# Abstract Algebra (3th ed.)/Chapter1/Exercises 1.1

## 05.03.2020

### 1)

1. This operator is not associative. For example we have that $(2 - 3) - 2 = -3$ and $2 - (3 - 2) = 1$.
2. This operator is ...

### 5)

We focus on the the existence of a inverse element for groups. If the statement were to be true, then for all $a \in \mathbb{Z} / n \mathbb{Z}$, where $n \gt 1$, there has to exists some $b \in \mathbb{Z} / n \mathbb{Z}$, such that $\bar{ab} = \bar{1}$. But no such $\bar{b}$ exists for $\bar{0}$.

### 7)

Note that $[x] = 0$ for every $x \in G$. First of all we want to prove that $\star$ is well defined. As stated on page 2 in the book this amounts to specify the following function, $f(x) = round(x)$. Furthermore the two subsets of $G$:

• $A_1 = \{ x \in G | x \lt 0.5 \}$
• $A_2 = \{ x \in G | x \geq 0.5 \}$

Next we want to prove that $\langle G , \star \rangle$ is a model for commutative group theory. This amounts to proofs for the following axioms:

• Associative: Too boring.
• Existence of unique neutral element: Choose $0$. Now $x + 0 - [x + 0] = x + 0 - 0 = x$ for all $x \in G$.
• Existence of $x^{-1}$ for all $x \in G$. Choose $x^{-1} = 1 - x$ for $x \gt 0$ and $x^{-1} = 0$ for $x = 0$.For $x \gt 0$ we have $x + 1 - x - [x + 1 - x] = 1 - 1 = 0$. And so on.
• The group is commutative: We go on: $$x \star y = x + y - [x + y] = y + x - [y + x] = y \star x$$ In general the normal + operator is commutative.

### 8)

Here we are given $G = \{ z \in \mathbb{Z} | z^n = 1 \text{ for some } n \in \mathbb{Z} \}$

(a). This one seems to have some less involved solutions. However the old fashioned way:

• Associative: Too boring
• Neutral element: Pick $1$.
• Inverse element: Pick $z^{-1} = z^{n - 1}$. Here I'm not sure of how to interpret the set $\mathbb{Z}^{+}$. But let it be $\{1 , 2 , ...\}$. Now we know that $z^n = z z^{n - 1} = z^{n - 1}z = 1$.

(b).

### 9)

Here we are given $G = \{ a + b \sqrt{2} \in \mathbb{R} | a, b \in \mathbb{Q} \}$.

(a). Prove that it is a group under addition.

• Associative: Sure.
• Neutral element: pick $a,b = 0$. Now $0 + 0 \sqrt{2} = 0$. And we have $0 + x = x + 0 = x$ for every $x \in G$.
• For every element in $G$ there is an inverse: Pick $a^{-1},b^{-1} \in \mathbb{Q}$. Now we have that $$a + b\sqrt{2} + a^{-1} + b^{-1} \sqrt{2} =$$ $$a + a^{-1} + b \sqrt{2} + b^{-1} \sqrt{2} =$$ $$0 + \sqrt{2}(b + b^{-1}) =$$ $$0 + 0 = 0$$

(b). Here we exclude the $0$ element. This is done since $0 \cdot a = a \cdot 0 \neq 1$ for any $a \in G$. We again prove:

• Associative: Sure.
• Neutral element: Pick $a + b \sqrt{2} = 1$. That is $a = 1$ and $b = 0$. I am somewhat sure that the equation do not have a solution for $a,b \in \mathbb{Q}$ when $b \neq 0$.
• For every element exists inverse: Pick $$x^{-1} = \frac{1}{a + b \sqrt{2}}$$ Now we just need to show that $x^{-1} \in G$. That is $x^{-1}$ is on the form $a + b \sqrt{2}$. Let's go: $$\frac{1}{a + b \sqrt{2}} = \frac{a - b \sqrt{2}}{a^{2} - 2 b^{2}} =$$ $$\frac{a}{a^{2} - 2b^{2}} - \frac{b}{a^{2} - 2b^{2}} \sqrt{2}$$ done!

### 15)

We can prove this by induction on $n$. So we get

• Base case $n = 1$: Sure we have that $(a)^{-1} = a^{-1}$ for any $a$ in $G$.
• Now let $(a_1 \cdots a_n)^{-1} = a_{n}^{-1} \cdots a_{1}^{-1}$, from which follows $$(a_1 \cdots a_n)^{-1} = a_{n}^{-1} \cdots a_{1}^{-1} \Rightarrow a_{n + 1}^{-1}(a_1 \cdots a_n)^{-1} = a_{n + 1}^{-1} a_{n}^{-1} \cdots a_{1}^{-1}$$ where $$a_{n + 1}^{-1}(a_1 \cdots a_n)^{-1} = (a_1 \cdots a_{n + 1})^{-1}$$ By prop. 1 part 5.

