Problems
1
By corollary 4 (page 80).
Generators of $\mathcal{Z}_6$ are $1,5$ since $gcd(1,6) = gcd(5,6) = 1$.
Generators of $\mathcal{Z}_8$ are $1,3,5,7$ since $gcd(1,8) = gcd(3,8) = gcd(5,8) = gcd(7,8) = 1$.
Generators of $\mathcal{Z}_{20}$ are $1,3,7,9,11,13,17,19$ since $gcd(1,20) = gcd(3,20) = gcd(7,20) = gcd(9,20) = gcd(11,20) = gcd(13,20) = gcd(17,20) = gcd(19,20) = 1$.
5
$\langle 3 \rangle = \\{ 3^0, 3^1, 3^2, 3^3, 3^4, 3^5, \dots\\} \cup \\{ 3^{-1}, 3^{-2}, 3^{-3}, 3^{-4}, 3^{-5}, \dots \\} = \\{0, 3, 9, 7 , 1, 3, \dots\\} \cup \\{ -3, 9, -7, 1, -3, \dots \\} = \\{0, 3, 9, 7 , 1, 3\\} \cup \\{ 17, 9, 13, 1, 17 \\} = \\{0, 1, 3, 7, 9, 13, 17\\}$.
$\langle 7 \rangle = \\{ 7^0, 7^1, 7^2, 7^3, 7^4, 7^5, \dots\\} \cup \\{ 7^{-1}, 7^{-2}, 7^{-3}, 7^{-4}, 7^{-5}, \dots \\} = \\{ 0, 7, 9, 3, 1, 7, \dots \\} \cup \\{ -7, 9, -3, 1, -7, \dots \\} = \\{ 0, 7, 9, 3, 1 \\} \cup \\{ 13, 1, 17, 1 \\} = \\{ 0, 1, 3, 7, 9, 13, 17 \\}$
10
By corollary (page 82). One generator is $\langle 24/8 \rangle = \langle 3 \rangle = \\{ 3^0, 3^1, 3^2, 3^3, 3^4, 3^5, 3^6, 3^7, 3^8, 3^9 \\} \cup \\{ 3^{-1}, 3^{-2}, 3^{-3}, 3^{-4}, 3^{-5}, 3^{-6}, 3^{-7}, 3^{-8}, 3^{-9} \\} = \\{ 0, 3, 6, 9, 12, 15, 18, 21, 0 \\} \cup \\{ 21, 6, 15, 12, 9, 18, 3, 0, 21 \\} = \\{ 0, 3, 6, 9, 12, 15, 18, 21 \\}$
Note any generator of that subgroup must be contained in it as $a = a^1 \in \langle a \rangle$.
By corollary 3 (page 80). Generators are $3^5 = 15$ and $3^7 = 21$, as $gcd(24,5) = gcd(24, 7) = 1$.
By corollary 3 (page 80). Generators of arbitrary $G$ are $1,5,7,11,13,17,19,23$ since $gcd(24,i) = 1$. Observe since $G$ is generated by $a$, Any candidate must be of the form $a^i$. So we covered all of them.
11
Follows trivially by corollary 3 (page 80), as $gcd(n, -1) = 1$.
27
We know given a positive integer $n$, there is a complex $z$ such that $z^n = 1$. Then $S_n = \\{ z^0, z^1, z^2, \dots \\} = \\{ z^0, z^1, \dots, z^{n-1} \\}$. Clearly it is a group.
For $z^{-i}$ observe $-i = n(m) + r$ where $0 \leq r < n$. Then $-i-r$ is divisable by $n$, and by theorem 4.1 (page 76), $z^{-i} = z^r$. Then $\\{ z^{-1}, z^{- 2}, \dots \\}$ is contained in $S_n$.
Thus we conclude $S_n = \langle z \rangle$ is a subgroup of order $n$.
30
We call a subgroup new if it is not $\{e\}$ or $G$. Observe constructing it contradicts a given hypothesis.
Select $a \neq e$. If $\langle a \rangle$ is of infinite order, then $\langle a^2 \rangle$ is a new subgroup. So $\langle a \rangle$ is of finite order $n$.
If $\langle a \rangle \neq G$ then $\langle a \rangle$ is a new subgroup. So $\langle a \rangle = G$.
If $n$ is not prime, i.e composite, then by theorem 4.3 (page 81), we can take divisor $k$ such that $\langle a^{n/k} \rangle$ is a new subgroup of order $k$. Note by divisibility $1 < k < n$.
It follows $G$ is a finite cyclic group of prime order $n$.
36
| 4 | 8 | 12 | 16 | |
|---|---|---|---|---|
| 4 | 16 | 12 | 8 | 4 |
| 8 | 12 | 4 | 16 | 8 |
| 12 | 8 | 16 | 4 | 12 |
| 16 | 4 | 8 | 12 | 16 |
All entries are contained in $\{ 4, 8, 12, 16 \}$, So closed. $16$ is the identity. Every row has an $16$ entry showing inverses existince. The group is cyclic.
Its generators are all its elements, $4, 8, 12$ and $16$. To see why you can trace the table. For example $8^1 = 8$, $8^2 = 4$, $8^3 = 8^2 \cdot 8 = 4 \cdot 8 = 12$, $8^4 = 8^3 \cdot 8 = 12 \cdot 8 = 16$.
48
$H$ is a subgroup by Theorem 3.2 (page 63). Given $a = 10 k_0 = 8 k_1$ and $b = 10 k_2 = 8 k_3$, Trivially $a+b = 10(k_0 + k_2) = 8(k_1 + k_3) \in H$. Also $-a = 10 (-k_0) = 8 (-k_1) \in H$.
$H$ is not a subgroup in case of "OR". Consider the counter-example $10 + 8 = 18$ as $18$ is neither divisible by $10$ nor $8$, Violating closeness property.
59
Partially Solved.
Let $G$ be a group with only $a$ and $b$ elements of order 2. We try to come-up with a contradiction.
By definition $a^2 = b^2 = e$, so $a^{-1} = a$ and $b^{-1} = b$. Clearly $ab \neq a, b$, or $e$. For example if $ab = e$ then $b = a^{-1} = a$ which is not true as $a$ and $b$ are given as distinct elements.
Case $ab = ba$. Then $(ab)^2 = (ab)(ba) = aea = a^2 = e$. Contradiction.
Case $ab \neq ba$. No solution found for that case.
61
Let $x \in \langle a \rangle \cap \langle b \rangle$. Then by corollary 1 (page 79), $|x|$ divides both $10$ and $21$. Since they are coprime, $|x| = 1$ and $x^1 = x = e$.