$\newcommand{\identity}{\mathrm{id}} \newcommand{\notdivide}{\nmid} \newcommand{\notsubset}{\not\subset} \newcommand{\lcm}{\operatorname{lcm}} \newcommand{\gf}{\operatorname{GF}} \newcommand{\inn}{\operatorname{Inn}} \newcommand{\aut}{\operatorname{Aut}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\cis}{\operatorname{cis}} \newcommand{\chr}{\operatorname{char}} \newcommand{\Null}{\operatorname{Null}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$

# Exercises1.3Exercises

###### Exercise1
Hint

(a) $A \cap B = \{ 2 \}\text{;}$ (b) $B \cap C = \{ 5 \}\text{.}$

###### Exercise2
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) $A \times D = \emptyset\text{.}$

###### Exercise6
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, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}$

###### Exercise10
Hint

$(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}$

###### Exercise14
Hint

$A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}$

###### Exercise17
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.

###### Exercise18
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. $f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}$

###### Exercise20
Hint

(a) $f(n) = n + 1\text{.}$

###### Exercise22
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.

###### Exercise23
Hint

$f^{-1}(x) = (x+1)/(x-1)\text{.}$

###### Exercise24
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 $f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}$

###### Exercise25
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.

###### Exercise28
Hint

Let $X = {\mathbb N} \cup \{ \sqrt{2}\, \}$ and define $x \sim y$ if $x + y \in {\mathbb N}\text{.}$

# Exercises2.3Exercises

###### Exercise1
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, \end{align*}

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

###### Exercise3
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 $n\text{.}$

###### Exercise8
Hint

Follow the proof in Example 2.4.

###### Exercise11
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 $n\text{.}$

###### Exercise17Fibonacci Numbers
Hint

For (a) and (b) use mathematical induction. (c) Show that $f_1 = 1\text{,}$ $f_2 = 1\text{,}$ and $f_{n + 2} = f_{n + 1} + f_n\text{.}$ (d) Use part (c). (e) Use part (b) and Exercise 2.3.16.

###### Exercise19
Hint

Use the Fundamental Theorem of Arithmetic.

###### Exercise23
Hint

Use the Principle of Well-Ordering and the division algorithm.

###### Exercise27
Hint

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

###### Exercise29
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 $6k + 5\text{.}$

# Exercises3.4Exercises

###### Exercise1
Hint

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

###### Exercise2
Hint

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

###### Exercise6
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*}
###### Exercise8
Hint

Pick two matrices. Almost any pair will work.

###### Exercise15
Hint

There is a nonabelian group containing six elements.

###### Exercise16
Hint

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

###### Exercise17
Hint

The are five different groups of order 8.

###### Exercise18
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.

###### Exercise25
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*}
###### Exercise31
Hint

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

###### Exercise35
Hint

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

###### Exercise41
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, $(a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}$

###### Exercise46
Hint

Look at $S_3\text{.}$

###### Exercise49
Hint

$b a = a^4 b = a^3 a b = ab$

# Exercises4.4Exercises

###### Exercise1
Hint

(a) False; (c) false; (e) true.

###### Exercise2
Hint

(a) 12; (c) infinite; (e) 10.

###### Exercise3
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) $\{ 1, -1, i, -i \}\text{.}$

###### Exercise4
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}. \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}. \end{equation*}
###### Exercise10
Hint

(a) $0\text{;}$ (b) $1, -1\text{.}$

###### Exercise11
Hint

1, 2, 3, 4, 6, 8, 12, 24.

###### Exercise15
Hint

(a) $-3 + 3i\text{;}$ (c) $43- 18i\text{;}$ (e) $i$

###### Exercise16
Hint

(a) $\sqrt{3} + i\text{;}$ (c) $-3\text{.}$

###### Exercise17
Hint

(a) $\sqrt{2} \cis( 7 \pi /4)\text{;}$ (c) $2 \sqrt{2} \cis( \pi /4)\text{;}$ (e) $3 \cis(3 \pi/2)\text{.}$

###### Exercise18
Hint

(a) $(1 - i)/2\text{;}$ (c) $16(i - \sqrt{3}\, )\text{;}$ (e) $-1/4\text{.}$

###### Exercise22
Hint

(a) 292; (c) 1523.

