# 512 Midterm

Problem  Let $f: \mathbb R_{\geq 0} \to \mathbb R$ be a continuous function such that for every $x \in \mathbb R_{\geq 0}$, $\lim_{n \to \infty} f(nx) = 0.$

Prove that $\lim_{x \to \infty} f(x) = 0$.

Solution  Let $\epsilon > 0$ be given. Define, for every $n \in \mathbb N$, the sets $A_n : = \{x \in \mathbb R_{\geq 0} \mid \forall m \geq n, \ f(mx) \leq \epsilon\}.$ We have that $\bigcup A_n = \mathbb R_{\geq 0}$ because of the limit hypothesis. The continuity assumption yields the closedness of each $A_n$. Since $\mathbb R_{> 0}$ is a complete metric space, the Baire Category theorem guarantees the existence of an interval $(a,b) \subset A_n$ for some $n$. For $N$ large enough, $Nb > (N+1)a$ and since $(ka,kb) \subset A_n$ for all $k \geq n$, $(Na,\infty) \subset A_n$.

# 501 Midterm

Problem. Show that $SO(2, \mathbb{R})^2$ cannot be a subgroup of the group $SO(3, \mathbb{R})$.

Solution. The group $SO(2,\mathbb{R})$ contains the rings of unity so it contains a subgroup isomorphic to $C_3$. But a subgroup of $SO(3,\mathbb{R})$ cannot be isomorphic to $C_3^2$, by the classification theorem for it.

Problem. Let $G$ be a group of odd order and $N$ a normal subgroup of order 5. Prove that $N \subset Z(G)$.

Solution. Let $\sigma : G \to \text{Aut}(N)$ be the action of $G$ on $N$ by conjugation. Since $N \cong C_5$, $\text{Aut}(N) \cong C_4$. Let $g \in G$. Then $|\sigma(g)|$ divides $|\text{Aut}(N)|$ and $|\sigma(g)|$ divides $|G|$ because $|\sigma(g)|$ divides $|g|$. So it must be that $|\sigma(g)| = 1$ which implies that $\sigma(g)$ acts trivially on $N$. That is, $gng ^{-1} = n$ for all $n \in N$.

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 $|Z(G)| = 3$, but 3 doesn’t divide 10. The third one can’t happen because 3 doesn’t divide 10. The fourth one implies that $|Z(G)| = 2$ which implies $G/Z(G) \cong C_5$. Since $G/Z(G)$ is cyclic, $G$ is abelian, but then the class equation should consist of all 1’s. The same argument for four works for the fifth one.

# Two-colored bracelets

Problem. A bracelet $b$ is the same as a bracelet $b'$ if and only if they have the same number of beads and $b$ can be made to look like $b'$ 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 $V := \mathbb{F}_2^p$. Let $\Phi: V \times D_p \to V$ be the action of $D_p$, the dihedral group of order $2p$, on $V$.

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 $x = (x_1, \ldots, x_p) \in V.$ If $e \in D_p$ is the identity, then $V^e = V$. If $r \in D_p$ is a rotation and $x \in V^r$, then $rx = \ldots = r^{p-1}x$ which implies $x_j = x_k$ for all $j, k$. So $V^r$ consists of the two one-colored bracelets. If $s \in D_p$ is a reflection which fixes the first bead and $x \in V^s$, then $x=sx$ so it is required that $x_2 = x_{p-1}, \ldots, x_{(p-1)/2} = x^{(p-1)/2+1}$. So $V^s$ consists of $2 \cdot 2^{(p-1)/2}$ possible bracelets. The argument is similar for the other reflections. There are $p-1$ rotations and $p$ reflections, so $\sum_{g \in G}|V^g| = 2^p + (p-1)2 + p2^{(p+1)}$. By Burnside’s lemma,

$|V \backslash D_p| = \frac{2^p + (p-1)2 + p2^{(p+1)}}{2p}.$

The same type of argument works for any number of beads, but the analysis for rotations becomes cumbersome.

# M511 Midterm 1

3. Let $E$ be a countable subset of $\mathbb{R}$. Prove that there is an $a \in \mathbb{R}$ such that $E \cap (E + a) = \emptyset$, where $E + a = \{x + a \mid x \in E\}.$

We are victorious if we can find an $a \notin E' = \{e_2 - e_1 \mid e_1, e_2 \in E \text{ and } e_1 \neq e_2\}.$ This can be done as $E'$ is a countable subset of $\mathbb{R}.$

6. Let $A_1, \cdots, A_n$ be Lebesgue measurable sets contained in $[0,1]$ such that $\sum_{k=1}^n |A_k| > n-1.$ Prove that $|\cap_{k=1}^n A_k| > 0.$

It’s clear $|A_1| > 0.$ We have $|A_1 \cap A_2| = |A_1| + |A_2| - |A_1 \cup A_2| > 0.$ Then $|\cap_{k=1}^{n+1} A_k| > 0$ by the same argument and the inductive hypothesis, which follows from $\sum_{k=1}^{n} |A_k| + 1 \geq \sum_{k=1}^{n+1} |A_k| > k$.

# S404 Final Exam

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 $A \in M_n(\mathbb{C})$. Suppose $A$ is similar to a single Jordan block with $\alpha$ along the diagonals. It’s clear $\alpha$ is the only eigenvalue of the Jordan block. Since $A$ and the Jordan block associated to $A$ have the same characteristic polynomial, $\alpha$ is also the only eigenvalue of $A$. Since $A$ is conjugate to a single Jordan block, the geometric multiplicity of $A$ is 1. That is, $\dim \ker (A - \alpha I) = 1$. This implies the dimension of the eigenspace of $A$ is 1.

