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Problem Let be a continuous function such that for every ,
Prove that .
Solution Let be given. Define, for every , the sets We have that because of the limit hypothesis. The continuity assumption yields the closedness of each . Since is a complete metric space, the Baire Category theorem guarantees the existence of an interval for some . For large enough, and since for all , .
Problem. Show that cannot be a subgroup of the group .
Solution. The group contains the rings of unity so it contains a subgroup isomorphic to . But a subgroup of cannot be isomorphic to , by the classification theorem for it.
Problem. Let be a group of odd order and a normal subgroup of order 5. Prove that .
Solution. Let be the action of on by conjugation. Since , . Let . Then divides and divides because divides . So it must be that which implies that acts trivially on . That is, for all .
Problem. Which of the following class equations cannot hold for a group of order 10?
1 + 1 + 1 + 2 + 5
1 + 2 + 2 + 5
1 + 2 + 3 + 4
1 + 1 + 2 + 2 + 2 + 2
1 + 1 + 1 + 1 + 1 + 5
Solution. The first one implies that , but 3 doesn’t divide 10. The third one can’t happen because 3 doesn’t divide 10. The fourth one implies that which implies . Since is cyclic, is abelian, but then the class equation should consist of all 1’s. The same argument for four works for the fifth one.
Problem. A bracelet is the same as a bracelet if and only if they have the same number of beads and can be made to look like after a rotation or reflection. If a bracelet has a prime number of beads and each bead can be one of two colors, how many different bracelets can be made?
Solution. We can consider bracelets as elements in the vector space . Let be the action of , the dihedral group of order , on .
To see how many bracelets are invariant under a reflection or rotation, it suffices to count the number of different orbits because each orbit is an equivalence class of bracelets. Let If is the identity, then . If is a rotation and , then which implies for all . So consists of the two one-colored bracelets. If is a reflection which fixes the first bead and , then so it is required that . So consists of possible bracelets. The argument is similar for the other reflections. There are rotations and reflections, so . By Burnside’s lemma,
The same type of argument works for any number of beads, but the analysis for rotations becomes cumbersome.
3. Let be a countable subset of . Prove that there is an such that , where
We are victorious if we can find an This can be done as is a countable subset of
6. Let be Lebesgue measurable sets contained in such that Prove that
It’s clear We have Then by the same argument and the inductive hypothesis, which follows from .
Problem: Prove that a complex matrix is conjugate to a single Jordan block if and only if all of its eigenvectors are scalar multiples of one another.
Solution: Let . Suppose is similar to a single Jordan block with along the diagonals. It’s clear is the only eigenvalue of the Jordan block. Since and the Jordan block associated to have the same characteristic polynomial, is also the only eigenvalue of . Since is conjugate to a single Jordan block, the geometric multiplicity of is 1. That is, . This implies the dimension of the eigenspace of is 1.
Conversely, suppose that the dimension of the eigenspace of is 1. Let be the eigenvalue corresponding to this eigenvector. We can get all other eigenvectors because any scalar multiple of the eigenvector which spans the eigenspace is again an eigenvector of So in fact, . That is, the number of Jordan blocks corresponding to is 1. Since every can be conjugated into Jordan form and since is the only eigenvalue, is conjugate to a single Jordan block.
Problem: Let be a map of a neighborhood of the origin in into and assume that and that for . Show that there is a neighborhood of the origin in and a positive number such that if and , then
Solution: Since we have Also, . Thus By the implicit function theorem, there exists an open neighborhood around for which every corresponds to a unique , where is and . Now take a neighborhood around with radius so small that its closure lies in . Since is a closed subset of , it is compact. So by the extreme value theorem, there exists a positive number for which for all Since is open and convex, it follows that Clearly holds when Suppose Then . That is, on
Problem: Let be a map of a neighborhood of the origin in into and assume that and that for Show that there is a neighborhood of the origin in and a constant such that, if and then
Solution: Define Then since for By the implicit function theorem, there exist open sets both containing 0 such that for each there is a unique, differentiable such that So on . Since is differentiable, is continuous. If we restrict to the closure of a neighborhood such that , then is uniformly continuous on So is Lipschitz continuous with on .
Problem: Let satisfy
for all What is the largest possible dimension for ? Find a basis for some of maximal dimension satisfying this condition.
Solution: Let be the canonical basis of . Define and Then it can be seen dim (ker ker So a basis for is
Problem: Let be a partition of a group such that for any and for any , the product set is contained in some Prove that the element of which contains the identity of is a normal subgroup of and that is the set of cosets of .
Solution: It’s clear is a surjective group homomorphism. Thus, is a group. Let Then So by the map .