Difference between revisions of "Math 360, Fall 2021, Assignment 6"

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(Other problems:)
 
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==Definitions:==
 
==Definitions:==
 
# Symmetry: the property that a mathematical object remains unchanged under a set of operations or transformations. Suppose $A \subseteq R^{n}$. A symmetry of $A$ is an isometry $\sigma : R^{n} \rightarrow R^{n}$ such that if $x \in A$ then also $\sigma (x) = A$. Claims: (1) The composition of two symmetries of $A$ is again a symmetry of $A$. (2) The identity map is always a symmetry of $A$ ("trivial symmetry"). (3) Any symmetry is invertible, and its inverse is again a symmetry.
 
# Symmetry: the property that a mathematical object remains unchanged under a set of operations or transformations. Suppose $A \subseteq R^{n}$. A symmetry of $A$ is an isometry $\sigma : R^{n} \rightarrow R^{n}$ such that if $x \in A$ then also $\sigma (x) = A$. Claims: (1) The composition of two symmetries of $A$ is again a symmetry of $A$. (2) The identity map is always a symmetry of $A$ ("trivial symmetry"). (3) Any symmetry is invertible, and its inverse is again a symmetry.
  +
# Symmetry group: Suppose $A\subseteq\mathbb{R}^n$. The ''symmetry group'' of $A$ is the set of symmetries of $A$ (see above), regarded as a group under composition. (Not to be confused with the ''symmetric group'' $\mathrm{Sym}(A)$, which is the group of all bijections from $A$ to $A$ and is usually a far larger group than the ''symmetry group'' defined here.)
# The symmetric group defined over any set is the group whose elements are all the bijections from the set to itself, and whose group operation is the composition of functions. (The set of symmetries of A forms a group under composition $(S(A), \circ)$)
 
 
# If $G$ is a group, then the order $|G|$ of $G$ is the number of elements in $G$. (Recall from Section 0 that, for any set S, $|S|$ is the cardinality of $S$.)
 
# If $G$ is a group, then the order $|G|$ of $G$ is the number of elements in $G$. (Recall from Section 0 that, for any set S, $|S|$ is the cardinality of $S$.)
 
# The symmetry group of a regular n-gon.
 
# The symmetry group of a regular n-gon.
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==Book Problems:==
 
==Book Problems:==
# 1. For $\mathbb R \in \mathbb C$, under addition, (1) $e = 0 \in \mathbb C$, $0 \in \mathbb Z, e \in \mathbb Z$. (2) If $a, b \in \mathbb R,a + b \in \mathbb R$. (3) If $a \in \mathbb R$, $a' = -a \ mathbb R$. Yes, it is a group.
+
# 1. For $\mathbb R \in \mathbb C$, under addition, (1) $e = 0 \in \mathbb C$, $0 \in \mathbb R, e \in \mathbb R$. (2) If $a, b \in \mathbb R,a + b \in \mathbb R$. (3) If $a \in \mathbb R$, $a' = -a \mathbb R$. Yes, it is a group.
 
# 2. No. (1) $0 \not\in \mathbb Q^{+}$, (3)For $a \in \mathbb Q^{+}, a' = -a < 0 \not\in \mathbb Q^{+}$.
 
# 2. No. (1) $0 \not\in \mathbb Q^{+}$, (3)For $a \in \mathbb Q^{+}, a' = -a < 0 \not\in \mathbb Q^{+}$.
# 8. No. Suppose this is a set $A$. (1)$det(e) = 1 \neq 2, e \not\in A$. Inverse not in it. (2)$det(M) * det(N) = 4 \neq 2, \forall M, N \in A$, not closed under matrix multiplication.
+
# 8. No. Suppose this is a set $A$. (1)$\det(e) = 1 \neq 2, e \not\in A$. Inverse not in it. (2)$det(M) * det(N) = 4 \neq 2, \forall M, N \in A$, not closed under matrix multiplication.
  +
# 9. Yes. $I=\mathrm{diag}(1,1,\dots,1)$ is a diagonal matrix without zeros on the diagonal. If $A=\mathrm{diag}(a_1,\dots,a_n)$ and $B=\mathrm{diag}(b_1,\dots,b_n)$ are two such matrices, then so it $AB=\mathrm{diag}(a_1b_1,\dots,a_nb_n)$, as is $A^{-1}=\mathrm{diag}(a_1^{-1},\dots,a_n^{-1})$. (Note that although $\{(1, 1), (1, 2) \}$ has determinant $0$, is it not a diagonal matrix.)
# 9. No. $\{(1, 1), (1, 2) \}$ has determinant $0$.
 