###### Exercise27
Hint

$|\langle g \rangle \cap \langle h \rangle| = 1\text{.}$

###### Exercise31
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 $G\text{.}$

###### Exercise37
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 $G\text{.}$

# Exercises5.3Exercises

###### Exercise1
Hint

(a) $(12453)\text{;}$ (c) $(13)(25)\text{.}$

###### Exercise2
Hint

(a) $(135)(24)\text{;}$ (c) $(14)(23)\text{;}$ (e) $(1324)\text{;}$ (g) $(134)(25)\text{;}$ (n) $(17352)\text{.}$

###### Exercise3
Hint

(a) $(16)(15)(13)(14)\text{;}$ (c) $(16)(14)(12)\text{.}$

###### Exercise4
Hint

$(a_1, a_2, \ldots, a_n)^{-1} = (a_1, a_{n}, a_{n-1}, \ldots, a_2)$

###### Exercise5
Hint

(a) $\{ (13), (13)(24), (132), (134), (1324), (1342) \}$ is not a subgroup.

###### Exercise8
Hint

$(12345)(678)\text{.}$

###### Exercise11
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 $A_5\text{.}$

###### Exercise17
Hint

Calculate $(123)(12)$ and $(12)(123)\text{.}$

###### Exercise25
Hint

Consider the cases $(ab)(bc)$ and $(ab)(cd)\text{.}$

###### Exercise30
Hint

For (a), show that $\sigma \tau \sigma^{-1 }(\sigma(a_i)) = \sigma(a_{i + 1})\text{.}$

# Exercises6.4Exercises

###### Exercise1
Hint

The order of $g$ and the order $h$ must both divide the order of $G\text{.}$

###### Exercise2
Hint

The possible orders must divide 60.

###### Exercise3
Hint

This is true for every proper nontrivial subgroup.

Hint

False.

###### Exercise5
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 $2 + 3 {\mathbb Z}\text{.}$

###### Exercise7
Hint

$4^{\phi(15)} \equiv 4^8 \equiv 1 \pmod{15}\text{.}$

###### Exercise12
Hint

Let $g_1 \in gH\text{.}$ Show that $g_1 \in Hg$ and thus $gH \subset Hg\text{.}$

###### Exercise19
Hint

Show that $g(H \cap K) = gH \cap gK\text{.}$

###### Exercise22
Hint

If $\gcd(m,n) = 1\text{,}$ then $\phi(mn) = \phi(m)\phi(n)$ (Exercise 2.3.26 in Chapter 2).

# Exercises7.3Exercises

Hint

LAORYHAPDWK

###### Exercise3
Hint

Hint: V = E, E = X (also used for spaces and punctuation), K = R.

###### Exercise4
Hint

$26! - 1$

###### Exercise7
Hint

(a) 2791; (c) 112135 25032 442.

Hint

(a) 31; (c) 14.

###### Exercise10
Hint

(a) $n = 11 \cdot 41\text{;}$ (c) $n = 8779 \cdot 4327\text{.}$

# Exercises8.5Exercises

###### Exercise2
Hint

This cannot be a group code since $(0000) \notin C\text{.}$

Hint

(a) 2; (c) 2.

Hint

(a) 3; (c) 4.

###### Exercise6
Hint

(a) $d_{\min} = 2\text{;}$ (c) $d_{\min} = 1\text{.}$

###### Exercise7
Hint

1. $(00000), (00101), (10011), (10110)$

\begin{equation*} G = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \end{equation*}
2. $(000000), (010111), (101101), (111010)$

\begin{equation*} G = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \end{equation*}
###### Exercise9
Hint

Multiple errors occur in one of the received words.

###### Exercise11
Hint

(a) A canonical parity-check matrix with standard generator matrix

\begin{equation*} G = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}. \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}. \end{equation*}
###### Exercise12
Hint

(a) All possible syndromes occur.

###### Exercise15
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.

###### Exercise19
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.

###### Exercise23
Hint

For 20 information positions, at least 6 check bits are needed to ensure an error-correcting code.

