condition automatically holds because
(id2g +2M)T Ω2g (id2g +2M) ≡ Ω2g + 2(M T Ω2g + Ω2g M) + 4M T Ω2g M
≡ Ω2g

(mod 4).

Nevertheless, note that the second condition is again satisfied whenever A2 = 0 and two of
A, B, and C are zero.
Choose an arbitrary lift of M in Mat2g×2g (Z/4Z), and by abuse of notation also refer to
it as M. By assumption, H(8) contains an element of the form id2g +2M + 4V for some
V ∈ Mat2g×2g (Z/2Z), which means that H(8) contains
(id2g +2M + 4V )2 ≡ id2g +4M

(mod 8).

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AARON LANDESMAN, ASHVIN A. SWAMINATHAN, JAMES TAO, AND YUJIE XU

Taking W ⊂ sp2g (Z/2Z) to be the trace-zero subspace as before, this implies that
id2g +4 · W ⊂ H(8) ∩ ker(Sp2g (Z/8Z) ։ Sp2g (Z/4Z)).
If the inclusion were strict, again the right-hand side would contain the kernel of reduction,
so the desired result follows. Therefore, suppose the inclusion is an equality, so that H(8) has
index 2 in Sp2g (Z/8Z). If g ≥ 3, Proposition 3 tells us that Sp2g (Z/8Z)ab is trivial, which is
a contradiction. If g = 2, the same proposition tells us that H(8) = [Sp4 (Z/8Z), Sp4 (Z/8Z)],
so since the image of this commutator under the abelianization map
Sp4 (Z/8Z) ։ Sp4 (Z/2Z) ։ Z/2Z
is trivial, H(8) cannot surject onto Sp4 (Z/2Z), which is again a contradiction.



3.4. Lifting from 2 to 4 and from 3 to 9.
Proposition 7. Fix g ≥ 2. The following statements hold:
(a) Take ℓ = 2. Then H(2) = Sp2g (Z/2Z) implies that H(4) = Sp2g (Z/4Z).
(b) Take ℓ = 3. Then H(3) = Sp2g (Z/3Z) implies that H(9) = Sp2g (Z/9Z).
Idea of Proof. The argument will proceed by induction on g. We start by verifying the base
case g = 2 in Lemma 8. We then inductively assume this holds for g − 1 and prove it
for g. We use the inductive hypothesis to construct a particular element lying in H(ℓ2 ) in
Lemma 9. Then, we use Lemma 10 to show that conjugates of this particular element generate sp4 (Z/ℓ2 Z), embedded as in (3.8). We finally translate around this copy of sp4 (Z/ℓ2 Z)
to obtain that H(ℓ2 ) = Sp2g (Z/ℓ2 Z).
The following lemma deals with the base case:
Lemma 8. Proposition 7 holds in the case that g = 2.
Proof. Let ℓ ∈ {2, 3}. We proceed by induction on g. The following Magma code verifies the
base cases for g = 2, that the only subgroup of Sp4 (Z/ℓ2 Z) surjecting onto Sp4 (Z/ℓZ) is all
of Sp4 (Z/ℓ2 Z).
for l in [2,3] do
Z := Integers();
G := GL(4,quo<Z|l*l>);
A := elt<G | 1,0,0,0, 1,-1,0,0, 0,0,1,1, 0,0,0,-1>;
B := elt<G| 0,0,-1,0, 0,0,0,-1, 1,0,1,0, 0,1,0,0>;
H := sub<G|A,B>;
maximals := SubgroupClasses(H: Al := "Maximal");
S := quo<Z|l>;
grp, f := ChangeRing(G, S);
for H in maximals do
if #f(H`subgroup) eq #Sp(4,l) then
assert false;
end if;
end for;
end for;
This concludes the proof of Lemma 8.



