Group $(\mathbb{Z}^*_{13}, \cdot)$, calculate order of the group, if it's a cyclic group, and the order of all its elements

Question

I have this exercise:

Considering the group $(\mathbb{Z}^*_{13}, \cdot)$ compute:
i) the order of the group.
ii) is it a cyclic group?
iii) establish the order of all its elements.

What I have tried to do is:

i)
I think that $(\mathbb{Z}^*_{13}, \cdot)$ means the group $\mathbb{Z}_{13} - \left \{ 0 \right \}$ under multiplication.
So the elements inside that set would be all the class modulo 13 from 1 to 12 without 0: $\mathbb{Z}^*_{13} = \left \{ [1], [2], \ldots , [12]\right \}$
hence,
the order of the group, i.e. the number of the elements inside the set, would be 12.

ii)
I have discovered by attempts that $2$ is one generator of the group.
Doing this:
$2^1 = 2 \ne 1 \\ 2^2 = 4 \\ 2^3 = 2^2 \cdot 2 = 4 \cdot 2 = 8 \\ 2^4 = 2^3 \cdot 2 = 8 \cdot 2 = 16 = 3 \\ 2^5 = 2^4 \cdot 2 = 3 \cdot 2 = 6 \\ 2^6 = 2^5 \cdot 2 = 6 \cdot 2 = 12 \\2^7 = 2^6 \cdot 2 = 12 \cdot 2 = 24 = 11 \\ 2^8 = 2^7 \cdot 2 = 11 \cdot 2 = 22 = 9 \\ 2^9 = 2^8 \cdot 2 = 9 \cdot 2 = 18 = 16 + 2 = 3 + 2 = 5 \\ 2^{10} = 2^9 \cdot 2 = 5 \cdot 2 = 10 \\ 2^{11} = 2^{10} \cdot 2 = 10 \cdot 2 = 20 = 18 + 2 = 5 + 2 = 7 \\ 2^{12} = 2^{11} \cdot 2 = 7 \cdot 2 = 14 = 1$

so the order of the element $2$ is 12, since $12$ is the power such that $2^{12} = e = 1$

and the elements of the subgroup generated by $2$ are $\left \langle 2 \right \rangle = \left \{ 2,4,8,3,6,12,11,9,5,10,7,1 \right \} = \left \{ 1,2,3,4,5,6,7,8,9,10,11,12 \right \}$,
i.e. all the element within the group. So the group it is generated by the element $2$ and it is cyclic.

Yes, but, ... I have done only the element $2$. I have to do the same for the other 11 elements in the set. It's hard, it's strenuous.
It's clear that it is labourious to find that for EVERY element in the set, even for a "small" set with 12 elements. Imagine a set with hundreds of elements... it is very long job following the development of the above.

I can see that in the exercise we have some data as:
$13$, that is a prime number;
$12$ are the elements inside the set;
the set contains remainder classes [1],...[12],
i.e. if we take the element [1]. This element represent a set that contains numbers that divided by 13 retrieve a remainder equal to 1, and each element inside [1] can be represented as a congruence as $a \equiv 1 \mbox{ (mod 13) }, a \in [1]$

So my question is:
given the data of the above, does exists any theorem or algorithm that ease the computing of the processes required in the exercise?

Please, can you help me? Many thanks!

Answer 1

Yes, unless from this point. I mean, given a cyclic group $G=\langle a\rangle$, you know that

$$k||G| \implies|a^k|=\frac{|G|}{k}$$

proof: first notice that $$(a^k)^{\frac{|G|}{k}}=a^{|G|}=e$$ since $m\rightarrow a^m$ is a homomorphism, so preserve power rules. Then we suppose it exists $m<\frac{|G|}{k}$ such that $(a^k)^m=e$, but in that case $km<|G|$ and satisfices $a^{km}=e$ which contradicts that $|a|=|G|$.

Otherwise, and actually as general case is

$$|a^k|=\frac{|G|}{\mathrm{gcd}(k,|G|)}$$

and the proof is similar. It's easy to follow that if $g:=\mathrm{gcd}(k,|G|)$, $(a^k)^{\frac{|G|}{g}}=e$. Now define $n:=\frac{|G|}{g}$ and suppose it exists $m<n$ s.t. $a^{mk}=(a^k)^m=e$, and take $m$ minimums with this property. Then as $|a^k|||G|\implies m||G|$, so $\frac{|G|}{n}=g<\frac{|G|}{m}$. If $\frac{|G|}{m}>k$, then $km<|G|$ which contradicts $a$ is generator. Is easy to see that $\forall r, a^r=a^{r\mod{|G|}}$ since $a^{r+l|G|}=a^r(a^{|G|})^l=a^r\dot{}e=a^r$, so this implies $a^l=e\iff l=0\mod{|G|}$, so applied to this case the consequence is that $km=0\mod{|G|}$, so $\frac{|G|}{m}|k$ and then $\frac{|G|}{m}\leq g$, which can't be.

The fact that $p=|G|$ is prime just implies that $(\mathbb{Z}_p^*,\dot{})$ is a commutative group.