# Exercises9.3Exercises

###### Exercise1
Hint

Every infinite cyclic group is isomorphic to ${\mathbb Z}$ by Theorem 9.7.

###### Exercise2
Hint

Define $\phi: {\mathbb C}^* \rightarrow GL_2( {\mathbb R})$ by

\begin{equation*} \phi(a + bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}. \end{equation*}
Hint

False.

###### Exercise6
Hint

Define a map from ${\mathbb Z}_n$ into the $n$th roots of unity by $k \mapsto \cis(2k\pi / n)\text{.}$

###### Exercise8
Hint

Assume that ${\mathbb Q}$ is cyclic and try to find a generator.

###### Exercise11
Hint

There are two nonabelian and three abelian groups that are not isomorphic.

Hint

(a) 12; (c) 5.

###### Exercise19
Hint

Draw the picture.

Hint

True.

Hint

True.

###### Exercise27
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 $H\text{.}$

###### Exercise38
Hint

Any automorphism of ${\mathbb Z}_6$ must send 1 to another generator of ${\mathbb Z}_6\text{.}$

###### Exercise45
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 $\phi(g_1) = \phi(g_2)\text{.}$

# Exercises10.3Exercises

###### Exercise1
Hint

(a)

\begin{equation*} \begin{array}{c|cc} & A_4 & (12)A_4 \\ \hline A_4 & A_4 & (12) A_4 \\ (12) A_4 & (12) A_4 & A_4 \end{array} \end{equation*}

(c) $D_4$ is not normal in $S_4\text{.}$

###### Exercise8
Hint

If $a \in G$ is a generator for $G\text{,}$ then $aH$ is a generator for $G/H\text{.}$

###### Exercise11
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 $i_g(H)\text{.}$

###### Exercise12
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 $(y x y^{-1}) g = g (y x y^{-1})\text{.}$

###### Exercise14
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}. \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, $ghg^{-1} = g h_1 \cdots h_n g^{-1} = (gh_1g^{-1})(gh_2g^{-1}) \cdots (gh_ng^{-1})\text{.}$

# Exercises11.3Exercises

###### Exercise2
Hint

(a) is a homomorphism with kernel $\{ 1 \}\text{;}$ (c) is not a homomorphism.

###### Exercise4
Hint

Since $\phi(m + n) = 7(m+n) = 7m + 7n = \phi(m) + \phi(n)\text{,}$ $\phi$ is a homomorphism.

###### Exercise5
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.

###### Exercise9
Hint

Let $a, b \in G\text{.}$ Then $\phi(a) \phi(b) = \phi(ab) = \phi(ba) = \phi(b)\phi(a)\text{.}$

###### Exercise17
Hint

Find a counterexample.

# Exercises12.3Exercises

###### Exercise1
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. \end{align*}
###### Exercise3
Hint

(a) is in $SO(2)\text{;}$ (c) is not in $O(3)\text{.}$

###### Exercise5
Hint

(a) $\langle {\mathbf x}, {\mathbf y} \rangle = \langle {\mathbf y}, {\mathbf x} \rangle\text{.}$

###### Exercise7
Hint

Use the unimodular matrix

\begin{equation*} \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}. \end{equation*}
###### Exercise10
Hint

Show that the kernel of the map $\det : O(n) \rightarrow {\mathbb R}^*$ is $SO(n)\text{.}$

Hint

True.

###### Exercise17
Hint

$p6m$

# Exercises13.3Exercises

###### Exercise1
Hint

There are three possible groups.

###### Exercise4
Hint

(a) $\{ 0 \} \subset \langle 6 \rangle \subset \langle 3 \rangle \subset {\mathbb Z}_{12}\text{;}$ (e) $\{ (1) \} \times \{ 0 \} \subset \{ (1), (123), (132) \} \times \{ 0 \} \subset S_3 \times \{ 0 \} \subset S_3 \times \langle 2 \rangle\subset S_3 \times {\mathbb Z}_4\text{.}$

###### Exercise7
Hint

Use the Fundamental Theorem of Finitely Generated Abelian Groups.

###### Exercise12
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 \}. \end{gather*}
###### Exercise16
Hint

Use the fact that $D_n$ has a cyclic subgroup of index 2.

