Let denote the – th harmonic number of weight . Prove that
Recall the generating function
Thus from to we get that
Integrating from to the result follows.
Let be nilpotent matrices such that . Evaluate the determinant
Lemma: If and are nilpotent matrices that commute and are scalars, then is nilpotent.
Proof: Since and commute, they are simultaneously triangularizable. Let be an invertible matrix such that and , where and are upper triangular. Note that since and are nilpotent, and must have zeros down the main diagonal. Hence is upper triangular with zeros along the main diagonal which means that it’s nilpotent. Finally and so is nilpotent.
We have . Then equating the two left hand sides and simplifying gives us . Thus by the lemma we know that is nilpotent, i.e., it’s eigenvalues are all zero. It follows that the eigenvalues of are all one and so .
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 suppose that is nilpotent. Show that if commute then
Since is algebraically closed and commute this means that are simultaneously triangularizable, there exists an invertible element such that both and are triangular. Since is both nilpotent and triangular, all its diagonal entries are zero and so the diagonal entries of are the same as the diagonal entries of . Thus,
because , are both triangular and the determinant of a triangular matrix is the product of its diagonal entries. So,
The exercise along its solution have been migrated from here.