1.4.1.
Hint.
(a) \(A \cap B = \{ 2 \}\text{;}\) (b) \(B \cap C = \{ 5 \}\text{.}\)
Appendix B Hints and Answers to Selected Exercises
1 Preliminaries
1.4 Exercises
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.
Observe that \(x \in A \cup B\) if and only if \(x \in A\) or \(x \in B\text{.}\) Equivalently, \(x \in B\) or \(x \in A\text{,}\) which is the same as \(x \in B \cup A\text{.}\) Therefore, \(A \cup B = B \cup A\text{.}\)
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.
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
1.4.17.
Hint.
(a) Not a map since \(f(2/3)\) is undefined; (b) this is a map; (c) not a map, since \(f(1/2) = 3/4\) but \(f(2/4)=3/8\text{;}\) (d) this is a map.
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.
1.4.23.
Hint.
\(f^{-1}(x) = (x+1)/(x-1)\text{.}\)
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 \(A_2\) 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 \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
1.4.25.
Hint.
(a) The relation fails to be symmetric. (b) The relation is not reflexive, since \(0\) is not equivalent to itself. (c) The relation is not transitive.
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.
Hint.
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.
Hint.
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 \(n\text{.}\)
2.4.8.
Hint.
Follow the proof in Example 2.4.
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 \(n\text{.}\)
2.4.17. Fibonacci Numbers.
Hint.
For (a) and (b) use mathematical induction. (c) Show that \(f_1 = 1\text{,}\) \(f_2 = 1\text{,}\) and \(f_{n + 2} = f_{n + 1} + f_n\text{.}\) (e) Use part (b) and Exercise 2.4.16.
2.4.19.
Hint.
Use the Fundamental Theorem of Arithmetic.
2.4.23.
Hint.
Use the Principle of Well-Ordering and the division algorithm.
2.4.27.
Hint.
Since \(\gcd(a,b) = 1\text{,}\) there exist integers \(r\) and \(s\) such that \(ar + bs = 1\text{.}\) Thus, \(acr + bcs = c\text{.}\)
2.4.29.
Hint.
Every prime must be of the form \(2\text{,}\) \(3\text{,}\) \(6n + 1\text{,}\) 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 \(a_2, \ldots\text{,}\) 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}\text{.}
\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.
\(H_1 = \{ \identity \}\text{,}\) \(H_2 = \{ \identity, \rho_1, \rho_2 \}\text{,}\) \(H_3 = \{ \identity, \mu_1 \}\text{,}\) \(H_4 = \{ \identity, \mu_2 \}\text{,}\) \(H_5 = \{ \identity, \mu_3 \}\text{,}\) \(S_3\text{.}\)
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.
\(b a = a^4 b = a^3 a b = ab\)
4 Cyclic Groups
4.5 Exercises
4.5.1.
Hint.
(a) False; (c) false; (e) true.
4.5.2.
Hint.
(a) \(12\text{;}\) (c) infinite; (e) \(10\text{.}\)
4.5.3.
Hint.
(a) \(7 {\mathbb Z} = \{ \ldots, -7, 0, 7, 14, \ldots \}\text{;}\) (b) \(\{ 0, 3, 6, 9, 12, 15, 18, 21 \}\text{;}\) (c) \(\{ 0 \}\text{,}\) \(\{ 0, 6 \}\text{,}\) \(\{ 0, 4, 8 \}\text{,}\) \(\{ 0, 3, 6, 9 \}\text{,}\) \(\{ 0, 2, 4, 6, 8, 10 \}\text{;}\) (g) \(\{ 1, 3, 7, 9 \}\text{;}\) (j) \(\{ 1, -1, i, -i \}\text{.}\)
4.5.4.
Hint.
(a)
\begin{equation*}
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix},
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}\text{.}
\end{equation*}
(c)
\begin{equation*}
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & -1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
-1 & 1 \\
-1 & 0
\end{pmatrix}, \\
\begin{pmatrix}
0 & 1 \\
-1 & 1
\end{pmatrix},
\begin{pmatrix}
0 & -1 \\
1 & -1
\end{pmatrix},
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}\text{.}
\end{equation*}
4.5.10.
Hint.
(a) \(0\text{;}\) (b) \(1, -1\text{.}\)
4.5.11.
Hint.
\(1, 2, 3, 4, 6, 8, 12, 24\text{.}\)
4.5.15.
Hint.
(a) \(-3 + 3i\text{;}\) (c) \(43- 18i\text{;}\) (e) \(i\)
4.5.16.
Hint.
(a) \(\sqrt{3} + i\text{;}\) (c) \(-3\text{.}\)
4.5.17.
Hint.
(a) \(\sqrt{2} \cis( 7 \pi /4)\text{;}\) (c) \(2 \sqrt{2} \cis( \pi /4)\text{;}\) (e) \(3 \cis(3 \pi/2)\text{.}\)
4.5.18.
Hint.
(a) \((1 - i)/2\text{;}\) (c) \(16(i - \sqrt{3}\, )\text{;}\) (e) \(-1/4\text{.}\)
4.5.22.
Hint.
(a) \(292\text{;}\) (c) \(1523\text{.}\)
4.5.27.
Hint.
\(|\langle g \rangle \cap \langle h \rangle| = 1\text{.}\)
4.5.31.
Hint.
The identity element in any group has finite order. Let \(g, h \in G\) have orders \(m\) and \(n\text{,}\) respectively. Since \((g^{-1})^m = e\) and \((gh)^{mn} = e\text{,}\) the elements of finite order in \(G\) form a subgroup of \(G\text{.}\)
4.5.37.
