University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (January2004)
Point of Post: This is the January 2004 part of the post started here.
NOTE: Instead of doing this post by post I thought it would make more sense to have it all consolidated. See here for the uptodate PDF of the parts of the exam I’ve finished.
Algebra Qual(Ph.D. Version)January 2004
Problem 1:
a) Let be a prime and a nontrivial finite abelian group. Let be the number of elements of of order . Show that .
b) Let be a prime and let be a finite group. Let be a Sylow subgroup of . Suppose that has order and commutes with all elements of . Show that .
c) Let the notation be as in b). Let be the set of elements of of order . Then, acts on by conjugation. Let . Show that the orbit of has only one element if and only if . Also, show that if the orbit of has more than one element, then the number of elements in the orbit is a multiple of .
d) Let be a finite group and let be a prime dividing . Let be the number of elements of of order . Show that .
Proof:
a) We know by the structure theorem that for some and some . Thus, it suffices to prove this for groups of that form. We begin by recalling that the order of has order . Thus, for the order of to have order it is necessary and sufficient that for each we have that and there exists at least one such that . Thus, of course we see that
What we claim though is that for each . Indeed, we note firstly that if is of order then and so . Thus, the only such possible are and evidently these are of order . Thus, the number of elements of of order is and thus the number of elements of of order or is as claimed. Thus, by we see that
We implicitly used that which we know is true since .
b) We claim that if then where which is by definition impossible. Indeed, by definition we have that
But, since commutes with everything in this may be rewritten as
What we claim though is that if where then and so . But, since is prime we know that if and thus if then contradictory to assumption. Thus, and consequently and . It evidently follows that which is a contradiction since is the maximal power of dividing . Thus, .
c) It’s evident that acts on since order is preserved under conjugation. The rest is simple. Namely, let . If (the orbit of under the action) has one element then evidently commutes with every element of and since we have by b) that . Recall though that by the orbitstabilizer theorem that and since we have that for some , and so if we must have that .
d) By the orbit decomposition theorem we may write where is the union of the distinct oneelement orbits of under the action described in 2) and is the union of the distinct orbits which have more than one element and thus
What we claim though is that . Indeed, if then by fact that it has a trivial orbit under the conjugation action tells us that it commutes with everything in and since as we proved it must be that , and since so that we have that . Conversely, if then evidently and so and by definition its orbit under the action is trivial and thus . Note though that is a nontrivial (since we know the center of groups is nontrivial) abelian group and so by a) we know that . And, since by b) we may conclude that
Problem 6: Let be a finite group and . Let be the distinct cosets of . Let be thefree complex vector space over the symbols . Then, for define on by if . The map is a representation of (you do not need to prove this). Show that in the decomposition of into irreps the trivial representation occurs only once.
Proof: We first note that if and then by definition and so and so and since is normal this implies that . Conversely, if then and . We thus note that if then for for all . Thus, if is the matrix of with respect to the ordered basis then the diagonal entries of are zero. Moreover, if then for every and thus where is the identity matrix. Thus, using the obvious fact that the number of times that the trivial representation shows up in the decomposition of is equal to where the inner product is taken in the group algebra. But, the above shows that
where we have evidently used Lagrange’s theorem.
References:
1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.
May 6, 2011  Posted by Alex Youcis  Algebra, Group Theory, Representation Theory, UMaryland Qualifying Exams  Algebra, Group Theory, January 2003, Qualifying Exams, Representation Theory, University of Maryland Qualifying Exams
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