###### Exercise21
Hint

$G/G'$ is abelian.

# Exercises14.4Exercises

###### Exercise1
Hint

Example 14.1: $0\text{,}$ ${\mathbb R}^2 \setminus \{ 0 \}\text{.}$ Example 14.2: $X = \{ 1, 2, 3, 4 \}\text{.}$

###### Exercise2
Hint

(a) $X_{(1)} = \{1, 2, 3 \}\text{,}$ $X_{(12)} = \{3 \}\text{,}$ $X_{(13)} = \{ 2 \}\text{,}$ $X_{(23)} = \{1 \}\text{,}$ $X_{(123)} = X_{(132)} = \emptyset\text{.}$ $G_1 = \{ (1), (23) \}\text{,}$ $G_2 = \{(1), (13) \}\text{,}$ $G_3 = \{ (1), (12)\}\text{.}$

###### Exercise3
Hint

(a) ${\mathcal O}_1 = {\mathcal O}_2 = {\mathcal O}_3 = \{ 1, 2, 3\}\text{.}$

###### Exercise6
Hint

The conjugacy classes for $S_4$ are

\begin{gather*} {\mathcal O}_{(1)} = \{ (1) \},\\ {\mathcal O}_{(12)} = \{ (12), (13), (14), (23), (24), (34) \},\\ {\mathcal O}_{(12)(34)} = \{ (12)(34), (13)(24), (14)(23) \},\\ {\mathcal O}_{(123)} = \{ (123), (132), (124), (142), (134), (143), (234), (243) \},\\ {\mathcal O}_{(1234)} = \{ (1234), (1243), (1324), (1342), (1423), (1432) \}. \end{gather*}

The class equation is $1 + 3 + 6 + 6 + 8 = 24\text{.}$

###### Exercise8
Hint

$(3^4 + 3^1 + 3^2 + 3^1 + 3^2 + 3^2 + 3^3 + 3^3)/8 = 21\text{.}$

###### Exercise11
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.

###### Exercise15
Hint

$(1 \cdot 2^6 + 3 \cdot 2^4 + 4 \cdot 2^3 + 2 \cdot 2^2 + 2 \cdot 2^1)/12 = 13\text{.}$

###### Exercise17
Hint

$(1 \cdot 2^8 + 3 \cdot 2^6 + 2 \cdot 2^4)/6 = 80\text{.}$

###### Exercise22
Hint

Use the fact that $x \in g C(a) g^{-1}$ if and only if $g^{-1}x g \in C(a)\text{.}$

# Exercises15.3Exercises

###### Exercise1
Hint

If $|G| = 18 = 2 \cdot 3^2\text{,}$ then the order of a Sylow 2-subgroup is 2, and the order of a Sylow 3-subgroup is 9.

###### Exercise2
Hint

The four Sylow 3-subgroups of $S_4$ are $P_1 = \{ (1), (123), (132) \}\text{,}$ $P_2 = \{ (1), (124), (142) \}\text{,}$ $P_3 = \{ (1), (134), (143) \}\text{,}$ $P_4 = \{ (1), (234), (243) \}\text{.}$

###### Exercise5
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.

###### Exercise8
Hint

Show that $G$ has a normal Sylow $p$-subgroup of order $p^2$ and a normal Sylow $q$-subgroup of order $q^2\text{.}$

Hint

False.

###### Exercise17
Hint

If $G$ is abelian, then $G$ is cyclic, since $|G| = 3 \cdot 5 \cdot 17\text{.}$ Now look at Example 15.14.

###### Exercise23
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.

###### Exercise26
Hint

Let $a G', b G' \in G/G'\text{.}$ Then $(a G')( b G') = ab G' = ab(b^{-1}a^{-1}ba) G' = (abb^{-1}a^{-1})ba G' = ba G'\text{.}$

# Exercises16.6Exercises

###### Exercise1
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.

###### Exercise3
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\}. \end{equation*}
###### Exercise4
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.