Hint.
If \(g\) is an element distinct from the identity in \(G\text{,}\) \(g\) must generate \(G\text{;}\) otherwise, \(\langle g \rangle\) is a nontrivial proper subgroup of \(G\text{.}\)
5 Permutation Groups
5.4 Exercises
5.4.1.
Hint.
(a) \((1 \, 2 \, 4 \, 5 \, 3)\text{;}\) (c) \((1 \, 3)(2 \, 5)\text{.}\)
5.4.2.
Hint.
(a) \((1 \, 3 \, 5)(2 \, 4)\text{;}\) (c) \((1 \, 4)(2 \, 3)\text{;}\) (e) \((1 \, 3 \, 2 \, 4)\text{;}\) (g) \((1 \, 3 \, 4)(2 \, 5)\text{;}\) (n) \((1 \, 7 \, 3 \, 5 \, 2)\text{.}\)
5.4.3.
Hint.
(a) \((1 \, 6)(1 \, 5)(1 \, 3)(1 \, 4)\text{;}\) (c) \((1 \, 6)(1 \, 4)(1 \, 2)\text{.}\)
5.4.4.
Hint.
\((a_1, a_2, \ldots, a_n)^{-1} = (a_1, a_{n}, a_{n-1}, \ldots, a_2)\)
5.4.5.
Hint.
(a) \(\{ (1 \, 3), (1 \, 3)(2 \, 4), (1 \, 3 \, 2), (1 \, 3 \, 4), (1 \, 3 \, 2 \, 4), (1 \, 3 \, 4 \, 2) \}\) is not a subgroup.
5.4.8.
Hint.
\((1 \, 2 \, 3 \, 4 \, 5)(6 \, 7 \, 8)\text{.}\)
5.4.11.
Hint.
Permutations of the form
\begin{equation*}
(1), (a_1, a_2)(a_3, a_4), (a_1, a_2, a_3), (a_1, a_2, a_3, a_4, a_5)
\end{equation*}
are possible for \(A_5\text{.}\)
5.4.17.
Hint.
Calculate \((1 \, 2 \, 3)(1 \, 2)\) and \((1 \, 2)(1 \, 2 \, 3)\text{.}\)
5.4.25.
Hint.
Consider the cases \((a,b)(b,c)\) and \((a,b)(c,d)\text{.}\)
5.4.29.
Hint.
Show that the center of \(D_n\) consists of the identity if \(n\) is odd and consists of the identity and a \(180^\circ\) rotation if \(n\) is even.
5.4.30.
Hint.
For (a), show that \(\sigma \tau \sigma^{-1 }(\sigma(a_i)) = \sigma(a_{i + 1})\text{.}\)
6 Cosets and Lagrange’s Theorem
6.5 Exercises
6.5.1.
Hint.
The order of \(g\) and the order \(h\) must both divide the order of \(G\text{.}\)
6.5.2.
Hint.
The possible orders must divide \(60\text{.}\)
6.5.3.
Hint.
This is true for every proper nontrivial subgroup.
6.5.4.
Hint.
False.
6.5.5.
Hint.
(a) \(\langle 8 \rangle\text{,}\) \(1 + \langle 8 \rangle\text{,}\) \(2 + \langle 8 \rangle\text{,}\) \(3 + \langle 8 \rangle\text{,}\) \(4 + \langle 8 \rangle\text{,}\) \(5 + \langle 8 \rangle\text{,}\) \(6 + \langle 8 \rangle\text{,}\) and \(7 + \langle 8 \rangle\text{;}\) (c) \(3 {\mathbb Z}\text{,}\) \(1 + 3 {\mathbb Z}\text{,}\) and \(2 + 3 {\mathbb Z}\text{.}\)
6.5.7.
Hint.
\(4^{\phi(15)} \equiv 4^8 \equiv 1 \pmod{15}\text{.}\)
6.5.12.
Hint.
Let \(g_1 \in gH\text{.}\) Show that \(g_1 \in Hg\) and thus \(gH \subset Hg\text{.}\)
6.5.19.
Hint.
Show that \(g(H \cap K) = gH \cap gK\text{.}\)
6.5.22.
Hint.
7 Introduction to Cryptography
7.4 Exercises
7.4.1.
Hint.
LAORYHAPDWK7.4.3.
Hint.
Hint:
V = E, E = X (also used for spaces and punctuation), K = R.7.4.4.
Hint.
\(26! - 1\)
7.4.7.
Hint.
(a) \(2791\text{;}\) (c) \(112135 25032 442\text{.}\)
7.4.9.
Hint.
(a) \(31\) (c) \(14\text{.}\)
7.4.10.
Hint.
(a) \(n = 11 \cdot 41\text{;}\) (c) \(n = 8779 \cdot 4327\text{.}\)
8 Algebraic Coding Theory
8.6 Exercises
8.6.2.
Hint.
This cannot be a group code since \((0000) \notin C\text{.}\)
8.6.3.
Hint.
(a) \(2\text{;}\) (c) \(2\text{.}\)
8.6.4.
Hint.
(a) \(3\text{;}\) (c) \(4\text{.}\)
8.6.6.
Hint.
(a) \(d_{\min} = 2\text{;}\) (c) \(d_{\min} = 1\text{.}\)
8.6.7.
Hint.
- \((00000), (00101), (10011), (10110)\)\begin{equation*} G = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \end{equation*}
- \((000000), (010111), (101101), (111010)\)\begin{equation*} G = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \end{equation*}
8.6.9.
Hint.
Multiple errors occur in one of the received words.
8.6.11.
Hint.