LIFTING SUBGROUPS OF SYMPLECTIC GROUPS OVER Z/ℓZ

7

In order to handle the inductive step of Proposition 7, we first introduce some notation.
Let φℓ,2g : Sp2g (Z/ℓ2 Z) ։ Sp2g (Z/ℓZ) denote the usual reduction map. Our next aim is
to define the maps π, ιℓ in the diagram
π

(3.2)

φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ))

Sp2g (Z/ℓ2 Z)

Sp2g−2 (Z/ℓ2 Z)
φℓ,2g−2

φℓ,2g

Sp2g−2 (Z/ℓZ)

ιℓ

Sp2g (Z/ℓZ)

The map ιℓ will be defined in (3.3) and the map π will be defined in (3.4) For the present
purpose, it is convenient to use a different Ω-matrix, which we shall denote by J2g , in the
definition of symplectic group Sp2g and its Lie algebra sp2g . Inductively define
#
"
#
"
J2g−2 0
0 1
.
and J2g ··=
J2 ··=
−1 0
0
J2
The matrix J2g is a block-diagonal matrix with each block being a copy of J2 .
Now let M ∈ Sp2g−2 (Z/ℓ2 Z). The map
#
"
M 0
M 7→
0 id2
gives an inclusion
ιℓ2 : Sp2g−2 (Z/ℓ2 Z) ֒→ Sp2g (Z/ℓ2 Z)
which, when taken modulo ℓ, reduces to an inclusion
(3.3)

ιℓ : Sp2g−2 (Z/ℓZ) ֒→ Sp2g (Z/ℓZ).

Thus, the group φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ)) (see the diagram in (3.2)) consists of matrices satisfying
"
# "
#
A(2g−2)×(2g−2) B(2g−2)×2
M 0
≡
(mod ℓ)
C2×(2g−2)
D2×2
0 id2
where M ∈ Sp2g−2 (Z/ℓZ) and the blocks A, B, C, D have the indicated sizes. For two such
matrices, we have
# "
# "
#
"
A2 B2
A1 A2 + B1 C2 A1 B2 + B1 D2
A1 B1
·
≡
C 1 D1
C 2 D2
C1 A2 + D1 C2 C1 B2 + D1 D2
"
#
A1 A2
A1 B2 + B1 D2
≡
(mod ℓ2 ),
C1 A2 + D1 C2
D1 D2
where the last step follows because B1 C2 ≡ C1 B2 ≡ 0 (mod ℓ2 ). Therefore, the map
#
"
B
A
2
(3.4)
π : φ−1
7→ A
ℓ,2g (Sp2g−2 (Z/ℓZ)) ։ Sp2g−2 (Z/ℓ Z) sending
C D
is a group homomorphism. This completes the definition of the maps ιℓ and π, and it is
apparent that the diagram in (3.2) commutes.

8

AARON LANDESMAN, ASHVIN A. SWAMINATHAN, JAMES TAO, AND YUJIE XU

With this notation set, we now continue our proof of Proposition 7. In Lemma 9, we show
via explicit matrix multiplication that one of two particular matrices lies in H(ℓ2 ).
2
Lemma 9. Suppose that π(H(ℓ2) ∩ φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ))) = Sp2g−2 (Z/ℓ Z) and H(ℓ) =
Sp2g (Z/ℓZ). Then, defining



(3.5)




·
Φ ·= id2g +ℓ · 






(3.6)




Ψ ··= id2g +ℓ · 






−1 −1 0 0
0

1

0 0

0

0

0 0

0

0

0 0

0(2g−4)×4

04×(2g−4)

0(2g−4)×(2g−4)

−1 −1 1 0
0

1

0 0

0

0

0 0

0

04×(2g−4)

−1 0 0
0(2g−4)×4

0(2g−4)×(2g−4)









and






,




we have that either Φ or Ψ lies in H(ℓ2 ).
2
Proof. Since we are assuming π(H(ℓ2 )∩φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ))) = Sp2g−2 (Z/ℓ Z), it follows that




id2g−2 +ℓ · 

0 0
1 0

02×(2g−4)

0(2g−4)×2 0(2g−4)×(2g−4)





2
lies in π(H(ℓ2) ∩ φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ))). It follows that H(ℓ ) contains an element of the form
M = id2g +ℓU for


0 0
02×(2g−4)
A2×2 

 1 0


,
U =

 0(2g−4)×2 0(2g−4)×(2g−4) B(2g−4)×2 
′
A′2×2
B2×(2g−4)
C2×2

where there is a linear relation between A2×2 and A′2×2 , as well as a linear relation between
′
B(2g−4)×2 and B2×(2g−4)
, imposed by the symplectic constraint M T J2g M = J2g . For any
M ∈ H(ℓ2), the group H(ℓ2 ) also contains
M −1 (id2g +ℓU)M = id2g +ℓM −1 UM,

where the right-hand-side only depends on the reduction of M modulo ℓ. Since H(ℓ2 ) surjects
onto Sp2g (Z/ℓZ), the matrix M (mod ℓ) ranges over all elements of Sp2g (Z/ℓZ). With this

