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

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 $\bar{0}$.

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.

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).

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!

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.

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 $x^n = 1$.

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

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

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

Let's roll:

- 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 $(x^{a})^{b} = x^{a \cdot b}$.
- 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.

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 $x^{n} x^{-n} = 1 x^{-n} = (x^{-1})^{n} = 1$.

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

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.

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

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

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

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 $hh^{-1} = 1$.

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 $n \in \mathbb{Z}$.

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.

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: $$ (a \star c,b \diamond d) = (c \star a, d \diamond d) \Rightarrow (a,b)(c,d) = (c,d)(a,b) $$

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.

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.

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 $x^n = 1$.

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

- 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.
- 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: $$ i = k \Rightarrow x^{2i} = 1 = x^{i} x^{-i} \Rightarrow x^i = x^{-i} $$

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.

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

Skipping this for now.

CommentsGuest Name:Comment: