F -groups
In this section, we study the relation between the sets Endo(N1) and Endo(N2)
for commensurable F -groups N1 and N2. The main result shows that most
injective group morphisms ϕ ∈ Endo(N1) ⊆ Aut(NQ) have some power ϕk such
that ϕk ∈ Endo(N2).
In the previous section we showed that if ϕQ∈ Aut(NQ) is an automorphism
with characteristic polynomial in Z[X], then there exists a full subgroup N of
NQ such that ϕQ(N ) ≤ N . Of course, this does not mean that ϕQwill induce
a group morphism on every full subgroup of NQ.
Example 6.4. The matrix
A = 5 2 1 2 1 2 1 2 !
has characteristic polynomial p(X) = X2− 3X + 1 and thus there exists a full
subgroup of Qn such that the matrix A induces a morphism on this subgroup. An example of such a subgroup is the lattice spanned by the vectors
v1= 1 0 , v2= 1 2 1 1 .
But Z2 is also a full subgroup of Qn and A(Z2) 6≤ Z2, so A doesn’t induce a group morphism on Z2.
Let us assume that ϕ : N → N is an automorphism and consider the extension
ϕlat of ϕ to the lattice hull Nlat. The morphism ϕlatis also an automorphism
of Nlat by Proposition 2.30 and by fixing a basis for log(Nlat
) as Z-module, we get an isomorphism from nQ to Qn as vector spaces such that this isomorphism
maps log(Nlat) to Zn. Since ϕlat is an automorphism, we get that under this isomorphism ϕQ(Zn
) = Zn. So every automorphism ϕ ∈ Aut(N ) has | det(ϕQ)| = 1. We conclude that for automorphisms ϕ : N → N the extension
ϕQ is integer-like.
Vice versa, Theorem 6.1 and Proposition 2.30 imply that every integer-like automorphism of Aut(NQ) induces an automorphism on some full subgroup of
NQ. As Example 6.4 above shows, an integer-like automorphism of NQdoes not
induce an automorphism on every full subgroup of NQ. The following result,
see [28, Theorem 3.4.], shows that by taking some power of the automorphism, the statement does hold.
Theorem 6.5. Let N be an F-group with corresponding radicable hull NQ. If
ϕ ∈ Aut(NQ)is integer-like then there exists some k > 0 such that ϕk(N ) = N.
So if N1and N2are two full subgroups of NQand ϕ : N1→ N1an automorphism,
then some power of ϕ induces an automorphism of N2. For example, the third
power of the matrix A above is equal to
A3=17 4 4 1
∈ GL(2, Z).
This theorem is crucial for the study of Anosov diffeomorphisms on infra- nilmanifolds, as we will explain in Part III. It forms a crucial ingredient in the proof of Theorem 3.36.
Unfortunately, a similar theorem cannot be true in the more general case of automorphisms ϕ ∈ Aut(NQ) with characteristic polynomial in Z[X], not even
in the abelian case, as we can see from the following example.
Example 6.6. Consider the matrix
B = 1 2 1 2 −3 2 5 2 !
with characteristic polynomial p(X) = X2−3X +2 and take the vector v =0 1
. A computation shows that
Bk(v) = v +1 2 k X i=1 2i−11 3
and thus Bk(v) /∈ Z2for all k > 0. This shows us that the condition | det(ϕ)| = 1
is necessary in Theorem 6.5.
To simplify notations, we will also denote by ϕ ∈ Aut(NQ) the unique extension
of an injective group morphism ϕ : N → N . The main result of this section shows that we can partially restore Theorem 6.5 by putting constraints on the determinant of ϕ.
Theorem 6.7. Let N1 and N2 be F-groups with identical radicable hull NQ.
Then there exists a finite number of primes p1, . . . , plsuch that for every injective
group morphism ϕ : N1→ N1 with pj- det(ϕ) for all j ∈ {1, . . . , l}, there exists
some k > 0 such that ϕk(N
GROUP MORPHISMS OF COMMENSURABLE F -GROUPS 99
Since the discussion above shows that every integer-like automorphism of NQ
induces an automorphism on some full subgroup of NQ, this theorem is really a
generalization of Theorem 6.5.
The proof of this theorem uses a few technical lemmas. The first one gives us information about the index of the intersection of subgroups.
Lemma 6.8. Let H be any group with finite index subgroups K1and K2of index
respectively k1 and k2. If gcd(k1, k2) = 1, then K1∩ K2 is a subgroup of index
k1k2 and if K10 is a normal subgroup of index k1 such that K1∩ K2= K10 ∩ K2,
then K1= K10.
Proof. For the first statement, note that [H : K1∩ K2] = [H : K1][K1 :
K1∩ K2] = [H : K2][K2: K1∩ K2] and thus kj | [H : K1∩ K2] for j ∈ {1, 2}.
Since gcd(k1, k2) = 1 and [H : K1∩ K2] ≤ [H : K1][H : K2], we have that
[H : K1∩ K2] = k1k2. Note that this is equivalent to the fact that [K1 :
K1∩ K2] = k2.
For the second statement, we assume that K1∩ K2= K10 ∩ K2 and we show
that K1∩ K10 is a subgroup of index 1 in K1. As K10 is a normal subgroup,
K1K10 is a subgroup of H. Since the index of K1∩ K10 in K1 is equal to the
index of K10 in K1K10, it divides k1. From the first statement we know that
K1∩ K2= K1∩ K10∩ K2is a subgroup of index k2in K1. Since K1∩ K10∩ K2≤
K1∩ K10 ≤ K1, the index of K1∩ K10 in K1divides k2. As the index of K1∩ K10
in K1 divides both k1 and k2, we conclude from gcd(k1, k2) = 1 that it’s equal
to 1.
Note that the first statement of Lemma 6.8 is equivalent to the fact that K1∩K2
has index k1in K2or K1∩ K2 has index k2in K1.
As a consequence of Lemma 6.8, we have the following result.
Lemma 6.9. Let H be any group with a finite number of normal subgroups of index i and ϕ : H → H be an injective group morphism such that ϕ(H) is a subgroup of finite index in H and gcd([H : ϕ(H)], i) = 1. If we write the distinct normal subgroups of index i as K1, . . . , Km, then there exists a permutation
π ∈ Smsuch that ϕ(Kj) = Kπ(j)∩ ϕ(H) for all j ∈ {1, . . . , m}.
Proof. By the previous lemma, we know that Kj∩ ϕ(H) is a normal subgroup of
index i in ϕ(H) and that if Kj 6= Kj0 also Kj∩ ϕ(H) 6= Kj0∩ ϕ(H). Since ϕ(H)
is isomorphic to H, there are exactly m different normal subgroups of index i in
ϕ(H) and thus they are all of the form Kj∩ ϕ(H). Since also ϕ(K1), . . . , ϕ(Km)
are m distinct normal subgroup of index i in ϕ(H), we find a permutation
Note that the first condition of Lemma 6.9 is satisfied for every index i in a finitely generated group, see Theorem 11.1. Also, as we mentioned in Section 2.4, the image of an injective group morphism is always of finite index for these groups and the index is given by the absolute value of the determinant, see Proposition 2.30.
Take the notations of Lemma 6.9 and π ∈ Sm the permutation corresponding
to ϕ : H → H and some fixed index i. Then for any k > 0, we have that
ϕk(Kj) = ϕk−1 Kπ(j)∩ ϕ(H) = ϕk−1 Kπ(j) ∩ ϕk(H)
since ϕ is injective and thus by induction we get that πk is the permutation
corresponding to ϕk
The last lemma we need is just the abelian version of Theorem 6.7. A uniform lattice of the vector space Qn
is a full subgroup of Qn
where we consider Qn as
the radicable hull of Zn.
Lemma 6.10. Let L be a uniform lattice in Qn, then there exists primes p1, . . . , pl such that for every A ∈ GL(n, Q) with A(Zn) ≤ Zn and pi - det(A),
there exists some k > 0 such that Ak(L) ≤ L.
Proof. Fix a basis for L as Z-module (which also forms a basis for Qn as vector
space) and denote by P the matrix of change of basis from this basis to the standard basis. Take m the product of all the denominators of the entries of P and P−1 and take pi all the primes dividing m. We claim that the primes pi
satisfy the conditions of the lemma.
So assume A ∈ GL(n, Q) with A(Zn) ≤ Zn and pi - det(A). The matrix representation of Ak for the chosen basis in L is given by P−1AkP and thus we
have to check that P−1AkP has integer entries for some k > 0.