LIFTING SUBGROUPS OF SYMPLECTIC GROUPS OVER Z/ℓZ

in mind, take a matrix M given by

1 1
02×(2g−4)
02×2

 0 1

M ··= 
 0(2g−4)×2 id2g−4 0(2g−4)×2
02×2
02×(2g−4)
id2

so that H(ℓ2 ) contains










9

(mod ℓ)

"

−1 −1

02×(2g−4)


0
1


(3.7) M −1 (id2g +ℓU)M = id2g +ℓ · 
0(2g−4)×2
0(2g−4)×(2g−4)


"
#

 A′ · 0 1
02×(2g−4)
2×2
0 0

0 −1
0



#

· A2×2 



.
0(2g−4)×2




0
0

2×2

Multiplying (3.7) on the left by (id2g +ℓU)−1 ≡ id2g −ℓU (mod ℓ2 ) shows that H(ℓ2 ) contains
N ··= (id2g +ℓU)−1 M −1 (id2g +ℓU)M

"
#
−1 −1
0 −1
02×(2g−4)
· A2×2


0
1
0
0


= id2g +ℓ · 
0(2g−4)×2
0(2g−4)×(2g−4)
0(2g−4)×2


"
#

 A′ · 0 1
02×(2g−4)
02×2
2×2
0 0

Conjugating N by any matrix of the form


id2
02×(2g−4)
02×2


P ··=  0(2g−4)×2 id2g−4 0(2g−4)×2 
02×2

02×(2g−4)







.





(mod ℓ),

M2×2

where M2×2 ∈ Sp2 (Z/ℓ2 Z), results in the matrix
N ′ ··= P −1 NP



−1 −1

02×(2g−4)


0
1


= id2g +ℓ · 
0(2g−4)×2
0(2g−4)×(2g−4)


"
#

 M −1 · A′ · 0 1
02×(2g−4)
2×2
2×2
0 0

By choosing M2×2 judiciously, we may arrange that
"
#
"
#
0 −1
1 0
· A2×2 · M2×2 =
0 0
0 0

"

0 −1
0

0



#

· A2×2 · M2×2 



.
0(2g−4)×2




0
2×2

or

"

0 0
0 0

#

,

10

AARON LANDESMAN, ASHVIN A. SWAMINATHAN, JAMES TAO, AND YUJIE XU

depending on whether the bottom row
gating by the matrix

id2


02×2

 0
 (2g−2)×2 0(2g−6)×2
id2

of A2×2 is nonzero or zero, respectively. Upon conju02×(2g−2)
02×(2g−6)

id2

id2g−6

0(2g−6)×2

02×(2g−6)

02×2

we conclude that H(ℓ2 ) contains either Φ or Ψ.





 ∈ Sp2g (Z/ℓZ),





Next, in Lemma 10, we show that we can conjugate the matrices Φ and Ψ from Lemma 9
to obtain all of Sp4 (Z/ℓZ).
Lemma 10. Let ℓ = 2 or 3. Let Φ and Ψ denote the upper-left 4 × 4 blocks of Φ and Ψ,
respectively. Then the sets
 −1
 −1
M ΦM : M ∈ Sp4 (Z/ℓZ)
and
M ΨM : M ∈ Sp4 (Z/ℓZ)
are both spanning sets for 1 + ℓ · sp4 (Z/ℓZ).

Proof. The following Magma code verifies that each of the sets defined in the lemma statement
span sp2g (Z/ℓZ).
for l in [2, 3] do
Z := Integers();
G := GL(4,quo<Z|l*l>);
A := elt<G| 1,0,0,0, 1,1,0,0, 0,0,1,0, 0,0,0,1>;
B := elt<G| 1,1,0,0, 0,1,1,0, 1,1,1,1, 1,1,0,1>;
H := sub<G|A,B>;
grp, f := ChangeRing(G, quo<Z|l>);
Lie := Kernel(f) meet H;
M := elt<G| 1-l,-l,0,0, 0,1+l,0,0, 0,0,1,0, 0,0,0,1>;
N := elt<H| 1-l,-l,l,0, 0,1 + l,0,0, 0,0,1,0, 0,-l,0,1>;
#sub<H|Conjugates(H,M)> eq #Lie;
#sub<H|Conjugates(H,N)> eq #Lie;
end for;
This concludes the proof of Lemma 10