(a) A canonical parity-check matrix with standard generator matrix
\begin{equation*}
G =
\begin{pmatrix}
1 \\ 1 \\ 0 \\ 0 \\ 1
\end{pmatrix}\text{.}
\end{equation*}
(c) A canonical parity-check matrix with standard generator matrix
\begin{equation*}
G =
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
1 & 1 \\
1 & 0
\end{pmatrix}\text{.}
\end{equation*}
8.6.12.
Hint.
(a) All possible syndromes occur.
8.6.15.
Hint.
(a) \(C\text{,}\) \((10000) + C\text{,}\) \((01000) + C\text{,}\) \((00100) + C\text{,}\) \((00010) + C\text{,}\) \((11000) + C\text{,}\) \((01100) + C\text{,}\) \((01010) + C\text{.}\) A decoding table does not exist for \(C\) since this is only a single error-detecting code.
8.6.19.
Hint.
Let \({\mathbf x} \in C\) have odd weight and define a map from the set of odd codewords to the set of even codewords by \({\mathbf y} \mapsto {\mathbf x} + {\mathbf y}\text{.}\) Show that this map is a bijection.
8.6.23.
Hint.
For \(20\) information positions, at least 6 check bits are needed to ensure an error-correcting code.
9 Isomorphisms
9.4 Exercises
9.4.1.
Hint.
Every infinite cyclic group is isomorphic to \({\mathbb Z}\) by Theorem 9.7.
9.4.2.
Hint.
Define \(\phi: {\mathbb C}^* \rightarrow GL_2( {\mathbb R})\) by
\begin{equation*}
\phi(a + bi) =
\begin{pmatrix}
a & b \\
-b & a
\end{pmatrix}\text{.}
\end{equation*}
9.4.3.
Hint.
False.
9.4.6.
Hint.
Define a map from \({\mathbb Z}_n\) into the \(n\)th roots of unity by \(k \mapsto \cis(2k\pi / n)\text{.}\)
9.4.8.
Hint.
Assume that \({\mathbb Q}\) is cyclic and try to find a generator.
9.4.11.
Hint.
There are two nonabelian and three abelian groups that are not isomorphic.
9.4.16.
Hint.
(a) \(12\text{;}\) (c) \(5\text{.}\)
9.4.19.
Hint.
Draw the picture.
9.4.20.
Hint.
True.
9.4.25.
Hint.
True.
9.4.27.
Hint.
Let \(a\) be a generator for \(G\text{.}\) If \(\phi :G \rightarrow H\) is an isomorphism, show that \(\phi(a)\) is a generator for \(H\text{.}\)
9.4.38.
Hint.
Any automorphism of \({\mathbb Z}_6\) must send 1 to another generator of \({\mathbb Z}_6\text{.}\)
9.4.45.
Hint.
To show that \(\phi\) is one-to-one, let \(g_1 = h_1 k_1\) and \(g_2 = h_2 k_2\) and consider \(\phi(g_1) = \phi(g_2)\text{.}\)
10 Normal Subgroups and Factor Groups
10.4 Exercises
10.4.1.
Hint.
(a)
\begin{equation*}
\begin{array}{c|cc}
& A_4 & (1 \, 2)A_4 \\ \hline
A_4 & A_4 & (1 \, 2) A_4 \\
(1 \, 2) A_4 & (1 \, 2) A_4 & A_4
\end{array}
\end{equation*}
(c) \(D_4\) is not normal in \(S_4\text{.}\)
10.4.8.
Hint.
If \(a \in G\) is a generator for \(G\text{,}\) then \(aH\) is a generator for \(G/H\text{.}\)
10.4.11.
Hint.
For any \(g \in G\text{,}\) show that the map \(i_g : G \to G\) defined by \(i_g : x \mapsto gxg^{-1}\) is an isomorphism of \(G\) with itself. Then consider \(i_g(H)\text{.}\)
10.4.12.
Hint.
Suppose that \(\langle g \rangle\) is normal in \(G\) and let \(y\) be an arbitrary element of \(G\text{.}\) If \(x \in C(g)\text{,}\) we must show that \(y x y^{-1}\) is also in \(C(g)\text{.}\) Show that \((y x y^{-1}) g = g (y x y^{-1})\text{.}\)
10.4.14.
Hint.
(a) Let \(g \in G\) and \(h \in G'\text{.}\) If \(h = aba^{-1}b^{-1}\text{,}\) then
\begin{align*}
ghg^{-1} & = gaba^{-1}b^{-1}g^{-1}\\
& = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\\
& = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}\text{.}
\end{align*}
We also need to show that if \(h = h_1 \cdots h_n\) with \(h_i = a_i b_i a_i^{-1} b_i^{-1}\text{,}\) then \(ghg^{-1}\) is a product of elements of the same type. However, \(ghg^{-1} = g h_1 \cdots h_n g^{-1} = (gh_1g^{-1})(gh_2g^{-1}) \cdots (gh_ng^{-1})\text{.}\)
11 Homomorphisms
11.4 Exercises
11.4.2.
Hint.
(a) is a homomorphism with kernel \(\{ 1 \}\text{;}\) (c) is not a homomorphism.
11.4.4.
Hint.
Since \(\phi(m + n) = 7(m+n) = 7m + 7n = \phi(m) + \phi(n)\text{,}\) \(\phi\) is a homomorphism.
11.4.5.
Hint.
For any homomorphism \(\phi : {\mathbb Z}_{24} \rightarrow {\mathbb Z}_{18}\text{,}\) the kernel of \(\phi\) must be a subgroup of \({\mathbb Z}_{24}\) and the image of \(\phi\) must be a subgroup of \({\mathbb Z}_{18}\text{.}\) Now use the fact that a generator must map to a generator.