###### Exercise7
Hint

Assume there is an isomorphism $\phi: {\mathbb C} \rightarrow {\mathbb R}$ with $\phi(i) = a\text{.}$

###### Exercise8
Hint

False. Assume there is an isomorphism $\phi: {\mathbb Q}(\sqrt{2}\, ) \rightarrow {\mathbb Q}(\sqrt{3}\, )$ such that $\phi(\sqrt{2}\, ) = a\text{.}$

###### Exercise13
Hint

(a) $x \equiv 17 \pmod{55}\text{;}$ (c) $x \equiv 214 \pmod{2772}\text{.}$

###### Exercise16
Hint

If $I \neq \{ 0 \}\text{,}$ show that $1 \in I\text{.}$

###### Exercise18
Hint

(a) $\phi(a) \phi(b) = \phi(ab) = \phi(ba) = \phi(b) \phi(a)\text{.}$

###### Exercise26
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 $ab =1\text{.}$

###### Exercise28
Hint

Compute $(a+b)^2$ and $(-ab)^2\text{.}$

###### Exercise34
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 $\gcd(bd,p) = 1\text{.}$

###### Exercise38
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 ${\mathbb M}_2({\mathbb R})\text{.}$

# Exercises17.4Exercises

###### Exercise2
Hint

(a) $9x^2 + 2x + 5\text{;}$ (b) $8x^4 + 7x^3 + 2x^2 + 7x\text{.}$

###### Exercise3
Hint

(a) $5 x^3 + 6 x^2 - 3 x + 4 = (5 x^2 + 2x + 1)(x -2) + 6\text{;}$ (c) $4x^5 - x^3 + x^2 + 4 = (4x^2 + 4)(x^3 + 3) + 4x^2 + 2\text{.}$

###### Exercise5
Hint

(a) No zeros in ${\mathbb Z}_{12}\text{;}$ (c) 3, 4.

###### Exercise7
Hint

Look at $(2x + 1)\text{.}$

###### Exercise8
Hint

(a) Reducible; (c) irreducible.

###### Exercise10
Hint

One factorization is $x^2 + x + 8 = (x + 2)(x + 9)\text{.}$

###### Exercise13
Hint

The integers $\mathbb Z$ do not form a field.

Hint

False.

###### Exercise16
Hint

Let $\phi : R \rightarrow S$ be an isomorphism. Define $\overline{\phi} : R[x] \rightarrow S[x]$ by $\overline{\phi}(a_0 + a_1 x + \cdots + a_n x^n) = \phi(a_0) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.}$

###### Exercise20Cyclotomic 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 $p\text{.}$

###### Exercise26
Hint

Find a nontrivial proper ideal in $F[x]\text{.}$

# Exercises18.3Exercises

###### Exercise1
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 $a = \pm 1, b = 0\text{.}$

###### Exercise2
Hint

(a) $5 = -i(1 + 2i)(2 + i)\text{;}$ (c) $6 + 8i = -i(1 + i)^2(2 + i)^2\text{.}$

Hint

True.

###### Exercise9
Hint

Let $z = a + bi$ and $w = c + di \neq 0$ be in ${\mathbb Z}[i]\text{.}$ Prove that $z/w \in {\mathbb Q}(i)\text{.}$

###### Exercise15
Hint

Let $a = ub$ with $u$ a unit. Then $\nu(b) \leq \nu(ub) \leq \nu(a)\text{.}$ Similarly, $\nu(a) \leq \nu(b)\text{.}$

###### Exercise16
Hint

Show that 21 can be factored in two different ways.

# Exercises19.4Exercises

Hint
Hint

False.

###### Exercise6
Hint

(a) $(a \vee b \vee a') \wedge a$

(c) $a \vee (a \wedge b)$

Hint

Not equivalent.

###### Exercise10
Hint

(a) $a' \wedge [(a \wedge b') \vee b] = a \wedge (a \vee b) \text{.}$

###### Exercise14
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 $R\text{.}$

Hint

(a) No.

