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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

I Basics
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) \(A \times D = \emptyset\text{.}\)
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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, \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
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1.4.10.
Hint.
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
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1.4.14.
Hint.
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
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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. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)
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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.
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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 \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
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1.4.28.
Hint.
Let \(X = {\mathbb N} \cup \{ \sqrt{2}\, \}\) and define \(x \sim y\) if \(x + y \in {\mathbb N}\text{.}\)
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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.
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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 \(n\text{.}\)
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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 \(n\text{.}\)
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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 \(n\text{.}\)
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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, \(m \mid n\text{.}\)
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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 \(c\text{.}\)
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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 \(6k + 5\text{.}\)
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II Algebra (and Runestone)
1 Groups
1.5 Exercises

1.5.1.

Hint.
(a) \(3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}\text{;}\) (c) \(18 + 26 \mathbb Z\text{;}\) (e) \(5 + 6 \mathbb Z\text{.}\)
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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*}
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1.5.15.

Hint.
There is a nonabelian group containing six elements.
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1.5.16.

Hint.
Look at the symmetry group of an equilateral triangle or a square.
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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.
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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}. \end{align*}
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1.5.31.

Hint.
Since \(abab = (ab)^2 = e = a^2 b^2 = aabb\text{,}\) we know that \(ba = ab\text{.}\)
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1.5.35.

Hint.
\(H_1 = \{ id \}\text{,}\) \(H_2 = \{ id, \rho_1, \rho_2 \}\text{,}\) \(H_3 = \{ id, \mu_1 \}\text{,}\) \(H_4 = \{ id, \mu_2 \}\text{,}\) \(H_5 = \{ id, \mu_3 \}\text{,}\) \(S_3\text{.}\)
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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, \((a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}\)
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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 \(ab = a^3 b a = ba\text{.}\)
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3 Runestone Testing
3.8 True/False Exercises

3.8.1. 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 \(n\text{?}\)
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3.9 Multiple Choice Exercises

3.10 Parsons Exercises

3.12 Matching Exercises

3.13 Clickable Area Exercises

3.19 Fill-In Exercises

3.20 Hodgepodge

3.21 Exercises that are Timed

Timed Exercises

3.21.1. 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 \(n\text{?}\)
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3.27 Group Exercises

3.29 Timed Chapter Exam

3.29.2. 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 \(n\text{?}\)
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