11.4.9.
Hint.
Let \(a, b \in G\text{.}\) Then \(\phi(a) \phi(b) = \phi(ab) = \phi(ba) = \phi(b)\phi(a)\text{.}\)
11.4.17.
Hint.
Find a counterexample.
12 Matrix Groups and Symmetry
12.4 Exercises
12.4.1.
Hint.
\begin{align*}
\frac{1}{2} \left[ \|{\mathbf x} + {\mathbf y}\|^2 + \|{\mathbf x}\|^2 - \| {\mathbf y}\|^2 \right] & = \frac{1}{2} \left[ \langle x + y, x + y \rangle - \|{\mathbf x}\|^2 - \| {\mathbf y}\|^2 \right]\\
& = \frac{1}{2} \left[ \| {\mathbf x}\|^2 + 2 \langle x, y \rangle + \| {\mathbf y}\|^2 - \|{\mathbf x}\|^2 - \| {\mathbf y}\|^2 \right]\\
& = \langle {\mathbf x}, {\mathbf y} \rangle\text{.}
\end{align*}
12.4.3.
Hint.
(a) is in \(SO(2)\text{;}\) (c) is not in \(O(3)\text{.}\)
12.4.5.
Hint.
(a) \(\langle {\mathbf x}, {\mathbf y} \rangle = \langle {\mathbf y}, {\mathbf x} \rangle\text{.}\)
12.4.7.
Hint.
Use the unimodular matrix
\begin{equation*}
\begin{pmatrix}
5 & 2 \\
2 & 1
\end{pmatrix}\text{.}
\end{equation*}
12.4.10.
Hint.
Show that the kernel of the map \(\det : O(n) \rightarrow {\mathbb R}^*\) is \(SO(n)\text{.}\)
12.4.13.
Hint.
True.
12.4.17.
Hint.
\(p6m\)
13 The Structure of Groups
13.4 Exercises
13.4.1.
Hint.
There are three possible groups.
13.4.4.
Hint.
(a) \(\{ 0 \} \subset \langle 6 \rangle \subset \langle 3 \rangle \subset {\mathbb Z}_{12}\text{;}\) (e) \(\{ (1) \} \times \{ 0 \} \subset \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \} \times \{ 0 \} \subset S_3 \times \{ 0 \} \subset S_3 \times \langle 2 \rangle\subset S_3 \times {\mathbb Z}_4\text{.}\)
13.4.7.
Hint.
Use the Fundamental Theorem of Finitely Generated Abelian Groups.
13.4.12.
Hint.
If \(N\) and \(G/N\) are solvable, then they have solvable series
\begin{gather*}
N = N_n \supset N_{n - 1} \supset \cdots \supset N_1 \supset N_0 = \{ e \}\\
G/N = G_n/N \supset G_{n - 1}/N \supset \cdots G_1/N \supset G_0/N = \{ N \}\text{.}
\end{gather*}
13.4.16.
Hint.
Use the fact that \(D_n\) has a cyclic subgroup of index \(2\text{.}\)
13.4.21.
Hint.
\(G/G'\) is abelian.
14 Group Actions
14.5 Exercises
14.5.1.
Hint.
Example 14.1: \(0\text{,}\) \({\mathbb R}^2 \setminus \{ 0 \}\text{.}\) Example 14.2: \(X = \{ 1, 2, 3, 4 \}\text{.}\)
14.5.2.
Hint.
(a) \(X_{(1)} = \{1, 2, 3 \}\text{,}\) \(X_{(1 \, 2)} = \{3 \}\text{,}\) \(X_{(1 \, 3)} = \{ 2 \}\text{,}\) \(X_{(2 \, 3)} = \{1 \}\text{,}\) \(X_{(1 \, 2 \, 3)} = X_{(1 \, 3 \, 2)} = \emptyset\text{.}\) \(G_1 = \{ (1), (2 \, 3) \}\text{,}\) \(G_2 = \{(1), (1 \, 3) \}\text{,}\) \(G_3 = \{ (1), (1 \, 2)\}\text{.}\)
14.5.3.
Hint.
(a) \({\mathcal O}_1 = {\mathcal O}_2 = {\mathcal O}_3 = \{ 1, 2, 3\}\text{.}\)
14.5.6.
Hint.
The conjugacy classes for \(S_4\) are
\begin{gather*}
{\mathcal O}_{(1)} = \{ (1) \},\\
{\mathcal O}_{(12)} = \{ (1 \, 2), (1 \, 3), (1 \, 4), (2 \, 3), (2 \, 4), (3 \, 4) \},\\
{\mathcal O}_{(1 \, 2)(3 \, 4)} = \{ (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \},\\
{\mathcal O}_{(123)} = \{ (1 \, 2 \, 3), (1 \, 3 \, 2), (1 \, 2 \, 4), (1 \, 4 \, 2), (1 \, 3 \, 4), (1 \, 4 \, 3), (2 \, 3 \, 4), (2 \, 4 \, 3) \},\\
{\mathcal O}_{(1234)} = \{ (1 \, 2 \, 3 \, 4), (1 \, 2 \, 4 \, 3), (1 \, 3 \, 2 \, 4), (1 \, 3 \, 4 \, 2), (1 \, 4 \, 2 \, 3), (1 \, 4 \, 3 \, 2) \}\text{.}
\end{gather*}
The class equation is \(1 + 3 + 6 + 6 + 8 = 24\text{.}\)
14.5.8.
Hint.