###### Exercise20
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 $a \vee b = b\text{.}$

# Exercises20.4Exercises

###### Exercise3
Hint

${\mathbb Q}(\sqrt{2}, \sqrt{3}\, )$ has basis $\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}$ over ${\mathbb Q}\text{.}$

###### Exercise5
Hint

The set $\{ 1, x, x^2, \ldots, x^{n-1} \}$ is a basis for $P_n\text{.}$

###### Exercise7
Hint

(a) Subspace of dimension 2 with basis $\{(1, 0, -3), (0, 1, 2) \}\text{;}$ (d) not a subspace

###### Exercise10
Hint

Since $0 = \alpha 0 = \alpha(-v + v) = \alpha(-v) + \alpha v\text{,}$ it follows that $- \alpha v = \alpha(-v)\text{.}$

###### Exercise12
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 $\alpha_0 v_0 + \cdots + \alpha_n v_n = 0\text{.}$

###### Exercise15Linear 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. \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 $u = v\text{.}$

###### Exercise17Direct 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. \end{align*}

# Exercises21.4Exercises

###### Exercise1
Hint

(a) $x^4 - (2/3) x^2 - 62/9\text{;}$ (c) $x^4 - 2 x^2 + 25\text{.}$

###### Exercise2
Hint

(a) $\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}\text{;}$ (c) $\{ 1, i, \sqrt{2}, \sqrt{2}\, i \}\text{;}$ (e) $\{1, 2^{1/6}, 2^{1/3}, 2^{1/2}, 2^{2/3}, 2^{5/6} \}\text{.}$

###### Exercise3
Hint

(a) ${\mathbb Q}(\sqrt{3}, \sqrt{7}\, )\text{.}$

###### Exercise5
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 $\alpha^3 + \alpha + 1 = 0\text{.}$

Hint

False.

###### Exercise14
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 $F(\beta_0, \ldots, \beta_n)\text{.}$

###### Exercise22
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, ${\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )\text{.}$

###### Exercise27
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. \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.

###### Exercise28
Hint

See the comments following Theorem 21.13.

# Exercises22.3Exercises

###### Exercise1
Hint

Make sure that you have a field extension.

###### Exercise4
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.

###### Exercise5
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.

###### Exercise7
Hint

(a) $x^5 -1 = (x+1)(x^4+x^3 + x^2 + x+ 1)\text{;}$ (c) $x^9 -1 = (x+1)( x^2 + x+ 1)(x^6+x^3+1)\text{.}$

Hint

True.

###### Exercise11
Hint

(a) Use the fact that $x^7 -1 = (x+1)( x^3 + x+ 1)(x^3+x^2+1)\text{.}$

Hint

False.

###### Exercise17
Hint

If $p(x) \in F[x]\text{,}$ then $p(x) \in E[x]\text{.}$

###### Exercise18
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 $(a_0, a_1, \ldots, a_{n-1})\text{.}$

###### Exercise24Wilson's Theorem
Hint

Factor $x^{p-1} - 1$ over ${\mathbb Z}_p\text{.}$

# Exercises23.4Exercises

###### Exercise1
Hint

(a) ${\mathbb Z}_2\text{;}$ (c) ${\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_2\text{.}$

###### Exercise2
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 $x^4 + x^2 + 1 = (x + 1)^2 (x + 2)^2 \text{.}$

###### Exercise3
Hint

If

\begin{equation*} [\gf(729): \gf(9)] = [\gf(729): \gf(3)] /[\gf(9): \gf(3)] = 6/2 = 3, \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 $\alpha \in \gf(729)\text{.}$

###### Exercise4
Hint

(a) $S_5\text{;}$ (c) $S_3\text{;}$ (g) see Example 23.10.

###### Exercise5
Hint

(a) ${\mathbb Q}(i)$

###### Exercise7
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. Since $G(E/F)$ is a subgroup of $S_3$ whose order is divisible by 3, conclude that this group must be isomorphic to ${\mathbb Z}_3$ or $S_3\text{.}$

###### Exercise9
Hint

$G$ is a subgroup of $S_n\text{.}$

Hint

True.

###### Exercise20
Hint

1. 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 $\Phi_p( \omega^i)\text{.}$

2. 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}, \end{equation*}

where $a_i \in {\mathbb Q}\text{.}$ Prove that $\phi_i$ is an isomorphism of fields. Show that $\phi_2$ generates $G({\mathbb Q}(\omega)/{\mathbb Q})\text{.}$

3. 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 $G( {\mathbb Q}( \omega ) / {\mathbb Q})\text{.}$