I’m collecting in this blog post a number of simple group-theoretic lemmas, all of the following flavour: if
is a subgroup of some product
of groups, then one of three things has to happen:
- (
too small)
is contained in some proper subgroup
of
, or the elements of
are constrained to some sort of equation that the full group
does not satisfy. - (
too large)
contains some non-trivial normal subgroup
of
, and as such actually arises by pullback from some subgroup of the quotient group
. - (Structure) There is some useful structural relationship between
and the groups
.
These sorts of lemmas show up often in ergodic theory, when the equidistribution of some orbit is governed by some unspecified subgroup
![{H}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BH%7D&bg=ffffff&fg=000000&s=0&c=20201002)
of a product group
![{G_1 \times \dots \times G_k}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_1+%5Ctimes+%5Cdots+%5Ctimes+G_k%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, and one needs to know further information about this subgroup in order to take the analysis further. In some cases only two of the above three options are relevant. In the cases where
![{H}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BH%7D&bg=ffffff&fg=000000&s=0&c=20201002)
is too “small” or too “large” one can reduce the groups
![{G_1,\dots,G_k}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_1%2C%5Cdots%2CG_k%7D&bg=ffffff&fg=000000&s=0&c=20201002)
to something smaller (either a subgroup or a quotient) and in applications one can often proceed in this case by some induction on the “size” of the groups
![{G_1,\dots,G_k}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_1%2C%5Cdots%2CG_k%7D&bg=ffffff&fg=000000&s=0&c=20201002)
(for instance, if these groups are Lie groups, one can often perform an induction on dimension), so it is often the structured case which is the most interesting case to deal with.
It is perhaps easiest to explain the flavour of these lemmas with some simple examples, starting with the
case where we are just considering subgroups
of a single group
.
Lemma 1 Let
be a subgroup of a group
. Then exactly one of the following hold:
- (i) (
too small) There exists a non-trivial group homomorphism
into a group
such that
for all
. - (ii) (
normally generates
)
is generated as a group by the conjugates
of
.
Proof: Let
be the group normally generated by
, that is to say the group generated by the conjugates
of
. This is a normal subgroup of
containing
(indeed it is the smallest such normal subgroup). If
is all of
we are in option (ii); otherwise we can take
to be the quotient group
and
to be the quotient map. Finally, if (i) holds, then all of the conjugates
of
lie in the kernel of
, and so (ii) cannot hold.
Here is a “dual” to the above lemma:
Lemma 2 Let
be a subgroup of a group
. Then exactly one of the following hold:
- (i) (
too large)
is the pullback
of some subgroup
of
for some non-trivial normal subgroup
of
, where
is the quotient map. - (ii) (
is core-free)
does not contain any non-trivial conjugacy class
.
Proof: Let
be the normal core of
, that is to say the intersection of all the conjugates
of
. This is the largest normal subgroup of
that is contained in
. If
is non-trivial, we can quotient it out and end up with option (i). If instead
is trivial, then there is no non-trivial element
that lies in the core, hence no non-trivial conjugacy class lies in
and we are in option (ii). Finally, if (i) holds, then every conjugacy class of an element of
is contained in
and hence in
, so (ii) cannot hold.
For subgroups of nilpotent groups, we have a nice dichotomy that detects properness of a subgroup through abelian representations:
Lemma 3 Let
be a subgroup of a nilpotent group
. Then exactly one of the following hold:
- (i) (
too small) There exists non-trivial group homomorphism
into an abelian group
such that
for all
. - (ii)
.
Informally: if
is a variable ranging in a subgroup
of a nilpotent group
, then either
is unconstrained (in the sense that it really ranges in all of
), or it obeys some abelian constraint
.
Proof: By definition of nilpotency, the lower central series
![\displaystyle G_2 := [G,G], G_3 := [G,G_2], \dots](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++G_2+%3A%3D+%5BG%2CG%5D%2C+G_3+%3A%3D+%5BG%2CG_2%5D%2C+%5Cdots&bg=ffffff&fg=000000&s=0&c=20201002)
eventually becomes trivial.