\((3^4 + 3^1 + 3^2 + 3^1 + 3^2 + 3^2 + 3^3 + 3^3)/8 = 21\text{.}\)
14.5.11.
Hint.
The group of rigid motions of the cube can be described by the allowable permutations of the six faces and is isomorphic to \(S_4\text{.}\) There are the identity cycle, 6 permutations with the structure \((abcd)\) that correspond to the quarter turns, 3 permutations with the structure \((ab)(cd)\) that correspond to the half turns, 6 permutations with the structure \((ab)(cd)(ef)\) that correspond to rotating the cube about the centers of opposite edges, and 8 permutations with the structure \((abc)(def)\) that correspond to rotating the cube about opposite vertices.
14.5.15.
Hint.
\((1 \cdot 2^6 + 3 \cdot 2^4 + 4 \cdot 2^3 + 2 \cdot 2^2 + 2 \cdot 2^1)/12 = 13\text{.}\)
14.5.17.
Hint.
\((1 \cdot 2^8 + 3 \cdot 2^6 + 2 \cdot 2^4)/6 = 80\text{.}\)
14.5.22.
Hint.
Use the fact that \(x \in g C(a) g^{-1}\) if and only if \(g^{-1}x g \in C(a)\text{.}\)
15 The Sylow Theorems
15.4 Exercises
15.4.1.
Hint.
If \(|G| = 18 = 2 \cdot 3^2\text{,}\) then the order of a Sylow \(2\)-subgroup is \(2\text{,}\) and the order of a Sylow \(3\)-subgroup is \(9\text{.}\)
15.4.2.
Hint.
The four Sylow \(3\)-subgroups of \(S_4\) are \(P_1 = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\text{,}\) \(P_2 = \{ (1), (1 \, 2 \, 4), (1 \, 4 \, 2) \}\text{,}\) \(P_3 = \{ (1), (1 \, 3 \, 4), (1 \, 4 \, 3) \}\text{,}\) \(P_4 = \{ (1), (2 \, 3 \, 4), (2 \, 4 \, 3) \}\text{.}\)
15.4.5.
Hint.
Since \(|G| = 96 = 2^5 \cdot 3\text{,}\) \(G\) has either one or three Sylow \(2\)-subgroups by the Third Sylow Theorem. If there is only one subgroup, we are done. If there are three Sylow \(2\)-subgroups, let \(H\) and \(K\) be two of them. Therefore, \(|H \cap K| \geq 16\text{;}\) otherwise, \(HK\) would have \((32 \cdot 32)/8 = 128\) elements, which is impossible. Thus, \(H \cap K\) is normal in both \(H\) and \(K\) since it has index \(2\) in both groups.
15.4.8.
Hint.
Show that \(G\) has a normal Sylow \(p\)-subgroup of order \(p^2\) and a normal Sylow \(q\)-subgroup of order \(q^2\text{.}\)
15.4.10.
Hint.
False.
15.4.17.
Hint.
If \(G\) is abelian, then \(G\) is cyclic, since \(|G| = 3 \cdot 5 \cdot 17\text{.}\) Now look at Example 15.14.
15.4.23.
Hint.
Define a mapping between the right cosets of \(N(H)\) in \(G\) and the conjugates of \(H\) in \(G\) by \(N(H) g \mapsto g^{-1} H g\text{.}\) Prove that this map is a bijection.
15.4.26.
Hint.
Let \(a G', b G' \in G/G'\text{.}\) Then \((a G')( b G') = ab G' = ab(b^{-1}a^{-1}ba) G' = (abb^{-1}a^{-1})ba G' = ba G'\text{.}\)
16 Rings
16.7 Exercises
16.7.1.
Hint.
(a) \(7 {\mathbb Z}\) is a ring but not a field; (c) \({\mathbb Q}(\sqrt{2}\, )\) is a field; (f) \(R\) is not a ring.
16.7.3.
Hint.
(a) \(\{1, 3, 7, 9 \}\text{;}\) (c) \(\{ 1, 2, 3, 4, 5, 6 \}\text{;}\) (e)
\begin{equation*}
\left\{
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix},
\right\}\text{.}
\end{equation*}
16.7.4.
Hint.
(a) \(\{0 \}\text{,}\) \(\{0, 9 \}\text{,}\) \(\{0, 6, 12 \}\text{,}\) \(\{0, 3, 6, 9, 12, 15 \}\text{,}\) \(\{0, 2, 4, 6, 8, 10, 12, 14, 16 \}\text{;}\) (c) there are no nontrivial ideals.
16.7.7.
Hint.
Assume there is an isomorphism \(\phi: {\mathbb C} \rightarrow {\mathbb R}\) with \(\phi(i) = a\text{.}\)
16.7.8.
Hint.
False. Assume there is an isomorphism \(\phi: {\mathbb Q}(\sqrt{2}\, ) \rightarrow {\mathbb Q}(\sqrt{3}\, )\) such that \(\phi(\sqrt{2}\, ) = a\text{.}\)
16.7.13.
Hint.
(a) \(x \equiv 17 \pmod{55}\text{;}\) (c) \(x \equiv 214 \pmod{2772}\text{.}\)
16.7.16.
Hint.
If \(I \neq \{ 0 \}\text{,}\) show that \(1 \in I\text{.}\)
16.7.18.
Hint.
(a) \(\phi(a) \phi(b) = \phi(ab) = \phi(ba) = \phi(b) \phi(a)\text{.}\)
16.7.26.
Hint.
Let \(a \in R\) with \(a \neq 0\text{.}\) Then the principal ideal generated by \(a\) is \(R\text{.}\) Thus, there exists a \(b \in R\) such that \(ab =1\text{.}\)
16.7.28.
