Proving product of subgroups
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Contents
Introduction
This article is about general strategies used to prove that a given group, is a product of two subgroups we have at hand. These techniques include situations where both subgroups are normal, where only one is normal, and where neither is normal.
The group action method
Further information: Group equals product of transitive subgroup and isotropy of a point Suppose we need to prove that a given group can be expressed as the product of two subgroups and . The group action method is as follows. We find an action of the group on some set such that:
- contains the isotropy subgroup of some point
- acts transitively on (the action being given by restriction from )
Under these circumstances, we can show that , as follows:
- Pick any
- Find a such that
- Thus, fixes , so . We'll thus get for some
Note that since is equivalent to , we can construct a group action with the roles of and reversed.
Described below are some applications of this method.
Frattini's argument
Further information: Frattini's argument
The setup here is as follows: is a group, is a normal subgroup, and is an automorph-conjugate subgroup of . We need to show that:
Here, is the normalizer of in .
The set here is the set of all conjugates of in . The -action is by conjugation.
- The isotropy at is : This is by definition of normalizer
- Since is normal in , any conjugate to in is an automorph of in . Moreover, was chosen to be automorph-conjugate, so any conjugate of in is actually conjugate to it in . Hence, acts transitively on .
Finding the Weyl group
Further information: Computing the Weyl group of the unitary group
This is the idea used to prove that in the unitary group , the normalizer of any torus is given by:
where is the symmetric group on letters, embedded as matrices inside .
Pick an element in with distinct diagonal entries. Elementary linear algebra tells us that any conjugate of it in , must have the same diagonal entries, possibly in a different order. Thus, its -orbit is precisely the set of different diagonal matrices with exactly those diagonal entries.
Let act on this finite set, by conjugation.
- The isotropy of any point in the set is exactly . This is based on the linear algebra fact that any matrix that commutes with a diagonal matrix with distinct diagonal entries, must also be diagonal.
- The symmetric group , viewed as permutation matrices, acts transitively by conjugation.
Another variant of the group action method
In this variant, instead of computing the isotropy at one point, we do the following: Find an action of on a set , such that:
- contains the kernel of the action (i.e. any element that acts trivially on , lives in )
- For any element of , there exists an element of that has exactly the same action on
Somewhere in between this and the previous approach is an approach where we look at the intersection of isotropies of some elements of .
Equivalence of definitions of central factor
Further information: Equivalence of definitions of central factor
There are two definitions of central factor:
- is a central factor of if (where is the centralizer of in )
- is a central factor of if every inner automorphism of restricts to an inner automorphism of
We'll show here that (2) implies (1):
The set is the set of elements of . The action is by conjugation. This action is well-defined because (2) in particular indicates that is normal in .
- The kernel of the action is precisely
- By condition (2), we have that for any element of , there exists an element in such that the action by conjugation on is the same for both.
This gives the result.
A similar technique is used to prove the equivalence of definitions of WC-subgroup.