Since
is a normal subgroup of
,
is also a subgroup of
. Suppose first that
is a proper subgroup of
, then the quotient map
is a non-trivial homomorphism to an abelian group
that annihilates
, and we are in option (i). Thus we may assume that
, and thus
![\displaystyle G_2 = [G,G] = [G, HG_2].](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++G_2+%3D+%5BG%2CG%5D+%3D+%5BG%2C+HG_2%5D.&bg=ffffff&fg=000000&s=0&c=20201002)
Note that modulo the normal group
![{G_3}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_3%7D&bg=ffffff&fg=000000&s=0&c=20201002)
,
![{G_2}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_2%7D&bg=ffffff&fg=000000&s=0&c=20201002)
commutes with
![{G}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, hence
![\displaystyle [G, HG_2] \subset [G,H] G_3 \subset H G_3](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5BG%2C+HG_2%5D+%5Csubset+%5BG%2CH%5D+G_3+%5Csubset+H+G_3&bg=ffffff&fg=000000&s=0&c=20201002)
and thus
![\displaystyle G = H G_2 \subset H H G_3 = H G_3.](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++G+%3D+H+G_2+%5Csubset+H+H+G_3+%3D+H+G_3.&bg=ffffff&fg=000000&s=0&c=20201002)
We conclude that
![{HG_3 = G}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BHG_3+%3D+G%7D&bg=ffffff&fg=000000&s=0&c=20201002)
. One can continue this argument by induction to show that
![{H G_i = G}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BH+G_i+%3D+G%7D&bg=ffffff&fg=000000&s=0&c=20201002)
for every
![{i}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bi%7D&bg=ffffff&fg=000000&s=0&c=20201002)
; taking
![{i}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bi%7D&bg=ffffff&fg=000000&s=0&c=20201002)
large enough we end up in option (ii). Finally, it is clear that (i) and (ii) cannot both hold.
Remark 4 When the group
is locally compact and
is closed, one can take the homomorphism
in Lemma 3 to be continuous, and by using Pontryagin duality one can also take the target group
to be the unit circle
. Thus
is now a character of
. Similar considerations hold for some of the later lemmas in this post. Discrete versions of this above lemma, in which the group
is replaced by some orbit of a polynomial map on a nilmanifold, were obtained by Leibman and are important in the equidistribution theory of nilmanifolds; see this paper of Ben Green and myself for further discussion.
Here is an analogue of Lemma 3 for special linear groups, due to Serre (IV-23):
Lemma 5 Let
be a prime, and let
be a closed subgroup of
, where
is the ring of
-adic integers. Then exactly one of the following hold:
- (i) (
too small) There exists a proper subgroup
of
such that
for all
. - (ii)
.
Proof: It is a standard fact that the reduction of
mod
is
, hence (i) and (ii) cannot both hold.
Suppose that (i) fails, then for every
there exists
such that
, which we write as
![\displaystyle h = g + O(p).](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++h+%3D+g+%2B+O%28p%29.&bg=ffffff&fg=000000&s=0&c=20201002)
We now claim inductively that for any
![{j \geq 0}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bj+%5Cgeq+0%7D&bg=ffffff&fg=000000&s=0&c=20201002)
and
![{g \in SL_2({\bf Z}_p)}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bg+%5Cin+SL_2%28%7B%5Cbf+Z%7D_p%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, there exists
![{h \in SL_2({\bf Z}_p)}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bh+%5Cin+SL_2%28%7B%5Cbf+Z%7D_p%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)
with
![{h = g + O(p^{j+1})}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bh+%3D+g+%2B+O%28p%5E%7Bj%2B1%7D%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)
; taking limits as
![{j \rightarrow \infty}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bj+%5Crightarrow+%5Cinfty%7D&bg=ffffff&fg=000000&s=0&c=20201002)
using the closed nature of
![{H}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BH%7D&bg=ffffff&fg=000000&s=0&c=20201002)
will then place us in option (ii).
The case
is already handled, so now suppose
. If
, we see from the
case that we can write
where
and
. Thus to establish the
claim it suffices to do so under the additional hypothesis that
.
