University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August2003)
Point of Post: This is the August 2003 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)August 2003
Problem 1:
a) Let be a group and be a cyclic normal subgroup. Show that if then .
b) Let be the dihedral group with presentation
Let be an odd prime dividing . Prove that contains a unique Sylow subgroup.
Proof:
a) Let then we know that is cyclic and so for some . So let be arbitrary, to show that (where is the inner automorphism associated to ) it obviously suffices to show that . To do this we first note that since and we have that . We then recall that order is invariant under conjugation and so in particular but since a cyclic group has precisely one subgroup of every order dividing it we may conclude that and so in particular . The conclusion follows.
b) Consider this is a cyclic subgroup of of order , moreover it’s normal since to which we can appeal to the common fact that a subgroup of index two is always normal (this is a corollary of a more general fact). But, we note that if the maximum power of that divides is that (since is an odd prime) and so by Sylow’s theorems that must contain a Sylow subgroup of order . But, by part a) we have that is a normal Sylow subgroup of and since for Sylow subgroup being normal and unique are equivalent the conclusion follows.
Problem 6: The character table of a group is given below.
where is the golden ratio. Suppose then that , , and .
a) Compute .
b) Use the character table of to show that is not solvable.
c) Use the character table of to show that there does not exist a homomorphism such that
for some .
Proof:
a) We know that (we know that is equal to the conjugacy class of since it is the only trivial conjugacy class). Thus, . But, by the orbitdecomposition theorem that and so doing the simple computation we find that . To find we recall (considering the inner product of and ) that
and thus . Then, considering we see that
and so .
b) We recall that the dual group has order equal the the abelianization . But, by inspection we see that and so and so (the commutator subgroup). It evidently follows that is not solvable.
c) Suppose that there were such a . We know then that there exists such that . But, this implies that and so in particular we know that . But, by assumption we also have that if is the element of that is in the problem statement then which is impossible.
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 1, 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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