From the choice of the primes pi, it follows that A projects to an element of
GL(n, Zm) and write this projection as π(A). Since GL(n, Zm) is a finite group,
there exists some k > 0 such that π(A)k= π(Ak) = π(I
n). This means that m
divides every entry of Ak− I
n. So we have that
P−1AkP − In = P−1(Ak− In)P
has integer entries because of our choice of m and thus also P−1AkP has integer
entries.
With the help of these lemmas, we are ready to give a proof of the general version of Theorem 6.7.
GROUP MORPHISMS OF COMMENSURABLE F -GROUPS 101
Proof of Theorem 6.7. Every injective group morphism ϕ : N1 → N1 also
induces an injective group morphism ϕlat : N1lat → N1lat with the same
determinant, so we can always assume that N1 is a lattice group. By fixing a
basis for log(N1) as Z-module (which is also a basis for the vector space nQ),
the vector space nQ is isomorphic to Qn such that log(N
1) is equal to Zn under
this isomorphism.
Under this isomorphisms, the injective group morphisms of N1 correspond to
Lie algebra automorphisms which maps Zn
into Zn. If we assume that also N
2
is a lattice group, then log(N2) is a uniform lattice in nQ' Qnand the theorem
follows from Lemma 6.10.
Next we show that if the theorem is true for N2, it is also true for every normal
subgroup K of N2. Let p1, . . . , pl be the finite number of primes corresponding
to N2 and add all the primes that divide the index of K in N2. If ϕ : N1→ N1
is a group morphism satisfying the conditions of the theorem, then we can assume ϕ(N2) ≤ N2 by taking some power of ϕ. Since ϕ is an injective group
morphism of N2, we can apply Lemma 6.9. So take K1= K, . . . , Kmall normal
subgroups of the same index as K in N2 and by Lemma 6.9, we know that there
exists π ∈ Smwith ϕ(Kj) = Kπ(j)∩ ϕ(N2). Now take k > 0 such that πk = 1,
then
ϕk(K) = ϕk(K1) = Kπk(1)∩ ϕk(N2) = K1∩ ϕk(N2) ≤ K1= K
and therefore ϕk(K) ≤ K as we want.
For the proof in the general case, we know that N2 ≤ N2lat is a subgroup of
finite index. Since N2latis a nilpotent group, Theorem 2.15 implies that N2 is a
subnormal subgroup. This means we can find subgroups
H0= N2≤ H1≤ . . . ≤ Hm= N2lat
with Hj normal in Hj+1. By applying the result of the previous paragraph on
the subgroups Hj we conclude that the theorem also holds for N2.
Remark 6.11. The proof of Theorem 6.7 also gives us a way of computing a set
of primes as in the statement of the theorem. Let N2be a subgroup of index
i in the lattice group Nlat
2 and let P be the matrix of change of basis from
Nlat
1 to N2lat in the vector space nQ. Write D =
a
b ∈ Q with a, b ∈ Z for the
determinant of P . Then the set of primes is given by all the prime divisors of the index i, the integer a and the denominators of the entries of P .
In Chapter 7 we start from an injective group morphism ϕ : N1 → N1 and
construct full subgroups N2 such that the some power of ϕ induces a group
morphism on N2. Under some conditions, the points of N2 will then project to
For the particular case of integer-like automorphisms of NQ, the technicalities
of Lemma 6.9 are avoided. So the proof of Theorem 6.5 is now immediate, avoiding the use of Jonquière groups as in the original paper [28].
Proof of Theorem 6.5. Let ϕ : N1→ N1 be an automorphism. From Lemma
6.10 it follows that there exists some k > 0 such that ϕk ∈ Aut(Nlat
2 ). Since ϕk
permutes the subgroups of index [Nlat
2 : N2] in N2lat, there is some power such
that N2 is mapped to itself.
From the simplified proof of Theorem 6.5, we can also compute the power k depending on the full subgroup N .
Example 6.12. The subgroup L generated by v1, v2of Example 6.4 contains
Z2 as a subgroup of index 2. Since L has three subgroups of index 2, the
automorphism induced by A induces a permutation of S3. Since the order of
S3 is 6, the matrix A6∈ GL(2, Z). In fact, A induces a transitive permutation
of S3 and so already A3∈ GL(2, Z).