### 16)

This follows directly from the definition of order. That is the smallest $n$ such that $x^n = 1$. Now assume that $x^{2} = 1$ for some $x$ of order larger than $2$. But now the order of this $x$ is not the smallest $n$ such that $x^n = 1$.

### 17)

Let $x \in G$. Let $|x| = n$ for $n$ positive integer. Now $$x^n = 1 \Rightarrow x^n x^{-1} = x^{-1} \Rightarrow x^{n - 1} = x^{-1}$$

### 18)

As far as I remember such a triple is proved by $p \Rightarrow q \Rightarrow r \Rightarrow p$.

• $$xy = yx \Rightarrow y^{-1} x y = y^{-1} y x \Rightarrow y^{-1} x y = x$$
• $$x^{-1} x y = x \Rightarrow x^{-1} y^{-1} x y = x^{-1} x = 1$$
• $$x^{-1} y^{-1} x y = 1 \Rightarrow x y = y x$$

### 19)

Let's roll:

1. By how the notation is defined, we have that $x^{a}$ is a series of $a$ $x$'s. The same for $x^{b}$. Sure $x^{a + b}$ is a series of $a + b$ $x$'s. For $(x^{a})^{b}$ we have pretty much the same notation argument: $x^{a}$ is a series of $a$ $x$'s. The whole thing is a series of $b$ $a$ $x$'s, hence we have $(x^{a})^{b} = x^{a \cdot b}$.
2. As far as I can tell we need to do this by induction on $a$. So:
• Base case, here $a = 1$: we sure have $(x)^{-1} = x^{-1}$
• Ind case: $$(x^{a})^{-1} = x^{-a} \Rightarrow (x^{a})^{-1} (x)^{-1} = x^{-a} x^{-1} \Rightarrow (x^{a + 1})^{-1} = x^{- (a + 1)}$$ Where $x^{-a} x^{-1} = x^{- (a + 1)}i$ follows from the notation of (2) thingy.

### 20)

Let $n$ be the order of $x$. Now $x^{n} = 1$. Furthermore $x^{n} x^{-n} = 1$ by what we have showed in (19). Finally $x^{n} x^{-n} = 1 x^{-n} = (x^{-1})^{n} = 1$.

### 21)

We write odd numbers as $2k + 1$, for $k \in \mathbb{N}$. Let $n$, the order of $x$, be odd. Now $$x^{2k + 1} = 1 \Rightarrow x = x^{2k} = (x^2)^{k}$$

### 22)

Let $x,g \in G$ with $|x| = n$. Now $$x^{n} = 1 \Rightarrow (g^{-1} x g)^{n} = g^{-1} x^{n} g = 1$$ The trick here is that $(g^{-1} x g)^{n}$ forms a chain where only the first $g^{-1}$ and the last $g$ is preserved. The rest eats each other.

### 23)

Let $x \in G$. Let $|x| = n \lt \infty$. Let $n = st$ for positive integers $s,t$. Now we have $$x^{st} = 1 \Rightarrow (x^{s})^{t} = 1 \Rightarrow |x^{s}| = t$$

### 24)

Let $a,b \in G$ which commutes. By induction we want to prove that $(ab)^n = a^{n} b^{n}$. First for positive $n$:

• Base case for $n = 1$: $ab = ab$
• Ind case: $$(ab)^n = a^n b^n \Rightarrow (ab)^n (ab) = a^n b^b ab \Rightarrow (ab)^{n + 1} = a^{n + 1} b^{n + 1}$$

Now $$(ab)^{-n} = ((ab)^{n})^{-1} = (a^n b^n)^{-1} = b^{-n} a^{-n} = a^{-n} b^{-n}$$

### 25)

Let $|x| = 2$ for all $x \in G$. Now let $a,b \in G$. First of all we have that $$aa = 1 \Rightarrow aa = aa^{-1} \Rightarrow a = a^{-1}$$ By prop. 2. So $$ab = (ab)^{-1} = b^{-1} a^{-1} = ba$$

### 26)

Here we know that for all $h \in H$ we have a $h^{-1} \in H$ due to closure. Furthermore $H$ has inherited associativity. Now $hh^{-1} = 1$.