Hint.
Compute \((a+b)^2\) and \((-ab)^2\text{.}\)
16.7.33.
Hint.
Let \(a/b, c/d \in {\mathbb Z}_{(p)}\text{.}\) Then \(a/b + c/d = (ad + bc)/bd\) and \((a/b) \cdot (c/d) = (ac)/(bd)\) are both in \({\mathbb Z}_{(p)}\text{,}\) since \(\gcd(bd,p) = 1\text{.}\)
16.7.37.
Hint.
Suppose that \(x^2 = x\) and \(x \neq 0\text{.}\) Since \(R\) is an integral domain, \(x = 1\text{.}\) To find a nontrivial idempotent, look in \({\mathbb M}_2({\mathbb R})\text{.}\)
17 Polynomials
17.5 Exercises
17.5.2.
Hint.
(a) \(9x^2 + 2x + 5\text{;}\) (b) \(8x^4 + 7x^3 + 2x^2 + 7x\text{.}\)
17.5.3.
Hint.
(a) \(5 x^3 + 6 x^2 - 3 x + 4 = (5 x^2 + 2x + 1)(x -2) + 6\text{;}\) (c) \(4x^5 - x^3 + x^2 + 4 = (4x^2 + 4)(x^3 + 3) + 4x^2 + 2\text{.}\)
17.5.5.
Hint.
(a) No zeros in \({\mathbb Z}_{12}\text{;}\) (c) \(3\text{,}\) \(4\text{.}\)
17.5.7.
Hint.
Look at \((2x + 1)\text{.}\)
17.5.8.
Hint.
(a) Reducible; (c) irreducible.
17.5.10.
Hint.
One factorization is \(x^2 + x + 8 = (x + 2)(x + 9)\text{.}\)
17.5.13.
Hint.
The integers \(\mathbb Z\) do not form a field.
17.5.14.
Hint.
False.
17.5.16.
Hint.
Let \(\phi : R \rightarrow S\) be an isomorphism. Define \(\overline{\phi} : R[x] \rightarrow S[x]\) by \(\overline{\phi}(a_0 + a_1 x + \cdots + a_n x^n) = \phi(a_0) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.}\)
17.5.20. Cyclotomic Polynomials.
Hint.
The polynomial
\begin{equation*}
\Phi_n(x) = \frac{x^n - 1}{x - 1} = x^{n - 1} + x^{n - 2} + \cdots + x + 1
\end{equation*}
is called the cyclotomic polynomial. Show that \(\Phi_p(x)\) is irreducible over \({\mathbb Q}\) for any prime \(p\text{.}\)
17.5.26.
Hint.
Find a nontrivial proper ideal in \(F[x]\text{.}\)
18 Integral Domains
18.4 Exercises
18.4.1.
Hint.
Note that \(z^{-1} = 1/(a + b\sqrt{3}\, i) = (a -b \sqrt{3}\, i)/(a^2 + 3b^2)\) is in \({\mathbb Z}[\sqrt{3}\, i]\) if and only if \(a^2 + 3 b^2 = 1\text{.}\) The only integer solutions to the equation are \(a = \pm 1, b = 0\text{.}\)
18.4.2.
Hint.
(a) \(5 = -i(1 + 2i)(2 + i)\text{;}\) (c) \(6 + 8i = -i(1 + i)^2(2 + i)^2\text{.}\)
18.4.4.
Hint.
True.
18.4.9.
Hint.
Let \(z = a + bi\) and \(w = c + di \neq 0\) be in \({\mathbb Z}[i]\text{.}\) Prove that \(z/w \in {\mathbb Q}(i)\text{.}\)
18.4.15.
Hint.
Let \(a = ub\) with \(u\) a unit. Then \(\nu(b) \leq \nu(ub) \leq \nu(a)\text{.}\) Similarly, \(\nu(a) \leq \nu(b)\text{.}\)
18.4.16.
Hint.
Show that 21 can be factored in two different ways.
19 Lattices and Boolean Algebras
19.5 Exercises
19.5.2.
Hint.
19.5.4.
Hint.
What are the atoms of \(B\text{?}\)
19.5.5.
Hint.
False.
19.5.6.
Hint.
(a) \((a \vee b \vee a') \wedge a\)
(c) \(a \vee (a \wedge b)\)
19.5.8.
Hint.
Not equivalent.
19.5.10.
Hint.
(a) \(a' \wedge [(a \wedge b') \vee b] = a \wedge (a \vee b) \text{.}\)
19.5.14.
Hint.
Let \(I, J\) be ideals in \(R\text{.}\) We need to show that \(I + J = \{ r + s : r \in I \text{ and } s \in J \}\) is the smallest ideal in \(R\) containing both \(I\) and \(J\text{.}\) If \(r_1, r_2 \in I\) and \(s_1, s_2 \in J\text{,}\) then \((r_1 + s_1) + (r_2 + s_2) = (r_1 + r_2) +(s_1 + s_2)\) is in \(I + J\text{.}\) For \(a \in R\text{,}\) \(a(r_1 + s_1) = ar_1 + as_1 \in I + J\text{;}\) hence, \(I + J\) is an ideal in \(R\text{.}\)
19.5.18.
Hint.
(a) No.
19.5.20.
Hint.