  +
# 11. No: the identity matrix does not belong to this set.
# 11. No.
 
  +
# 12. Yes: the identity matrix belongs to this set, and if $\det(A)=\pm1$ and $\det(B)=\pm1$ then also $\det(AB)=\pm1$ and $\det(A^{-1})=\pm1$.
# 12. Yes.
 
 
# 21. (a) $\{ -50, -25, 0, 25, 50 \}$. (b) $\{4, 2, 1, \frac{1}{2}, \frac{1}{4} \}$. (c) $\{ \frac{1}{\pi ^2}, \frac{1}{\pi}, 1, \pi, \pi^2 \}$.
 
# 21. (a) $\{ -50, -25, 0, 25, 50 \}$. (b) $\{4, 2, 1, \frac{1}{2}, \frac{1}{4} \}$. (c) $\{ \frac{1}{\pi ^2}, \frac{1}{\pi}, 1, \pi, \pi^2 \}$.
# 22. $\{ \{(0, -1), (-1, 0)\}, \{(1, 0), (0, 1)\}\}$
+
# 22. $\left\{\begin{bmatrix}0&-1\\-1&0\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix}\right\}$.
  +
# 23. $\left\{\left.\begin{bmatrix}1&n\\0&1\end{bmatrix}\,\right\rvert\,n\in\mathbb{Z}\right\}$.
# 23. $\{\{(n, -1),(0, 1)\}, \cdots \{(1, -1), (0, 1)\}, \{(1, 0), (0, 1)\}, \{(1,2), (0,1)\}, \{(1, 3), (0, 1)\} \cdots \{(1, n), (0,1)\}\}$
 
# 24. $\{\{3^n, 0),(0, 2^n)\}$
+
# 24. $\left\{\left.\begin{bmatrix}3^n&0\\0&2^n\end{bmatrix}\,\right\rvert\,n\in\mathbb{Z}\right\}$.
  +
# 25. $\left\{\dots,\begin{bmatrix}0&-1/2\\-1/2&0\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}0&-2\\-2&0\end{bmatrix},\begin{bmatrix}4&0\\0&4\end{bmatrix},\begin{bmatrix}0&-8\\-8&0\end{bmatrix},\dots\right\}$.
# 25. $\{\{(0, (-2)^n), ((-2)^n, 0)\}\}$
 
 
# 36. (a) see http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/modular.html (b) $\{0\}, \mathbb Z_6, \{0, 2, 4\}, \{0, 3\}, \{0, 2, 4\}, \mathbb Z_6$. (c)$1, 5$ (d)See https://drive.google.com/file/d/1ZU7k9peVMdC_mY5XCtIT5-Ld8Z4k_bjy/view?usp=sharing
 
# 36. (a) see http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/modular.html (b) $\{0\}, \mathbb Z_6, \{0, 2, 4\}, \{0, 3\}, \{0, 2, 4\}, \mathbb Z_6$. (c)$1, 5$ (d)See https://drive.google.com/file/d/1ZU7k9peVMdC_mY5XCtIT5-Ld8Z4k_bjy/view?usp=sharing
   
 
==Other problems:==
 
==Other problems:==
  +
By the classification of cyclic groups, and cyclic group must be isomorphic with either $(\mathbb{Z},+)$ or with $(\mathbb{Z}_n,+_n)$. In particular, any cyclic group must be equinumerous either with $\mathbb{Z}$ or with $\mathbb{Z}_n$. But $\mathbb{R}$ is an uncountable set, so $(\mathbb{R},+)$ cannot be cyclic.
For a cyclic group, $\exists g, m, m > 0 , g^m = e = 0$. However, here, $g^m = m \cdot g, m \cdot g = 0$ only when $g = 0$, $0 + 0 = 0$, no chance to generate $\mathbb R$.
 