First suppose that
for some
with
. By the
case, we can find
of the form
for some
. Raising to the
power and using
and
, we note that
![\displaystyle h^p = 1 + \binom{p}{1} X + \binom{p}{1} pY + \binom{p}{2} X pY + \binom{p}{2} pY X](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+h%5Ep+%3D+1+%2B+%5Cbinom%7Bp%7D%7B1%7D+X+%2B+%5Cbinom%7Bp%7D%7B1%7D+pY+%2B+%5Cbinom%7Bp%7D%7B2%7D+X+pY+%2B+%5Cbinom%7Bp%7D%7B2%7D+pY+X+&bg=ffffff&fg=000000&s=0&c=20201002)
![\displaystyle + \binom{p}{3} X pY X + O(p^2)](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%2B+%5Cbinom%7Bp%7D%7B3%7D+X+pY+X+%2B+O%28p%5E2%29&bg=ffffff&fg=000000&s=0&c=20201002)
![\displaystyle = 1 + pX + O(p^2),](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%3D+1+%2B+pX+%2B+O%28p%5E2%29%2C&bg=ffffff&fg=000000&s=0&c=20201002)
giving the claim in this case.
Any
matrix of trace zero with coefficients in
is a linear combination of
,
,
and is thus a sum of matrices that square to zero. Hence, if
is of the form
, then
for some matrix
of trace zero, and thus one can write
(up to
errors) as the finite product of matrices of the form
with
. By the previous arguments, such a matrix
lies in
up to
errors, and hence
does also. This completes the proof of the
case.
Now suppose
and the claim has already been proven for
. Arguing as before, it suffices to close the induction under the additional hypothesis that
, thus we may write
. By induction hypothesis, we may find
with
. But then
, and we are done.
We note a generalisation of Lemma 3 that involves two groups
rather than just one:
Lemma 6 Let
be a subgroup of a product
of two nilpotent groups
. Then exactly one of the following hold:
- (i) (
too small) There exists group homomorphisms
,
into an abelian group
, with
non-trivial, such that
for all
, where
is the projection of
to
. - (ii)
for some subgroup
of
.
Proof: Consider the group
. This is a subgroup of
. If it is all of
, then
must be a Cartesian product
and option (ii) holds. So suppose that this group is a proper subgroup of
. Applying Lemma 3, we obtain a non-trivial group homomorphism
into an abelian group
such that
whenever
. For any
in the projection
of
to
, there is thus a unique quantity
such that
whenever
. One easily checks that
is a homomorphism, so we are in option (i).
Finally, it is clear that (i) and (ii) cannot both hold, since (i) places a non-trivial constraint on the second component
of an element
of
for any fixed choice of
.
We also note a similar variant of Lemma 5, which is Lemme 10 of this paper of Serre:
Lemma 7 Let
be a prime, and let
be a closed subgroup of
. Then exactly one of the following hold:
- (i) (
too small) There exists a proper subgroup
of
such that
for all
. - (ii)
.
Proof: As in the proof of Lemma 5, (i) and (ii) cannot both hold. Suppose that (i) does not hold, then for any
there exists
such that
. Similarly, there exists
with
. Taking commutators of
and
, we can find
with
. Continuing to take commutators with
and extracting a limit (using compactness and the closed nature of
), we can find
with
. Thus, the closed subgroup
of
does not obey conclusion (i) of Lemma 5, and must therefore obey conclusion (ii); that is to say,
contains
. Similarly
contains
; multiplying, we end up in conclusion (ii).
The most famous result of this type is of course the Goursat lemma, which we phrase here in a somewhat idiosyncratic manner to conform to the pattern of the other lemmas in this post:
Lemma 8 (Goursat lemma) Let
be a subgroup of a product
of two groups
. Then one of the following hold:
- (i) (
too small)
is contained in
for some subgroups
,
of
respectively, with either
or
(or both). - (ii) (
too large) There exist normal subgroups
of
respectively, not both trivial, such that
arises from a subgroup
of
, where
is the quotient map. - (iii) (Isomorphism) There is a group isomorphism
such that
is the graph of
. In particular,
and
are isomorphic.
Here we almost have a trichotomy, because option (iii) is incompatible with both option (i) and option (ii). However, it is possible for options (i) and (ii) to simultaneously hold.
Proof: If either of the projections
,
from
to the factor groups
(thus
and
fail to be surjective, then we are in option (i). Thus we may assume that these maps are surjective.
Next, if either of the maps
,
fail to be injective, then at least one of the kernels
,
is non-trivial. We can then descend down to the quotient
and end up in option (ii).
The only remaining case is when the group homomorphisms
are both bijections, hence are group isomorphisms. If we set
we end up in case (iii).
