A few questions about cyclic groups

Question

I cannot understand how is $\langle 1 \rangle$ the creator of $(\mathbb{Z},+)$. I know that this element can create all-natural numbers, simply by adding-1 a certain amount of times, but the negative numbers cannot be created. So what I thought is that each element has a negative number, but still, I can't see how $\langle 1 \rangle$ can create by itself all elements of $\mathbb{Z}$, only both -1 and 1 can create together all elements.

In addition, I have seen a couple of times that for example - the group $Z^{*}_8$ if defined as $Z^{*}_8=\{[1],[3],[5],[7]\}$. In order to check the elements that a certain element create, which means the order, we check the power of the element until we start returning on the elements created, let say $\langle 5 \rangle=\{[5]^{1},[5]^{2},[5]^{3}\}=\{[5],[1],[5]\}$ so the order is 2. Now, I know that power represents the amount of time we run the method $*$ of the group, but if so, why is here with $Z^{*}_8$ we see (*) as ($\cdot $), and in the group - $(\mathbb{Z},+)$ we check as $\langle 2 \rangle=\{2,2+2,2+2+2,...\}$ and so on?

I hope you can help me understand this. Thanks!

Answer 1

Notice that $\langle 1\rangle$ contains $-1$, since without it it isn't a group ($-1$ is the inverse element of $1$, because $1+(-1)=0$ in $(\mathbb{Z},+)$). Hence, you should understand that $\langle 1\rangle$ being a group implies $-1\in \langle 1\rangle$ so then you can easily see how it generates all integers.

In the second question, in $\mathbb{Z}_8^\times$ you're referring to the multiplicative group of $\mathbb{Z}_8$, to be said, the $\mathbb{Z}_8$ elements that have inverse for multiplication (Notice this is NOT the same as inverse for addition, the operation you use in integer groups). So in $\mathbb{Z}_8^\times$ the operation you're using is multiplication, while in $\mathbb{Z}$ or $\mathbb{Z}_8$ you're using addition, and powers in addition and multiplication are different though both can share the same $a^n$ notation:

• In multiplication, $a^n=a\cdot a\cdots a$
• In addition, $a^n=a+a+\cdots+a$

Answer 2

A group $G$, written multiplicatively, is called cyclic if there is an elemet $a \in G$ with $G = \langle a \rangle =\{a^n \mid n \in \mathbb{Z}\}$. Note that $n$ spans all the integers. If the group is written additively, this becomes $$ G = \langle a \rangle =\{na \mid n \in \mathbb{Z}\}. $$ So in exponential notation we get terms like $$ \ldots, a^{-2}, a^{-1}, a^0=e, a, a^2, \ldots $$ In additive notation this would read $$ \ldots, -2a, -a, 0, a, 2a, \ldots $$ With $a=1$ this explains why $1$ generates all of $\mathbb{Z}$ (as does $-1$).