Problems

4

Clearly, For arbitrary $a,c \in G$ and $b,d \in H$

$$ \begin{aligned} &ac = ca \wedge bd = db \\ \leftrightarrow &(ac, bd) = (ca, db) \\ \leftrightarrow &(a,b)(c,d) = (c,d)(a,b) \end{aligned} $$

I guess the general case is any group-theoretic property on both $G$ and $H$ is also on $G \oplus H$, and vice verca.

5

Assume for the sake of contradiction $Z \oplus G$ is cyclic. Then by definition there is a generator $(a,b)$. Then necessarily $\langle a \rangle = Z$ and $\langle b \rangle = G$ as by definition we have $(a,b)^k = (a^k,b^k)$. Observe $\langle a \rangle$ is of infinite order. Fix $c \in Z$, Then we know $a^k = c$ for some $k$. Compute $(a,b)^k = (a^k,b^k) = (c,d)$. Let $h$ be the element other than $d$ in $G$. Now we can’t generate $(c,h)$. By theorem 4.1 (page 76) if $a^i = a^k$ then $i = k$. In other words, $k$ is the only integer that yields $a^k = c$.

6

Clearly $(1,1) \in Z_8 \oplus Z_2$ is of order $8$. We claim no element of $Z_4 \oplus Z_4$ is of order $8$, Which suffices to solve the problem.

From Theorem 4.3 (page 81) we know any element of $Z_4$ is of order, which divides $4$. In other words, For any element $a$, there is $k \leq 4$ such that $k |a| = 4$. Similarly for another element $b$ we have $k' |b| = 4$.

So for any $(a,b) \in Z_4 \oplus Z_4$, Observe $(a,b)^4 = (a^4,b^4) = (a^{k |a|},b^{k' |b|}) = ((a^{|a|})^k, (b^{|b|})^{k'}) = (0^k, 0^{k'}) = (0, 0)$. So order of $(a,b)$ is at most $4$.

15

Let $\phi: C \rightarrow R \oplus R$ where $\phi(a+bi) = (a,b)$.

  • Injective. $\phi(a+bi) = \phi(c+di)$ implies $(a,b) = (c,d)$, and in turn $a = c$ and $b = d$.
  • Surjective. For any $(a,b)$ we have $\phi(a+bi) = (a,b)$.
  • Preserves Operation. $\phi(a+bi) \phi(c+di) = (a,b) (c,d) = (a+c, b+d) = \phi( (a+c)+(b+d)i ) = \phi ( (a+bi) + (c+di) )$.

17

Since $G \oplus H$ is cyclic, it has a generator $(a,b)$. It follows $\langle a \rangle = G$

and $\langle b \rangle = H$. If that is not the case, Then we can select an element from $G$ or $H$ whereby $(a,b)^k = (a^k, b^k)$ won’t cover it, on it corresponding index.

21

Denote the equivalence $\langle (g,h) \rangle = \langle g \rangle \oplus \langle h \rangle$ by (1).

Recall theorem 8.1 (page 158).

By definition we know $(g,h)^k = (g^k,h^k)$ where $g^k \in \langle g \rangle$ and $h^k \in \langle h \rangle$.

The condition is $|g|$ and $|h|$ are coprime. Observe it is equivalent to $lcm(|g|,|h|) = |g| |h|$.

(Necessity) We show given (1), The condition holds. Since sets are equal, and cardinality of L.H.S is $|g| \cdot |h|$, Then $|(g,h)| = |g| \cdot |h|$. By thm 8.1, The condition is satisfied.

(Sufficent) We show given the condition, (1) holds. By thm 8.1, $|(g,h)| = |g| \cdot |h|$. So its cardinality is the same as R.H.S, and it is a subset of it. It follows (1) holds.

23

Any element in $Z_3$ is of order 3, except the identity $0$. Consider an arbitrary non-identity element

$(x_1, x_2, \dots, x_k) \neq e = \underbrace{(0, .., 0)}_{k \text{ times}}$

in $\underbrace{Z_3 \oplus .. \oplus Z_3}_{k \text{ times}}$. We claim $| (x_1, \dots, x_k) | = 3$.

Following the fact all non-identity elements are of order 3, and we have some $x_i \neq 0$,

$$ \begin{aligned} (x_1, x_2, \dots, x_k)^1 &= (x_1^1, x_2^1, \dots, x_k^1) \neq e \\ (x_1, x_2, \dots, x_k)^2 &= (x_1^2, x_2^2, \dots, x_k^2) \neq e \\ (x_1, x_2, \dots, x_k)^3 &= (0, 0, \dots,0) = e \\ \end{aligned} $$

Therefore we have $3^k - 1$ elements of order 3 in $\underbrace{Z_3 \oplus \dots \oplus Z_3}_{k \text{ times}}$.

35

Recall the square root of any complex number $z$ exists. Observe $C^*$ is closed under the square root operation.

Assume for the sake of contradiction, there is an isomorphism $\phi : C^* \rightarrow R^* \oplus R^*$. Then by surjectivity there is some complex $z$ where $\phi(z) = (-1,-1)$. It follows

$$ \begin{aligned} \phi(\sqrt{z} \cdot \sqrt{z}) &= (-1,-1) \\ \phi(\sqrt{z}) \cdot \phi(\sqrt{z}) &= \\ ( \phi(\sqrt{z}) )^2 &= \\ (a,b)^2 &= \\ (a^2, b^2) &= \end{aligned} $$

In other words $a^2 = -1$ and $b^2 = -1$, but either of these leads to a contradiction, as no square of a real number is negative.

46

The infinite group is $\mathcal{Z} \oplus D_4 \oplus A_4$. Clearly $\{ (e_Z, x, e_{A_4}) \mid x \in D_4 \}$ and $\{ (e_Z, e_{D_4}, x) \mid x \in A_4 \}$ are both subgroups.

48

Claim. It is all permutations on $\mathcal{Z}_2 \oplus \mathcal{Z}_2$ which maps $(0,0)$ to itself.

Note. Our characterization is consistent with the fact the identity is always mapped to itself, and that isomorphism is a bijection.

Fact. In any group, fixing element $a_0$, then for any elements $b_0 \neq b_1$, we have $a_0 b_0 \neq a_0 b_1$.

Lemma. For any $(a,b) \in \mathcal{Z}_2 \oplus \mathcal{Z}_2$, $(a, b)^2 = (a^2, b^2) = (0,0) = e$, As $0^2 = 0$ and $1^2 = 0$.

Lemma. Any two elements of $X = \{ (0,1), (1,0), (1,1) \}$ multiplies to the third.

For distinct $a,b,c \in X$, $ab \neq (0,0)$ since $aa = (0,0)$. Also $ab \neq a$ since $a (0,0) = a$. Similarly $ab \neq b$. Therefore the only remaining choice is $ab = c$.

Theorem. Our permutations preserve the operation.

We know for distinct elements $a, b, c \in X$, we have $ab = c$. As $\phi$ is a permutation on these, We have $X = \{ \phi(a), \phi(b), \phi(c) \}$. It follows $\phi(a) \phi(b) = \phi(c)$. That concludes $\phi(c) = \phi(ab) = \phi(a) \phi(b)$.