Latest revision as of 15:11, 12 November 2021

I tell them that if they will occupy themselves with the study of mathematics, they will find in it the best remedy against the lusts of the flesh.

- Thomas Mann, The Magic Mountain

Read:[edit]

  1. Section 5.

Carefully define the following terms, then give one example and one non-example of each:[edit]

  1. Symmetry (of a subset $A\subseteq\mathbb{R}^n$).
  2. Symmetry group (of a subset $A\subseteq\mathbb{R}^n$).
  3. Order (of a group; see Definition 5.3 on page 50 of the text).
  4. $D_n$ (the dihedral group with order $2n$).
  5. Subgroup (of a group).
  6. Trivial subgroup (see Definition 5.5 on page 51 of the text).
  7. Improper subgroup (see Definition 5.5 on page 51 of the text).
  8. $\left\langle S\right\rangle$ (the subgroup generated by the subset $S$).
  9. $\left\langle g\right\rangle$ (the cyclic subgroup generated by the element $g$; see Definition 5.18 on page 54).

Carefully state the following theorems (you do not need to prove them):[edit]

  1. Theorem concerning unions and intersections of subgroups.

Solve the following problems:[edit]

  1. Section 5, problems 1, 2, 8, 9, 11, 12, 21, 22, 23, 24, 25, and 36.
  2. Prove that $(\mathbb{R},+)$ is not a cyclic group. (Hint: $\mathbb{R}$ is an uncountable set. Now look again at the list of elements of a cyclic subgroup. What can you conclude about the cardinality of a cyclic group?)
--------------------End of assignment--------------------

Questions:[edit]

Solutions:[edit]

Definitions:[edit]

  1. Symmetry: the property that a mathematical object remains unchanged under a set of operations or transformations. Suppose $A \subseteq R^{n}$. A symmetry of $A$ is an isometry $\sigma : R^{n} \rightarrow R^{n}$ such that if $x \in A$ then also $\sigma (x) = A$. Claims: (1) The composition of two symmetries of $A$ is again a symmetry of $A$. (2) The identity map is always a symmetry of $A$ ("trivial symmetry"). (3) Any symmetry is invertible, and its inverse is again a symmetry.
  2. Symmetry group: Suppose $A\subseteq\mathbb{R}^n$. The symmetry group of $A$ is the set of symmetries of $A$ (see above), regarded as a group under composition. (Not to be confused with the symmetric group $\mathrm{Sym}(A)$, which is the group of all bijections from $A$ to $A$ and is usually a far larger group than the symmetry group defined here.)
  3. If $G$ is a group, then the order $|G|$ of $G$ is the number of elements in $G$. (Recall from Section 0 that, for any set S, $|S|$ is the cardinality of $S$.)
  4. The symmetry group of a regular n-gon.
  5. Subgroup: Suppose $(G, \triangle )$ is a group, and $H \subseteq G$. We say that $H$ is a subgroup of $G$ if (1) $e \in H$, (2) if $h_{1}, h_{2} \in H$, then $h_{1} \triangle h_{2} \in H$, (3) if $h \in H$, then $h' \in H$. (This ensure that $(H, \triangle )$ is also a group).
  6. If $G$ is a group, The subgroup $\{e\}$ is the trivial subgroup of G. All other subgroups are nontrivial.
  7. If $G$ is a group, then the subgroup consisting of $G$ itself is the improper subgroup of $G$. All other subgroups are proper subgroups.
  8. Suppose $(G, \triangle )$ is a group, and $S \subseteq G$ (subset, not necessarily a subgroup). define $\left\langle S \right\rangle$ to be the intersection of all subgroups of $G$ that contain $S$.
  9. let $G$ be a group and let $g \in G$. then the subgroup $\{ g^n | n\in \mathbb Z \}$ of $G$, characterized in Theorem 5.17, is called the cyclic subgroup of $G$ generated by $g$, and denote by $\left\langle g \right\rangle$. Theorem 5.17: $H = \{ g^n | n \in \mathbb Z \}$ is a subgroup of $G$ and is the smallest subgroup go $G$ that contain $g$, that is, every subgroup containing $g$ contains $H$.

