potentiation

Potentiation in a group: $(G,\circ ),ea\in G,n\in \mathbb{Z}$

Potentiation for a group works just as we are used to from highschool:

$$\begin{array}{r}{a}^n:=\{\begin{array}{l}en=0\\ an=1\\ a^{n-1}\circ an>1\\ (a^{-1}{)}^{-n}n<0\end{array}\end{array}$$

Note: It’s ${}^{-n}$ to turn the negative $n$ into a positive number.

Theorem: $a^{n+m}=a^n\circ a^m,a^{nm}=(a^n{)}^m$, $\forall n,m\in \mathbb{Z}$

In the additive group $(\mathbb{Z},+)$, potentiation becomes repeated addition (= multiplication) rather than repeated multiplication:

$$\begin{array}{rl}{a}^n& =\underset{n\text{ times}}{\underbrace{a+a+\cdots +a}}=n\cdot a\\ a^{-n}& =\underset{n\text{ times}}{\underbrace{(-a)+(-a)+\cdots +(-a)}}=-n\cdot a\end{array}$$

→ In an abelian group $(G,+)$ (additive context), we write $-a$ instead of $a^{-1}$ and $n\cdot a$ instead of $a^n$.

Why is $n^0=1$?

$$\begin{array}{rlr}8& =2^3& ÷2\\ 4& =2^2& ÷2\\ 2& =2^1& ÷2\\ 1& =2^0& ÷2\\ 1/2& =2^{-1}& ÷2\end{array}$$

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$$\begin{array}{rl}\forall n\in \mathbb{Z},n\ne 0:& \\ n^0& =n^{m-m}=n^m\cdot n^{-m}\\ & =\frac{n^m}{n^m}=1\end{array}$$

And $0^0$? If you are working in areas like combinatorics, set theory, abstract algebra, or computer science, $0^0$ is generally taken to be $1$. This definition simplifies many formulas and theorems.

The set of all powers of $a$: $⟨a⟩:=\{a^n\mid n\in \mathbb{Z}\}$

Let $(G,\circ )$ be a group and $\exists a\in G:⟨a⟩=G$, then $⟨a⟩$, is a cyclic subgroup of $G$, generated by $a$ (“von $a$ erzeugte Untergruppe”).

$a\le G$, … is it actually a subgroup?
Note empty: $a^0=e\in ⟨a⟩⟹⟨a⟩\ne \varnothing$
Closed: $x,y\in ⟨a⟩⟹\exists n,m\in \mathbb{Z}:x=a^n,y=a^m⟹x\circ y=a^n{a}^m=a^{n+m}\in ⟨a⟩$ → $⟨a⟩$ abelian group
Invertible: $x\in ⟨a⟩⟹\exists n\in \mathbb{Z}:x=a^n⟹x^{-1}=a^{-n}⟹a^{-n}\in ⟨a⟩$ (because $a^n{a}^{-n}=a^0=e)$
$✓$

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