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Suppose that has this property that if is an eigenvalue of then is not an eigenvalue of . Show that if and only if for any . In other words the centralizer of equals the centralizer of .
It is clear that implies for any . Now suppose that for some and set . We want to prove that . We have
and so . It now follows that for any integer and thus for any and any integer we have
where is the identity matrix. Now let be a generalized eigenvector corresponding to an eigenvalue of . Then for some integer and thus, by we have . Therefore, since we are assuming that is not an eigenvalue of , we must have . So, since every element of is a linear combination of some generalized eigenvectors of , we get for all , i.e. and hence .
The exercise can also be found here.
Let . Define the group
Prove that where is the dihedral group.
Using or equivalently we can write each element of in the form where . Using we may assume that . Using we may also assume that . It is easy to prove inductively that .
Let . We prove that . Obviously . Furthermore, since
If (such that ) then hence and therefore or . If then hence which is a contradiction since .
which is precisely the dihedral group with elements.
Let be a finite group and suppose that , are two subgroups of such that and . Show that
Recall that and thus . Hence and so
where and .
Now, since and we have and that is and . So if we let and then and thus
due to .
The exercise along its solution have been migrated from here .
Let and let
Find all irreducible factors of .
Let and let be an -th primitive root of unity. Prove that
where denotes the Euler’s totient function.
where since is the minimal polynomial of over .
The result follows.