Finding conjugate subgroups
This is a survey article related to:conjugacy
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This article describes techniques to solve problems of the following nature:
- Prove that two subgroups of a group are conjugate. In other words, given a group
and two subgroups
and
of
, we need to show the existence of a
suchthat
- Prove a domination condition: Show that a given subgroup is conjugate to some subgroup of another given subgroup. In other words, given a group
and subgroups
and
, we want to show that there exists
such that
- Prove the existence of another conjugate subgroup, with somewhat nicer properties
Move the big, not the small
It is helpful to keep in mind that the following are equivalent for subgroups of
:
- A conjugate of
is contained in
-
is contained in a conjugate of
It is usually the second form that is more useful. In other words, it is usually more helpful to try to find a conjugate of containing
, rather than try to find a conjugate of
contained in
. In some sense, the bigger groups are easier to move around than the smaller groups.
A formulation equivalent to (2) is:
- The action of
on the coset space
has a fixed point
Proving the domination in Sylow's theorem
Further information: Sylow's theorem#Proof of domination
One part of Sylow's theorem states if for for a finite group , a
-Sylow subgroup
of
and a
-subgroup
of
,
is conjugate to a subgroup of
.
We now consider the action of on the quotient space
, by left multiplication. Since the order of
is relatively prime to
, there is some orbit whose order is not a multiple of
, but since
is a
-group, this orbit must have size one. Thus, the action of
on the quotient space
has a fixed point, so
is contained in the conjugate of
that fixes that point.
Proving that any finite subgroup is contained in the orthogonal group
Further information: Orthogonal group is finite-dominating in general linear group
Another example is the following: given a finite subgroup of (the general linear group), there exists a conjugate of it contained in
(the orthogonal group). In other words, the orthogonal group dominates all finite groups.
Again, it is easier instead to prove the following: given any finite subgroup of , there exists a conjugate of the orthogonal group containing it. Saying that a group is contained in a conjugate of the orthogonal group, is equivalent to saying that there exists a positive definite symmetric bilinear form invariant under the action of the group. Thus, the problem reduces to the following: given any finite subgroup of
, find a positive-definite symmetric bilinear form invariant under the group.
This is achieved by taking any positive-definite symmetric bilinear form, and then averaging it over the group.
Proving that any finite subgroup of diffeomorphism group has an invariant Riemannian metric
We can take this somewhat further: given any finite subgroup of the self-diffeomorphism group of a differential manifold, there exists a Riemannian metric on the differential manifold that is invariant under the finite group. In other words, there is a Riemannian metric such that the elements of the finite group are isometries for that Riemannian metric.
The proof is exactly the same as the previous one: we start with any Riemannian metric, and average under the action of the finite group. The key difference from the preceding case is that in the previous case, all the isometry groups for positive-define symmetric bilinear forms were conjugate. Now, it is no longer true that isometry groups of Riemannian metrics are all conjugate, so we cannot obtain a single group that is finite-dominating.
PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]Move the big, and Sylow's theorem
We often need to combine the basic results of Sylow theory, with the idea that it's the bigger subgroup that needs to be moved around. Below are some examples.
Moving to nice p-subgroups
Further information: Every p-subgroup is conjugate to a p-subgroup whose normalizer in the Sylow is Sylow in its normalizer
The setup is like this: is a finite group,
is a
-Sylow subgroup, and
is a subgroup of
contained inside
. Then, we need to find a subgroup
of
such that
and
are conjugate inside
, and such that
is a
-Sylow subgroup inside
.
The key trick here is again, to focus attention on trying to move the big rather than the small. So, instead of chasing , we look at the subgroup
. Let
be a
-Sylow subgroup of
. Now, by the domination part of Sylow's theorem,
is contained in some conjugate
of
(i.e. in another Sylow subgroup). Let us try to determine what
looks like.
Clearly, . This contains
, because
is contained in both subgroups. Also, it cannot be strictly bigger than
, because
is Sylow inside
.
So what we've done is: instead of moving to a good subgroup of
, we've found a Sylow subgroup
in which
is good. Now, of course, we can do a conjugation taking
to
and the image of
under that conjugation is the required
.