\(( \Rightarrow)\text{.}\) \(a = b \Rightarrow (a \wedge b') \vee (a' \wedge b) = (a \wedge a') \vee (a' \wedge a) = O \vee O = O\text{.}\) \(( \Leftarrow)\text{.}\) \(( a \wedge b') \vee (a' \wedge b) = O \Rightarrow a \vee b = (a \vee a) \vee b = a \vee (a \vee b) = a \vee [I \wedge (a \vee b)] = a \vee [(a \vee a') \wedge (a \vee b)] = [a \vee (a \wedge b')] \vee [a \vee (a' \wedge b)] = a \vee [(a \wedge b') \vee (a' \wedge b)] = a \vee 0 = a\text{.}\) A symmetric argument shows that \(a \vee b = b\text{.}\)
20 Vector Spaces
20.5 Exercises
20.5.3.
Hint.
\({\mathbb Q}(\sqrt{2}, \sqrt{3}\, )\) has basis \(\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}\) over \({\mathbb Q}\text{.}\)
20.5.5.
Hint.
The set \(\{ 1, x, x^2, \ldots,
x^{n-1} \}\) is a basis for \(P_n\text{.}\)
20.5.7.
Hint.
(a) Subspace of dimension \(2\) with basis \(\{(1, 0, -3), (0, 1, 2) \}\text{;}\) (d) not a subspace
20.5.10.
Hint.
Since \(0 = \alpha 0 = \alpha(-v + v) = \alpha(-v) + \alpha v\text{,}\) it follows that \(- \alpha v = \alpha(-v)\text{.}\)
20.5.12.
Hint.
Let \(v_0 = 0, v_1, \ldots,
v_n \in V\) and \(\alpha_0 \neq 0, \alpha_1, \ldots, \alpha_n \in F\text{.}\) Then \(\alpha_0 v_0 + \cdots + \alpha_n v_n = 0\text{.}\)
20.5.15. Linear Transformations.
Hint.
(a) Let \(u, v \in \ker(T)\) and \(\alpha \in F\text{.}\) Then
\begin{gather*}
T(u +v) = T(u) + T(v) = 0\\
T(\alpha v) = \alpha T(v) = \alpha 0 = 0\text{.}
\end{gather*}
Hence, \(u + v, \alpha v \in \ker(T)\text{,}\) and \(\ker(T)\) is a subspace of \(V\text{.}\)
(c) The statement that \(T(u) = T(v)\) is equivalent to \(T(u-v) = T(u) - T(v) = 0\text{,}\) which is true if and only if \(u-v = 0\) or \(u = v\text{.}\)
20.5.17. Direct Sums.
Hint.
(a) Let \(u, u' \in U\) and \(v, v' \in V\text{.}\) Then
\begin{align*}
(u + v) + (u' + v') & = (u + u') + (v + v') \in U + V\\
\alpha(u + v) & = \alpha u + \alpha v \in U + V\text{.}
\end{align*}
21 Fields
21.5 Exercises
21.5.1.
Hint.
(a) \(x^4 - (2/3) x^2 - 62/9\text{;}\) (c) \(x^4 - 2 x^2 + 25\text{.}\)
21.5.2.
Hint.
(a) \(\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}\text{;}\) (c) \(\{ 1, i, \sqrt{2}, \sqrt{2}\, i \}\text{;}\) (e) \(\{1, 2^{1/6}, 2^{1/3}, 2^{1/2}, 2^{2/3}, 2^{5/6} \}\text{.}\)
21.5.3.
Hint.
(a) \({\mathbb Q}(\sqrt{3}, \sqrt{7}\, )\text{.}\)
21.5.5.
Hint.
Use the fact that the elements of \({\mathbb Z}_2[x]/ \langle x^3 + x + 1 \rangle\) are 0, 1, \(\alpha\text{,}\) \(1 + \alpha\text{,}\) \(\alpha^2\text{,}\) \(1 + \alpha^2\text{,}\) \(\alpha + \alpha^2\text{,}\) \(1 + \alpha + \alpha^2\) and the fact that \(\alpha^3 + \alpha + 1 = 0\text{.}\)
21.5.8.
Hint.
False.
21.5.14.
Hint.
Suppose that \(E\) is algebraic over \(F\) and \(K\) is algebraic over \(E\text{.}\) Let \(\alpha \in K\text{.}\) It suffices to show that \(\alpha\) is algebraic over some finite extension of \(F\text{.}\) Since \(\alpha\) is algebraic over \(E\text{,}\) it must be the zero of some polynomial \(p(x) = \beta_0 + \beta_1 x + \cdots + \beta_n x^n\) in \(E[x]\text{.}\) Hence \(\alpha\) is algebraic over \(F(\beta_0, \ldots, \beta_n)\text{.}\)
21.5.22.
Hint.
Since \(\{ 1, \sqrt{3}, \sqrt{7}, \sqrt{21}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, )\) over \({\mathbb Q}\text{,}\) \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) \supset {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )\text{.}\) Since \([{\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) : {\mathbb Q}] = 4\text{,}\) \([{\mathbb Q}( \sqrt{3} + \sqrt{7}\, ) : {\mathbb Q}] = 2\) or 4. Since the degree of the minimal polynomial of \(\sqrt{3} +\sqrt{7}\) is 4, \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )\text{.}\)
21.5.27.
Hint.
Let \(\beta \in F(\alpha)\) not in \(F\text{.}\) Then \(\beta = p(\alpha)/q(\alpha)\text{,}\) where \(p\) and \(q\) are polynomials in \(\alpha\) with \(q(\alpha) \neq 0\) and coefficients in \(F\text{.}\) If \(\beta\) is algebraic over \(F\text{,}\) then there exists a polynomial \(f(x) \in F[x]\) such that \(f(\beta) = 0\text{.}\) Let \(f(x) = a_0 + a_1 x + \cdots + a_n x^n\text{.}\) Then
\begin{equation*}
0 = f(\beta) = f\left( \frac{p(\alpha)}{q(\alpha)} \right) = a_0 + a_1 \left( \frac{p(\alpha)}{q(\alpha)} \right) + \cdots + a_n \left( \frac{p(\alpha)}{q(\alpha)} \right)^n\text{.}
\end{equation*}
Now multiply both sides by \(q(\alpha)^n\) to show that there is a polynomial in \(F[x]\) that has \(\alpha\) as a zero.