Theorems:[edit]

  1. Suppose $(G, \triangle )$ is a group, and $H \leq G$ (is subgroup of) and $K \leq G$. Then $H \cap K$ is also a subgroup (however, $H \cup K$ is usually not).

Book Problems:[edit]

  1. 1. For $\mathbb R \in \mathbb C$, under addition, (1) $e = 0 \in \mathbb C$, $0 \in \mathbb R, e \in \mathbb R$. (2) If $a, b \in \mathbb R,a + b \in \mathbb R$. (3) If $a \in \mathbb R$, $a' = -a \mathbb R$. Yes, it is a group.
  2. 2. No. (1) $0 \not\in \mathbb Q^{+}$, (3)For $a \in \mathbb Q^{+}, a' = -a < 0 \not\in \mathbb Q^{+}$.
  3. 8. No. Suppose this is a set $A$. (1)$\det(e) = 1 \neq 2, e \not\in A$. Inverse not in it. (2)$det(M) * det(N) = 4 \neq 2, \forall M, N \in A$, not closed under matrix multiplication.
  4. 9. Yes. $I=\mathrm{diag}(1,1,\dots,1)$ is a diagonal matrix without zeros on the diagonal. If $A=\mathrm{diag}(a_1,\dots,a_n)$ and $B=\mathrm{diag}(b_1,\dots,b_n)$ are two such matrices, then so it $AB=\mathrm{diag}(a_1b_1,\dots,a_nb_n)$, as is $A^{-1}=\mathrm{diag}(a_1^{-1},\dots,a_n^{-1})$. (Note that although $\{(1, 1), (1, 2) \}$ has determinant $0$, is it not a diagonal matrix.)
  5. 11. No: the identity matrix does not belong to this set.
  6. 12. Yes: the identity matrix belongs to this set, and if $\det(A)=\pm1$ and $\det(B)=\pm1$ then also $\det(AB)=\pm1$ and $\det(A^{-1})=\pm1$.
  7. 21. (a) $\{ -50, -25, 0, 25, 50 \}$. (b) $\{4, 2, 1, \frac{1}{2}, \frac{1}{4} \}$. (c) $\{ \frac{1}{\pi ^2}, \frac{1}{\pi}, 1, \pi, \pi^2 \}$.
  8. 22. $\left\{\begin{bmatrix}0&-1\\-1&0\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix}\right\}$.
  9. 23. $\left\{\left.\begin{bmatrix}1&n\\0&1\end{bmatrix}\,\right\rvert\,n\in\mathbb{Z}\right\}$.
  10. 24. $\left\{\left.\begin{bmatrix}3^n&0\\0&2^n\end{bmatrix}\,\right\rvert\,n\in\mathbb{Z}\right\}$.
  11. 25. $\left\{\dots,\begin{bmatrix}0&-1/2\\-1/2&0\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}0&-2\\-2&0\end{bmatrix},\begin{bmatrix}4&0\\0&4\end{bmatrix},\begin{bmatrix}0&-8\\-8&0\end{bmatrix},\dots\right\}$.
  12. 36. (a) see http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/modular.html (b) $\{0\}, \mathbb Z_6, \{0, 2, 4\}, \{0, 3\}, \{0, 2, 4\}, \mathbb Z_6$. (c)$1, 5$ (d)See https://drive.google.com/file/d/1ZU7k9peVMdC_mY5XCtIT5-Ld8Z4k_bjy/view?usp=sharing

Other problems:[edit]

By the classification of cyclic groups, and cyclic group must be isomorphic with either $(\mathbb{Z},+)$ or with $(\mathbb{Z}_n,+_n)$. In particular, any cyclic group must be equinumerous either with $\mathbb{Z}$ or with $\mathbb{Z}_n$. But $\mathbb{R}$ is an uncountable set, so $(\mathbb{R},+)$ cannot be cyclic.