We can combine the Goursat lemma with Lemma 3 to obtain a variant:
Corollary 9 (Nilpotent Goursat lemma) Let
be a subgroup of a product
of two nilpotent groups
. Then one of the following hold:
- (i) (
too small) There exists
and a non-trivial group homomorphism
such that
for all
. - (ii) (
too large) There exist normal subgroups
of
respectively, not both trivial, such that
arises from a subgroup
of
. - (iii) (Isomorphism) There is a group isomorphism
such that
is the graph of
. In particular,
and
are isomorphic.
Proof: If Lemma 8(i) holds, then by applying Lemma 3 we arrive at our current option (i). The other options are unchanged from Lemma 8, giving the claim.
Now we present a lemma involving three groups
that is known in ergodic theory contexts as the “Furstenberg-Weiss argument”, as an argument of this type arose in this paper of Furstenberg and Weiss, though perhaps it also implicitly appears in other contexts also. It has the remarkable feature of being able to enforce the abelian nature of one of the groups once the other options of the lemma are excluded.
Lemma 10 (Furstenberg-Weiss lemma) Let
be a subgroup of a product
of three groups
. Then one of the following hold:
- (i) (
too small) There is some proper subgroup
of
and some
such that
whenever
and
. - (ii) (
too large) There exists a non-trivial normal subgroup
of
with
abelian, such that
arises from a subgroup
of
, where
is the quotient map. - (iii)
is abelian.
Proof: If the group
is a proper subgroup of
, then we are in option (i) (with
), so we may assume that
![\displaystyle \{ h_3 \in G_3: (1,h_2,h_3) \in H \} = G.](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5C%7B+h_3+%5Cin+G_3%3A+%281%2Ch_2%2Ch_3%29+%5Cin+H+%5C%7D+%3D+G.&bg=ffffff&fg=000000&s=0&c=20201002)
Similarly we may assume that
![\displaystyle \{ h_3 \in G_3: (h_1,1,h_3) \in H \} = G.](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5C%7B+h_3+%5Cin+G_3%3A+%28h_1%2C1%2Ch_3%29+%5Cin+H+%5C%7D+%3D+G.&bg=ffffff&fg=000000&s=0&c=20201002)
Now let
![{g_3,g'_3}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bg_3%2Cg%27_3%7D&bg=ffffff&fg=000000&s=0&c=20201002)
be any two elements of
![{G}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG%7D&bg=ffffff&fg=000000&s=0&c=20201002)
. By the above assumptions, we can find
![{h_1 \in G_1, h_2 \in G_2}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bh_1+%5Cin+G_1%2C+h_2+%5Cin+G_2%7D&bg=ffffff&fg=000000&s=0&c=20201002)
such that
![\displaystyle (1, h_2, g_3) \in H](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%281%2C+h_2%2C+g_3%29+%5Cin+H&bg=ffffff&fg=000000&s=0&c=20201002)
and
![\displaystyle (h_1,1, g'_3) \in H.](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28h_1%2C1%2C+g%27_3%29+%5Cin+H.&bg=ffffff&fg=000000&s=0&c=20201002)
Taking commutators to eliminate the
![{h_1,h_2}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7Bh_1%2Ch_2%7D&bg=ffffff&fg=000000&s=0&c=20201002)
terms, we conclude that
![\displaystyle (1, 1, [g_3,g'_3]) \in H.](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%281%2C+1%2C+%5Bg_3%2Cg%27_3%5D%29+%5Cin+H.&bg=ffffff&fg=000000&s=0&c=20201002)
Thus the group
![{\{ h_3 \in G_3: (1,1,h_3) \in H \}}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7B%5C%7B+h_3+%5Cin+G_3%3A+%281%2C1%2Ch_3%29+%5Cin+H+%5C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
contains every commutator
![{[g_3,g'_3]}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7B%5Bg_3%2Cg%27_3%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, and thus contains the entire group
![{[G_3,G_3]}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7B%5BG_3%2CG_3%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
generated by these commutators. If
![{G_3}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_3%7D&bg=ffffff&fg=000000&s=0&c=20201002)
fails to be abelian, then
![{[G_3,G_3]}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7B%5BG_3%2CG_3%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
is a non-trivial normal subgroup of
![{G_3}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_3%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, and
![{H}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BH%7D&bg=ffffff&fg=000000&s=0&c=20201002)
now arises from
![{G_1 \times G_2 \times G_3/[G_3,G_3]}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_1+%5Ctimes+G_2+%5Ctimes+G_3%2F%5BG_3%2CG_3%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
in the obvious fashion, placing one in option (ii). Hence the only remaining case is when
![{G_3}](https://webcf.waybackmachine.org/web/20210812035634im_/https://s0.wp.com/latex.php?latex=%7BG_3%7D&bg=ffffff&fg=000000&s=0&c=20201002)
is abelian, giving us option (iii).