### 27)

We want to prove that $H = \{ x^n | x \in \mathbb{Z} \}$ for $x \in G$ is a sub group of $G$:

• Closed under operator of $G$: This follows from the definition of $H$. That is $x \in H \Rightarrow x^{n} \in H$ for all $n \in \mathbb{Z}$.
• Closed under inverse: Quite similar to the one above. $x^n \in H \Rightarrow x^{-n} \in H$ for all $n \in \mathbb{Z}$.

### 28)

Let $(A,\star)$, $(B,\diamond)$, and let their direct product $A \times B$ as defined in example (6). Now

• $A \times B$ has $(1,1)$ as identity: $$(a,b)(1,1) = (a \star 1, b \diamond 1) = (a,b)$$ This goes the other way as well, that is $(1,1)$ commutes.
• An element $(a,b)$ has $(a^{-1},b^{-1})$ as inverse: $$(a,b)(a^{-1},b^{-1}) = (a \star a^{-1},b \diamond b^{-1}) = (1,1)$$ The other way as well.

### 29)

We want to prove that $A \times B$ is abelien if and only if $A$ and $B$ are.

• Proof of the right direction: $$(a,b)(c,d) = (c,d)(a,b) \Rightarrow (a \star c,b \diamond d) = (c \star a, d \diamond d)$$
• Proof of the left direction: $$(a \star c,b \diamond d) = (c \star a, d \diamond d) \Rightarrow (a,b)(c,d) = (c,d)(a,b)$$

### 30)

We want to prove that $(a, 1)$ and $(1,b)$ commutes: $$(a,1) (1, b) = (a \star 1,1 \diamond b) = (a,b) = (1 \star a,b \diamond 1) = (1,a) (b,1)$$ Furthermore we want to deduce that $|(a,b)|$ is the least common multiple of $|a|$ and $|b|$. Let $|a| = n$ and $|b| = m$. We know that $$(a^{n \cdot k_0},b^{m \cdot k_1}) = (1,1)$$ for $k_0 , k_1 \in \mathbb{Z}^{+}$. In order for this to work we must have $n \cdot k_0 = m \cdot k_1$. Furthermore $k_0$ and $k_1$ should be chosen as small as possible since the order requires so.

### 31)

I'm not fully aware of how to reason about finite groups. There must be some correspondence between group order and element order that I have missed. I'll be back.

### 32)

Let $x \in G$ of finite order $n$. Now assume in order to reach a contradiction that $x^m = x^l$ for $0 \leq m \lt l \lt n$. There must be some $x^{p}$ for which $x^{l} x^{p} = x^{l + p} = 1$. That is $l + p = n$. But this must also hold for $x^m$, that is $x^m x^p = 1$. Here we have that $m + p \lt n$ contradicting that $n$ is the smallest for which $x^n = 1$.

### 33)

Let $x \in G$ of finite order. Prove

1. Let $n$ be odd. Assume in order to reach a contradiction that $x^i = x^{-i}$ for some $i : 1 \leq i \leq n - 1$. Now we have that $$x^i x^{-i} = 1 \Rightarrow x^{2i} = 1 = x^{n} \Rightarrow 2i = n k_0$$ Since $i \in [1,n - 1]$, we have that $k_0 = 1$. Which implies $2i = n$, a contradiction since we have assumed $n$ to be odd.
2. Let $n = 2k$ and $1 \leq i \lt n$. We want to prove that $x^i = x^{-i} \iff i = k$.
• Right direction: $$x^i = x^{-i} \Rightarrow x^{2i} = 1 = x^n \Rightarrow 2i = nk_0$$ Since $1 \leq i \lt n$, we have that $k_0 = 1$, and we can finish: $$2i = n \Rightarrow 2i = 2k \Rightarrow i = k$$
• Left direction: $$i = k \Rightarrow x^{2i} = 1 = x^{i} x^{-i} \Rightarrow x^i = x^{-i}$$

### 34)

Let $x \in G$ of infinite order. We want to prove that $x^n , n \in \mathbb{Z}$ are all distinct. In order to reach a contradiction let $x^a = x^b$. We split into cases

• Let $a \lt b$, and let $a,b \gt 0$. Now $$x^a x^{-a} = 1 = x^{b} x^{-a} = x^{b - a}$$ But since $a \lt b$, we have that $b - a \gt 0$ contradicting that no positive power of $x$ is the identity.
• Let $a \lt b$, and let $a = 0$. Now $$x^a = 1 = x^b$$ But again we have that $b \gt 0$, and hence a contradiction since no positive power of $x$ is the identity.
• The rest of the cases are variations of the above.

### 35)

Involves the division algorithm which I'm not reading up on right now.

### 36)

Skipping this for now.