 # PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

## AppendixDHints and Answers to Selected Exercises

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### IBasics1Preliminaries1.4Exercises

#### Warm-up

##### 1.4.2.
Hint.
(a) $$A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}\text{;}$$ (d) $$A \times D = \emptyset\text{.}$$
##### 1.4.6.
Hint.
If $$x \in A \cup (B \cap C)\text{,}$$ then either $$x \in A$$ or $$x \in B \cap C\text{.}$$ Thus, $$x \in A \cup B$$ and $$A \cup C\text{.}$$ Hence, $$x \in (A \cup B) \cap (A \cup C)\text{.}$$ Therefore, $$A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)\text{.}$$ Conversely, if $$x \in (A \cup B) \cap (A \cup C)\text{,}$$ then $$x \in A \cup B$$ and $$A \cup C\text{.}$$ Thus, $$x \in A$$ or $$x$$ is in both $$B$$ and $$C\text{.}$$ So $$x \in A \cup (B \cap C)$$ and therefore $$(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)\text{.}$$ Hence, $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}$$
##### 1.4.10.
Hint.
$$(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}$$
##### 1.4.14.
Hint.
$$A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}$$

#### More Exercises

##### 1.4.18.
Hint.
(a) $$f$$ is one-to-one but not onto. $$f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}\text{.}$$ (c) $$f$$ is neither one-to-one nor onto. $$f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}$$
##### 1.4.20.
Hint.
(a) $$f(n) = n + 1\text{.}$$
##### 1.4.22.
Hint.
(a) Let $$x, y \in A\text{.}$$ Then $$g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))\text{.}$$ Thus, $$f(x) = f(y)$$ and $$x = y\text{,}$$ so $$g \circ f$$ is one-to-one. (b) Let $$c \in C\text{,}$$ then $$c = (g \circ f)(x) = g(f(x))$$ for some $$x \in A\text{.}$$ Since $$f(x) \in B\text{,}$$ $$g$$ is onto.
##### 1.4.24.
Hint.
(a) Let $$y \in f(A_1 \cup A_2)\text{.}$$ Then there exists an $$x \in A_1 \cup A_2$$ such that $$f(x) = y\text{.}$$ Hence, $$y \in f(A_1)$$ or $$f(A_2) \text{.}$$ Therefore, $$y \in f(A_1) \cup f(A_2)\text{.}$$ Consequently, $$f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\text{.}$$ Conversely, if $$y \in f(A_1) \cup f(A_2)\text{,}$$ then $$y \in f(A_1)$$ or $$f(A_2)\text{.}$$ Hence, there exists an $$x \in A_1$$ or there exists an $$x \in A_2$$ such that $$f(x) = y\text{.}$$ Thus, there exists an $$x \in A_1 \cup A_2$$ such that $$f(x) = y\text{.}$$ Therefore, $$f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\text{,}$$ and $$f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}$$
##### 1.4.28.
Hint.
Let $$X = {\mathbb N} \cup \{ \sqrt{2}\, \}$$ and define $$x \sim y$$ if $$x + y \in {\mathbb N}\text{.}$$

### 2The Integers2.4Exercises

#### 2.4.1.

The base case, $$S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2$$ is true.
Assume that $$S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6$$ is true. Then
\begin{align*} 1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 & = [k(k + 1)(2k + 1)]/6 + (k + 1)^2\\ & = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6\text{,} \end{align*}
and so $$S(k + 1)$$ is true. Thus, $$S(n)$$ is true for all positive integers $$n\text{.}$$

#### 2.4.3.

The base case, $$S(4): 4! = 24 \gt 16 =2^4$$ is true. Assume $$S(k): k! \gt 2^k$$ is true. Then $$(k + 1)! = k! (k + 1) \gt 2^k \cdot 2 = 2^{k + 1}\text{,}$$ so $$S(k + 1)$$ is true. Thus, $$S(n)$$ is true for all positive integers $$n\text{.}$$

#### 2.4.11.

Hint.
The base case, $$S(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x$$ is true. Assume $$S(k): (1 + x)^k -1 \geq kx$$ is true. Then
\begin{align*} (1 + x)^{k + 1} - 1 & = (1 + x)(1 + x)^k -1\\ & = (1 + x)^k + x(1 + x)^k - 1\\ & \geq kx + x(1 + x)^k\\ & \geq kx + x\\ & = (k + 1)x\text{,} \end{align*}
so $$S(k + 1)$$ is true. Therefore, $$S(n)$$ is true for all positive integers $$n\text{.}$$

#### 2.4.19.

Hint.
Use the Fundamental Theorem of Arithmetic.

