Problems
1
$\phi(n) = 2n$. If $2a = 2b$ then $a = b$. For each $2k$ we have $\phi(k) = 2k$. Observe $\phi(ab) = 2(a+b) = 2a + 2b = \phi(a)\phi(b)$, Following by usual properties of integers.
2
We Follow the same proof approach of Example 15 (page 130). Let $\phi \in Aut(Z)$ be arbitrary. Then by the usual properties of integers and isomorphisms, $\phi(k) = \phi(1 + 1 + \dots + 1) = \phi(1) + \dots + \phi(1) = k \cdot \phi(1)$. But by definition $\phi(1) = c$ for some integer $c$. Therefore $\phi(k) = kc$. In other words, $Aut(Z) = \{ \phi \mid \exists c, \; \forall k \; \phi(k) = kc \}$
4
Caylay table of $U(8)$:
| 1 | 3 | 5 | 7 | |
|---|---|---|---|---|
| 1 | 1 | 3 | 5 | 7 |
| 3 | 3 | 1 | 7 | 5 |
| 5 | 5 | 7 | 1 | 3 |
| 7 | 7 | 5 | 3 | 1 |
Caylay table of $U(10)$:
| 1 | 3 | 7 | 9 | |
|---|---|---|---|---|
| 1 | 1 | 3 | 7 | 9 |
| 3 | 3 | 9 | 1 | 7 |
| 7 | 7 | 1 | 9 | 3 |
| 9 | 9 | 7 | 3 | 1 |
Recall from theorem 6.2 (page 126), Any $\phi$ maps the identity to the identity of the other group.
In $U(8)$ we have $3 \cdot 3 = 1$. Then $\phi(3 \cdot 3) = \phi(3) \cdot \phi(3) = \phi(1) = 1$. The only non-identity element in $U(10)$ satisfying that is $9$. Hence $\phi(3) = 9$.
Similarly $5 \cdot 5 = 1$. Then we must have some $a \in U(10)$ such that $a \cdot a = 1$ where $a \notin \{ 1, 9 \}$. Contradiction.
8
Injective. Given $\log_{10} a = \log_{10} b$, we get $10^{\log_{10} a} = 10^{\log_{10} b}$, and $a = b$.
Surjective. Given $x \in \mathcal{R}$, take $a = 10^x \in \mathcal{R}^+$. Then $\log_{10} a = \log_{10} 10^x = x$.
Group Operation. Observe $\phi(ab) = \log_{10} ab = \log_{10} a + \log_{10} b = \phi(a) + \phi(b)$.
11
Observe $\phi(a^3 b^{-2}) = \phi(a^3) + \phi(b^{-2}) = [\phi(a)]^3 + [\phi(b)]^{-2} = (\overline{a})^3 + (\overline{b})^{-2}$. We used theorem 6.2 (2).
12
$(\rightarrow)$. For any $a,b \in G$, We have:
$$ \begin{aligned} \alpha(a^{-1}b^{-1}) &= \alpha(a^{-1}) \alpha(b^{-1}) \\ (a^{-1} b^{-1})^{-1} &= \\ ba &= ab \end{aligned} $$$(\leftarrow)$. Symmetrically, If we have $b^{-1} a^{-1} = a^{-1} b^{-1}$, Then $\alpha(ab) = \alpha(a) \alpha(b)$. Bijection is clear by properties of inverses.
14
By theorem 6.5 (page 131), $Aut(Z_3) \approx U(3)$ and $Aut(Z_4) \approx U(4)$, so $Aut(Z_3) \approx Aut(Z_4)$ by the transitivity of isomorphism. But $Z_3 \not\approx Z_4$ as the two groups have different orders, so no bijection exists.
21
Clearly groups $H$ and $K$ are isomorphic to $S_4$. By transitivity $H \approx K$.
22
For every $c = 2, 3, 4 \dots$, Consider the subset $H_c = \{ ck \mid k \in \mathcal{Z} \}$. It is a subgroup, As it has the identity $c(0)$, inverses $c(-k)$, and closed $ck_1 + ck_2 = c(k_1 + k_2)$.
It remains to show those subgroups are distinct. For any $c_1$ and $c_2$ where $c_1 < c_2$ we have $c_1(1) \in H_{c_1}$ but $c_1(1) \not\in H_{c_2}$. Therefore $H_{c_1} \neq H_{c_2}$.
24
We use theorem 3.2 (page 63). If $\phi(a) = a$ then $\phi(a^{-1}) = (\phi(a))^{-1} = a^{-1}$. Also, If $\phi(a) = a$ and $\phi(b) = b$ then $\phi(ab) = \phi(a) \phi(b) = ab$.
34
Let $K$ be a subgroup of $G$. We use theorem 3.2 (page 63).
Inverse. For any $\phi(k) \in \phi(K)$, $(\phi(k))^{-1} = \phi(k^{-1})$. But $k^{-1} \in K$, So $\phi(k^{-1}) \in \phi(K)$.
Closed. For $\phi(k_1), \phi(k_2) \in \phi(K)$, We have $\phi(k_1) \phi(k_2) = \phi(k_1 k_2)$. But $k_1 k_2 \in K$, So $\phi(k_1 k_2) \in \phi(K)$.