21.5.28.
Hint.
See the comments following Theorem 21.13.
22 Finite Fields
22.4 Exercises
22.4.1.
Hint.
Make sure that you have a field extension.
22.4.4.
Hint.
There are eight elements in \({\mathbb Z}_2(\alpha)\text{.}\) Exhibit two more zeros of \(x^3 + x^2 + 1\) other than \(\alpha\) in these eight elements.
22.4.5.
Hint.
Find an irreducible polynomial \(p(x)\) in \({\mathbb Z}_3[x]\) of degree \(3\) and show that \({\mathbb Z}_3[x]/ \langle p(x) \rangle\) has \(27\) elements.
22.4.7.
Hint.
(a) \(x^5 -1 = (x+1)(x^4+x^3 + x^2 + x+ 1)\text{;}\) (c) \(x^9 -1 = (x+1)( x^2 + x+ 1)(x^6+x^3+1)\text{.}\)
22.4.8.
Hint.
True.
22.4.11.
Hint.
(a) Use the fact that \(x^7 - 1 = (x + 1)( x^3 + x + 1)(x^3 + x^2 + 1)\text{.}\)
22.4.12.
Hint.
False.
22.4.17.
Hint.
If \(p(x) \in F[x]\text{,}\) then \(p(x) \in E[x]\text{.}\)
22.4.18.
Hint.
Since \(\alpha\) is algebraic over \(F\) of degree \(n\text{,}\) we can write any element \(\beta \in F(\alpha)\) uniquely as \(\beta = a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1}\) with \(a_i \in F\text{.}\) There are \(q^n\) possible \(n\)-tuples \((a_0, a_1, \ldots,
a_{n - 1})\text{.}\)
22.4.24. Wilson’s Theorem.
Hint.
Factor \(x^{p-1} - 1\) over \({\mathbb Z}_p\text{.}\)
23 Galois Theory
23.5 Exercises
23.5.1.
Hint.
(a) \({\mathbb Z}_2\text{;}\) (c) \({\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\)
23.5.2.
Hint.
(a) Separable over \(\mathbb Q\) since \(x^3 + 2 x^2 - x - 2 = (x - 1)(x + 1)(x + 2)\text{;}\) (c) not separable over \(\mathbb Z_3\) since \(x^4 + x^2 + 1 = (x + 1)^2 (x + 2)^2 \text{.}\)
23.5.3.
Hint.
If
\begin{equation*}
[\gf(729): \gf(9)] = [\gf(729): \gf(3)] /[\gf(9): \gf(3)] = 6/2 = 3\text{,}
\end{equation*}
then \(G(\gf(729)/ \gf(9)) \cong {\mathbb Z}_3\text{.}\) A generator for \(G(\gf(729)/ \gf(9))\) is \(\sigma\text{,}\) where \(\sigma_{3^6}( \alpha) = \alpha^{3^6} = \alpha^{729}\) for \(\alpha \in \gf(729)\text{.}\)
23.5.4.
Hint.
23.5.5.
Hint.
(a) \({\mathbb Q}(i)\)
23.5.7.
Hint.
Let \(E\) be the splitting field of a cubic polynomial in \(F[x]\text{.}\) Show that \([E:F]\) is less than or equal to \(6\) and is divisible by \(3\text{.}\) Since \(G(E/F)\) is a subgroup of \(S_3\) whose order is divisible by \(3\text{,}\) conclude that this group must be isomorphic to \({\mathbb Z}_3\) or \(S_3\text{.}\)
23.5.9.
Hint.
\(G\) is a subgroup of \(S_n\text{.}\)
23.5.16.
Hint.
True.
23.5.20.
Hint.
- Clearly \(\omega, \omega^2, \ldots, \omega^{p - 1}\) are distinct since \(\omega \neq 1\) or 0. To show that \(\omega^i\) is a zero of \(\Phi_p\text{,}\) calculate \(\Phi_p( \omega^i)\text{.}\)
- The conjugates of \(\omega\) are \(\omega, \omega^2, \ldots, \omega^{p - 1}\text{.}\) Define a map \(\phi_i: {\mathbb Q}(\omega) \rightarrow {\mathbb Q}(\omega^i)\) by\begin{equation*} \phi_i(a_0 + a_1 \omega + \cdots + a_{p - 2} \omega^{p - 2}) = a_0 + a_1 \omega^i + \cdots + c_{p - 2} (\omega^i)^{p - 2}\text{,} \end{equation*}where \(a_i \in {\mathbb Q}\text{.}\) Prove that \(\phi_i\) is an isomorphism of fields. Show that \(\phi_2\) generates \(G({\mathbb Q}(\omega)/{\mathbb Q})\text{.}\)
- Show that \(\{ \omega, \omega^2, \ldots, \omega^{p - 1} \}\) is a basis for \({\mathbb Q}( \omega )\) over \({\mathbb Q}\text{,}\) and consider which linear combinations of \(\omega, \omega^2, \ldots, \omega^{p - 1}\) are left fixed by all elements of \(G( {\mathbb Q}( \omega ) / {\mathbb Q})\text{.}\)
