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PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

Thomas W. Judson, Isaac Newton (Editor)

Appendix D Hints and Answers to Selected Exercises

1 Preliminaries
1.4 Exercises

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 \(x + y \in {\mathbb N}\text{.}\)

2 The Integers
2.4 Exercises

2.4.1.

Answer.
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, \end{align*}
and so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)

2.4.3.

Answer.
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, \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 \(6k + 5\text{.}\)

3 Groups
3.5 Exercises

3.5.1.

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

3.5.2.

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

3.5.6.

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*}

3.5.8.

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

3.5.15.

Hint.
There is a nonabelian group containing six elements.

3.5.16.

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

3.5.17.

Hint.
The are five different groups of order 8.

3.5.18.

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.

3.5.25.

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}. \end{align*}

3.5.31.

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

3.5.35.

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

3.5.41.

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, \((a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}\)

3.5.46.

Hint.
Look at \(S_3\text{.}\)

3.5.49.

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 \(ab = a^3 b a = ba\text{.}\)

3.5.55.

Answer.
\(1\)

3.5.56.

Answer.
\(2\)

3.5.57.

Answer.
\(n\)

3.5.58.

Answer.
\(n+1\)

3.5.59.

3.5.59.a
Answer.
\(2\)
3.5.59.b
3.5.59.b.i
Answer.
\(6\)
3.5.59.b.ii
Answer.
\(10\)

3.5.60.

3.5.60.a
Answer.
\(4\)
3.5.60.b
3.5.60.b.i
Answer.
\(8\)
3.5.60.b.ii
Answer.
\(12\)

5 Runestone Testing
5.7 True/False Exercises

5.7.1. True/False.

Hint.
\(P_n\text{,}\) the vector space of polynomials with degree at most \(n\text{,}\) has dimension \(n+1\) by Theorem 3.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)

5.8 Multiple Choice Exercises

5.8.1. 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?

5.8.2. Multiple-Choice, Not Randomized, Multiple Answers.

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

5.8.3. Multiple-Choice, Randomized, One Answer.

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

5.8.4. Multiple-Choice, Randomized, Multiple Answers.

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

5.8.5. 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.

5.9 Parsons Exercises

5.9.1. Parsons Problem, Mathematical Proof.

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

5.9.4. Parsons Problem, Mathematical Proof, Numbered Blocks.

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

5.11 Matching Exercises

5.11.3. Matching Problem, Linear Algebra.

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

5.12 Clickable Area Exercises

5.12.3. Clickable Areas, Text in a Table.

Hint.
Python boolean variables begin with capital latters.

5.16 Hodgepodge>

5.16.1. With Tasks in an Exercises Division.

5.16.1.a True/False.
Hint.
\(P_n\text{,}\) the vector space of polynomials with degree at most \(n\text{,}\) has dimension \(n+1\) by Theorem 3.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)

5.17 Exercises that are Timed

Timed Exercises

5.17.1. True/False.
Hint.
\(P_n\text{,}\) the vector space of polynomials with degree at most \(n\text{,}\) has dimension \(n+1\) by Theorem 3.2.16. [Cross-reference is just a demo, content is not relevant.] What happens if we relax the defintion and remove the parameter \(n\text{?}\)
5.17.2. 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?
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