#### 2.4.23.

Hint.
Let $$S = \{s \in {\mathbb N} : a \mid s\text{,}$$ $$b \mid s \}\text{.}$$ Then $$S \neq \emptyset\text{,}$$ since $$|ab| \in S\text{.}$$ By the Principle of Well-Ordering, $$S$$ contains a least element $$m\text{.}$$ To show uniqueness, suppose that $$a \mid n$$ and $$b \mid n$$ for some $$n \in {\mathbb N}\text{.}$$ By the division algorithm, there exist unique integers $$q$$ and $$r$$ such that $$n = mq + r\text{,}$$ where $$0 \leq r \lt m\text{.}$$ Since $$a$$ and $$b$$ divide both $$m\text{,}$$ and $$n\text{,}$$ it must be the case that $$a$$ and $$b$$ both divide $$r\text{.}$$ Thus, $$r = 0$$ by the minimality of $$m\text{.}$$ Therefore, $$m \mid n\text{.}$$

#### 2.4.27.

Hint.
Since $$\gcd(a,b) = 1\text{,}$$ there exist integers $$r$$ and $$s$$ such that $$ar + bs = 1\text{.}$$ Thus, $$acr + bcs = c\text{.}$$ Since $$a$$ divides both $$bc$$ and itself, $$a$$ must divide $$c\text{.}$$

#### 2.4.29.

Hint.
Every prime must be of the form 2, 3, $$6n + 1\text{,}$$ or $$6n + 5\text{.}$$ Suppose there are only finitely many primes of the form $$6k + 5\text{.}$$

### IIAlgebra1Groups1.5Exercises

#### 1.5.1.

Hint.
(a) $$3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}\text{;}$$ (c) $$18 + 26 \mathbb Z\text{;}$$ (e) $$5 + 6 \mathbb Z\text{.}$$

#### 1.5.2.

Hint.
(a) Not a group; (c) a group.

#### 1.5.6.

Hint.
\begin{equation*} \begin{array}{c|cccc} \cdot & 1 & 5 & 7 & 11 \\ \hline 1 & 1 & 5 & 7 & 11 \\ 5 & 5 & 1 & 11 & 7 \\ 7 & 7 & 11 & 1 & 5 \\ 11 & 11 & 7 & 5 & 1 \end{array} \end{equation*}

#### 1.5.8.

Hint.
Pick two matrices. Almost any pair will work.

#### 1.5.15.

Hint.
There is a nonabelian group containing six elements.

#### 1.5.16.

Hint.
Look at the symmetry group of an equilateral triangle or a square.

#### 1.5.17.

Hint.
The are five different groups of order 8.

#### 1.5.18.

Hint.
Let
\begin{equation*} \sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ a_1 & a_2 & \cdots & a_n \end{pmatrix} \end{equation*}
be in $$S_n\text{.}$$ All of the $$a_i$$s must be distinct. There are $$n$$ ways to choose $$a_1\text{,}$$ $$n-1$$ ways to choose $$a_2\text{,}$$ $$\ldots\text{,}$$ 2 ways to choose $$a_{n - 1}\text{,}$$ and only one way to choose $$a_n\text{.}$$ Therefore, we can form $$\sigma$$ in $$n(n - 1) \cdots 2 \cdot 1 = n!$$ ways.

#### 1.5.25.

Hint.
\begin{align*} (aba^{-1})^n & = (aba^{-1})(aba^{-1}) \cdots (aba^{-1})\\ & = ab(aa^{-1})b(aa^{-1})b \cdots b(aa^{-1})ba^{-1}\\ & = ab^na^{-1}\text{.} \end{align*}

#### 1.5.31.

Hint.
Since $$abab = (ab)^2 = e = a^2 b^2 = aabb\text{,}$$ we know that $$ba = ab\text{.}$$

#### 1.5.35.

Hint.
$$H_1 = \{ id \}\text{,}$$ $$H_2 = \{ id, \rho_1, \rho_2 \}\text{,}$$ $$H_3 = \{ id, \mu_1 \}\text{,}$$ $$H_4 = \{ id, \mu_2 \}\text{,}$$ $$H_5 = \{ id, \mu_3 \}\text{,}$$ $$S_3\text{.}$$

#### 1.5.41.

Hint.
The identity of $$G$$ is $$1 = 1 + 0 \sqrt{2}\text{.}$$ Since $$(a + b \sqrt{2}\, )(c + d \sqrt{2}\, ) = (ac + 2bd) + (ad + bc)\sqrt{2}\text{,}$$ $$G$$ is closed under multiplication. Finally, $$(a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}$$

#### 1.5.46.

Hint.
Look at $$S_3\text{.}$$

#### 1.5.49.

Hint.
Since $$a^4b = ba\text{,}$$ it must be the case that $$b = a^6 b = a^2 b a\text{,}$$ and we can conclude that $$ab = a^3 b a = ba\text{.}$$

#### 1.5.55.

$$1$$

#### 1.5.56.

$$2$$

#### 1.5.57.

$$n$$

#### 1.5.58.

$$n+1$$

#### 1.5.59.

##### 1.5.59.a
$$2$$
##### 1.5.59.b
###### 1.5.59.b.i
$$6$$
###### 1.5.59.b.ii
$$10$$

#### 1.5.60.

##### 1.5.60.a
$$4$$
##### 1.5.60.b
###### 1.5.60.b.i
$$8$$
###### 1.5.60.b.ii
$$12$$

### 3Runestone Testing3.7True/False Exercises

#### 3.7.1.True/False.

Hint.
$$P_n\text{,}$$ the vector space of polynomials with degree at most $$n\text{,}$$ has dimension $$n+1$$ by Theorem 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter $$n\text{?}$$

### 3.8Multiple Choice Exercises

#### 3.8.1.Multiple-Choice, Not Randomized, One Answer.

Hint 1.
What did you see last time you went driving?
Hint 2.
Maybe go out for a drive?

#### 3.8.2.Multiple-Choice, Not Randomized, Multiple Answers.

Hint.
Do you know the acronym…ROY G BIV for the colors of a rainbow, and their order?

Hint 1.
What did you see last time you went driving?
Hint 2.
Maybe go out for a drive?

Hint.
Do you know the acronym…ROY G BIV for the colors of a rainbow, and their order?

#### 3.8.5.Mathematical Multiple-Choice, Not Randomized, Multiple Answers.

Hint.
You can take a derivative on any one of the choices to see if it is correct or not, rather than using techniques of integration to find a single correct answer.

### 3.9Parsons Exercises

#### 3.9.1.Parsons Problem, Mathematical Proof.

Hint.
Dorothy will not be much help with this proof.

#### 3.9.4.Parsons Problem, Mathematical Proof, Numbered Blocks.

Hint.
Dorothy will not be much help with this proof.

### 3.11Matching Exercises

#### 3.11.3.Matching Problem, Linear Algebra.

Hint.
For openers, a basis for a subspace must be a subset of the subspace.

### 3.12Clickable Area Exercises

#### 3.12.3.Clickable Areas, Text in a Table.

Hint.
Python boolean variables begin with capital latters.

### 3.16Hodgepodge>

#### 3.16.1.With Tasks in an Exercises Division.

##### 3.16.1.aTrue/False.
Hint.
$$P_n\text{,}$$ the vector space of polynomials with degree at most $$n\text{,}$$ has dimension $$n+1$$ by Theorem 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter $$n\text{?}$$

### 3.17Exercises that are Timed

#### Timed Exercises

##### 3.17.1.True/False.
Hint.
$$P_n\text{,}$$ the vector space of polynomials with degree at most $$n\text{,}$$ has dimension $$n+1$$ by Theorem 1.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter $$n\text{?}$$
##### 3.17.2.Multiple-Choice, Not Randomized, One Answer.
Hint 1.
What did you see last time you went driving?
Hint 2.
Maybe go out for a drive?