We now have the tools to prove Proposition 7.
Proof of Proposition 7. The base case g = 2 is the content of Lemma 8. Now take g ≥ 3, and
suppose the result holds for g − 1. We shall now use the inductive hypothesis to show that
ker φℓ,2g ⊂ H(ℓ2 ), which would imply that H(ℓ2) = Sp2g (Z/ℓ2 Z). Since H(ℓ2 ) surjects onto
Sp2g (Z/ℓZ), we have that the group H(ℓ2) ∩ φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ)) surjects onto Sp2g−2 (Z/ℓZ),
−1
2
and therefore so does the group π(H(ℓ ) ∩φℓ,2g (Sp2g−2 (Z/ℓZ))). By the inductive hypothesis,
2
π(H(ℓ2 )∩φ−1
ℓ,2g (Sp2g−2 (Z/ℓZ))) = Sp2g−2 (Z/ℓ Z), so upon applying Lemma 9, we deduce that
either Φ or Ψ lies in H(ℓ2 ). Now, conjugating by matrices of the form
#
"
04×(2g−4)
M4×4
(mod ℓ)
0(2g−4)×4 id2g−4

LIFTING SUBGROUPS OF SYMPLECTIC GROUPS OVER Z/ℓZ

11

where M4×4 ∈ Sp4 (Z/ℓZ) serves to conjugate the upper-left 4 × 4 of Φ and Ψ by M4×4 . By
Lemma 10, we have that sp4 (Z/ℓ2 Z) ⊂ H(ℓ2 ), embedded as
#
"
sp4 (Z/ℓZ)
04×(2g−4)
.
(3.8)
id2g +ℓ ·
0(2g−4)×4 0(2g−4)×(2g−4)
Construct Q ∈ Sp2g (Z/ℓZ) by taking any g ×g permutation matrix and replacing each 1 with
an id2 -block. Conjugating the subspace in (3.8) by various such Q shows that H(ℓ2) contains
ker φℓ,2g . This can be seen by a straightforward argument involving choosing a basis for
sp2g (Z/ℓZ) whose elements are elementary matrices or sums of two elementary matrices. 
3.5. Finishing the Proof. We are now in position to complete the proof of Theorem 1
Proof of Theorem 1. We split into three cases:
• Suppose ℓ ≥ 5. Then Lemma 5 implies H(ℓ2 ) = Sp2g (Z/ℓ2 Z).
• Suppose ℓ = 3. Then Lemma 7 implies H(9) = Sp2g (Z/9Z).
• Suppose ℓ = 2. Then Lemma 7 implies H(4) = Sp2g (Z/4Z), and Lemma 6 implies
H(8) = Sp2g (Z/8Z).
Combining the above results with Lemma 4 gives the desired conclusion.

Acknowledgments
This research was supervised by Ken Ono and David Zureick-Brown at the Emory University
Mathematics REU and was supported by the National Science Foundation (grant number
DMS-1557960). We would like to thank David Zureick-Brown for suggesting the problem
that led to the present article and for offering us his invaluable advice and guidance. We
would also like to thank David Zureick-Brown for providing the intuition behind the proof of
Lemma 5. We would like to thank Jackson Morrow, Ken Ono, and David Zureick-Brown for
making several helpful comments regarding the composition of this article. We would like
to acknowledge Michael Aschbacher and Nick Gill for their helpful advice. We used Magma
and Mathematica for explicit calculations.
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Department of Mathematics, Stanford University, Stanford, CA 94305
E-mail address, Aaron Landesman: aaronlandesman@stanford.edu
Department of Mathematics, Harvard College, Cambridge, MA 02138
E-mail address, Ashvin A. Swaminathan: aaswaminathan@college.harvard.edu
Department of Mathematics, Harvard College, Cambridge, MA 02138
E-mail address, James Tao: jamestao@college.harvard.edu
Department of Mathematics, Caltech, Pasadena, CA 91126
E-mail address, Yujie Xu: yujiex@caltech.edu

