Monday
- 10am Office hours
Tuesday
- 2pm OA seminar
Monday
Tuesday
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
.