Conversely, suppose that the dimension of the eigenspace of $A$ is 1. Let $\alpha$ 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 $\alpha.$ So in fact, $\dim \ker(A -\alpha I) = 1$. That is, the number of Jordan blocks corresponding to $\alpha$ is 1. Since every $A \in M_n(\mathbb{C})$ can be conjugated into Jordan form and since $\alpha$ is the only eigenvalue, $A$ is conjugate to a single Jordan block.

# S414 Final Exam

Problem: Let $f$ be a $\mathcal{C}^1$ map of a neighborhood $V$ of the origin in $\mathbb{R}^2$ into $\mathbb{R}$ and assume that $f(0,0) = 0$ and that $f(x,y) \geq -2x$ for $(x,y) \in V$. Show that there is a neighborhood $U$ of the origin in $\mathbb{R}^2$ and a positive number $\epsilon$ such that if $(x_1,y_1), (x_2,y_2) \in U$ and $f(x_1,y_1) = f(x_2,y_2) = 0$, then $|y_2 - y_1| \geq \epsilon |x_2 - x_1|.$

Solution: Since $f(x,y) \geq -2x,$ we have $\frac{\partial f}{\partial x}(0,0) \geq -2.$ Also, $\frac{\partial f}{\partial x} : = \lim_{t^- \to 0} \frac{f(t,0) - f(0,0)}{t} \leq \lim_{t^- \to 0} \frac{-2t}{t} = -2$. Thus $\frac{\partial f}{\partial x} (0,0) = -2.$ By the implicit function theorem, there exists an open neighborhood $W$ around $(0,0)$ for which every $(x,y) \in W$ corresponds to a unique $g(x)$, where $g: \mathbb{R}^2 \to \mathbb{R}$ is $\mathcal{C}^1$ and $f(x,g(x)) = 0$. Now take a neighborhood $U$ around $(0,0)$ with radius so small that its closure lies in $W$. Since $\overline U$ is a closed subset of $\mathbb{R}^2$, it is compact. So by the extreme value theorem, there exists a positive number $\epsilon \in \mathbb{R}$ for which $|g'(x,y)| \leq \epsilon$ for all $(x,y) \in \overline U.$ Since $U$ is open and convex, it follows that $|y_2 - y_1| = |g(x_2) - g(x_1)| \leq \epsilon|x_2 - x_1|.$ Clearly $|y_2 - y_1| \geq \epsilon |x_2 - x_1|$ holds when $x_1 = x_2.$ Suppose $x_1 \neq x_2.$ Then $\frac{|y_2 - y_1|}{|x_2 - x_1|} = \frac{|g(x_2) - g(x_1)|}{|x_2 - x_1|} \geq \epsilon$. That is, $|y_2 - y_1| \geq \epsilon |x_2 - x_1|$ on $U.$

# S414 Exam 2

Problem: Let $f$ be a $C^1$ map of a neighborhood $V$ of the origin in $\mathbb{R}^2$ into $\mathbb{R}$ and assume that $f(0,0) = 0$ and that $f(x,y) \geq 2y$ for $(x,y) \in V.$ Show that there is a neighborhood $U$ of the origin in $\mathbb{R}^2$ and a constant $L$ such that, if $(x_1,y_1), (x_2,y_2) \in U$ and $f(x_1,y_1) = f(x_2,y_2) = 0,$ then $|y_2 - y_1| \leq L |x_2 - x_1|.$

Solution: Define $M = f'(0,0).$ Then $M_y \geq 2$ since $f(x,y) \geq 2y$ for $(x,y) \in V.$ By the implicit function theorem, there exist open sets $A, B \subset \mathbb{R}$ both containing 0 such that for each $x \in A$ there is a unique, differentiable $g(x) \in B$ such that $f(x,g(x)) = 0.$ So $f(x_1,y_1) = f(x_2,y_2) = f(x_1,g(x_1)) = f(x_2,g(x_2)) = 0$ on $A$. Since $g$ is differentiable, $g$ is continuous. If we restrict $g$ to the closure of a neighborhood $U$ such that $U \subset A$, then $g$ is uniformly continuous on $\overline U.$ So $g$ is Lipschitz continuous with $L = 1$ on $U$.

# S403 Final

Problem: Let $V \subset \mathbb{R}^4$ satisfy

$v^t \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}w = 0$ for all $v, w\in V.$ What is the largest possible dimension for $V$? Find a basis for some $V$ of maximal dimension satisfying this condition.

Solution: Let $e_i$ be the canonical basis of $\mathbb{R}^4$. Define $S(w) = (e_1 - e_3) \cdot w$ and $T(w) = (e_2 - e_4) \cdot w.$ Then it can be seen dim (ker $S \ \cap$ ker $T) = 2.$ So a basis for $V$ is $\{e_1 + e_3, e_2 + e_4 \}.$

# S403 Midterm

Problem: Let $P$ be a partition of a group $G$ such that for any $A \in P$ and for any $B \in P$, the product set $AB$ is contained in some $C \in P.$ Prove that the element of $P$ which contains the identity of $G$ is a normal subgroup $H$ of $G$ and that $P$ is the set of cosets of $H$.

Solution: It’s clear $\varphi: G \to P$ is a surjective group homomorphism. Thus, $P$ is a group. Let $a \in G.$ Then $\varphi(a) \varphi(e) \varphi(a^{-1}) = \varphi(aea^{-1}) = \varphi(aa^{-1}) = \varphi(e) = E.$ So $G/E \cong P$ by the map $\bar \varphi: G/E \to P$.