As before, we can combine this with previous lemmas to obtain a variant in the nilpotent case:
Lemma 11 (Nilpotent Furstenberg-Weiss lemma) Let
be a subgroup of a product
of three nilpotent groups
. Then one of the following hold:
- (i) (
too small) There exists
and group homomorphisms
,
for some abelian group
, with
non-trivial, such that
whenever
, where
is the projection of
to
. - (ii) (
too large) There exists a non-trivial normal subgroup
of
, such that
arises from a subgroup
of
. - (iii)
is abelian.
Informally, this lemma asserts that if
is a variable ranging in some subgroup
, then either (i) there is a non-trivial abelian equation that constrains
in terms of either
or
; (ii)
is not fully determined by
and
; or (iii)
is abelian.
Proof: Applying Lemma 10, we are already done if conclusions (ii) or (iii) of that lemma hold, so suppose instead that conclusion (i) holds for say
. Then the group
is not of the form
, since it only contains those
with
. Applying Lemma 6, we obtain group homomorphisms
,
into an abelian group
, with
non-trivial, such that
whenever
, placing us in option (i).
The Furstenberg-Weiss argument is often used (though not precisely in this form) to establish that certain key structure groups arising in ergodic theory are abelian; see for instance Proposition 6.3(1) of this paper of Host and Kra for an example.
One can get more structural control on
in the Furstenberg-Weiss lemma in option (iii) if one also broadens options (i) and (ii):
Lemma 12 (Variant of Furstenberg-Weiss lemma) Let
be a subgroup of a product
of three groups
. Then one of the following hold:
The ability to encode an abelian additive relation in terms of group-theoretic properties is vaguely reminiscent of the group configuration theorem.
Proof: We apply Lemma 10. Option (i) of that lemma implies option (i) of the current lemma, and similarly for option (ii), so we may assume without loss of generality that
is abelian. By permuting we may also assume that
are abelian, and will use additive notation for these groups.
We may assume that the projections of
to
and
are surjective, else we are in option (i). The group
is then a normal subgroup of
; we may assume it is trivial, otherwise we can quotient it out and be in option (ii). Thus
can be expressed as a graph
for some map
. As
is a group,
must be a homomorphism, and we can write it as
for some homomorphisms
,
. Thus elements
of
obey the constraint
.
If
or
fails to be injective, then we can quotient out by their kernels and end up in option (ii). If
fails to be surjective, then the projection of
to
also fails to be surjective (since for
,
is now constrained to lie in the range of
) and we are in option (i). Similarly if
fails to be surjective. Thus we may assume that the homomorphisms
are bijective and thus group isomorphisms. Setting
to the identity, we arrive at option (iii).
Combining this lemma with Lemma 3, we obtain a nilpotent version:
Corollary 13 (Variant of nilpotent Furstenberg-Weiss lemma) Let
be a subgroup of a product
of three groups
. Then one of the following hold:
Here is another variant of the Furstenberg-Weiss lemma, attributed to Serre by Ribet (see Lemma 3.3):
Lemma 14 (Serre’s lemma) Let
be a subgroup of a finite product
of groups
with
. Then one of the following hold:
- (i) (
too small) There is some proper subgroup
of
for some
such that
whenever
. - (ii) (
too large) One has
. - (iii) One of the
has a non-trivial abelian quotient
.
Proof: The claim is trivial for
(and we don’t need (iii) in this case), so suppose that
. We can assume that each
is a perfect group,
, otherwise we can quotient out by the commutator and arrive in option (iii). Similarly, we may assume that all the projections of
to
,
are surjective, otherwise we are in option (i).
We now claim that for any
and any
, one can find
with
for
and
. For
this follows from the surjectivity of the projection of
to
. Now suppose inductively that
and the claim has already been proven for
. Since
is perfect, it suffices to establish this claim for
of the form
for some
. By induction hypothesis, we can find
with
for
and
. By surjectivity of the projection of
to
, one can find
with
and
. Taking commutators of these two elements, we obtain the claim.
Setting
, we conclude that
contains
. Similarly for permutations. Multiplying these together we see that
contains all of
, and we are in option (ii).
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