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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!

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    Fortunately, you can reuse the calculations you already have to find the orders of the other elements. Note that if you know the order of $a$ then you also know the order of $a^n$ for any $n$. – 2017-02-13
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    It will remain cyclic when 13 replaced by any other prime (only in such cases) – 2017-02-13
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    @TobiasKildetoft Please, can you explain me better? – 2017-02-13
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    @larry01 Does exists any theorem that prove what you're saying? – 2017-02-13
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    @larry01 It is correct that it is cyclic for all primes. It is not correct that these are the only cases where it will be cyclic. – 2017-02-14
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    sorry for my mistake. here is a vast article, if needed https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n – 2017-02-14
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    @larry01 Maybe you are referring to the theorem: If $G$ is a group with a prime number $p$ of elements, then $G$ is a cyclic group. Furthermore, any element $a \ne e \in G$ is a generator of $G$. But, in the exercise there isn't a prime number of elements, they are $12$, not prime, since we are considering $\mathbb{Z}_{13} - \left \{ 0 \right \}$. Or am I wrong? – 2017-02-15

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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 $mk$, 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.

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    what is $k$? is it the order of the element? Hence, if the order of the group is $G=\left | 12 \right |$, and the order of the element $2$ is $k=12$, $12$ divides $|G| = 12$, i.e. $k| \left | G \right | \iff G = kq$ and, what is $|a^k|$? The order of $a^k$?. In this case $\left | 2^{12} \right | = \frac{\left | G \right |}{k} = \frac{12}{12} = 1$, but we have said that the order of $2^{12}$ is $12$ and not $1$. Please, can you explain me better? And what is the name of the theorem that says that if the order of the group is prime, then the group is commutative? – 2017-02-13
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    @JB-Franco Sorry, I thought the notation was universal. What I mean is that, if $a$ is a generator, and $k$ any divisor of $|G|$ (of course $<|G|$ also), so $k$ is the power s.c., given $b\in G$, $b=a^k$, considering $a$ is a generator. Yes, $|b|$ is the order of the element $b\in G$, or in other words $|b|=|\langle b\rangle|$. I don't know if it's a theorem, I saw it as a property, but we know that $\dot{}$ is commutative in any $\mathbb{Z}_n$, the point was that only when $n$ is prime every element has inverse for $\dot{}$, so in other case $(\mathbb{Z}_n^*,\dot{})$ isn't a group – 2017-02-14
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    @JB-Franco something else, the case $k=12$ works, since $2^{12}=1$ and $|1|=|\langle 1\rangle |=|\lbrace 1\rbrace|=1$ – 2017-02-14
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    Since you assume that $G$ is cyclic, it means that, first, I have to search one generator between $1$ and $12$, and from informations found inside this one generator I can calculate the order of the other elements. So, the process I have done to calculate the powers $2^n$ is right, but, I don't have to do it for the other elements because using the results from that process, I can calculate the order of the other elements. Am I right? – 2017-02-14
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    @JB-Franco exactly that's what I mean – 2017-02-14
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    Hence, A way to proceeding is the following: $$\begin{array}{rcl} 2^1 \mbox{ and } 1 | \left | G\right | & \Rightarrow & \left | 2^1 \right | = \left | 2 \right | = \frac{\left | G \right |}{1} = \frac{12}{1} = 12 \\ 2^2 \mbox{ and } 2| \left | G\right | & \Rightarrow & \left | 2^2 \right | = \left | 4 \right | = \frac{\left | G \right |}{2} = \frac{12}{2} = 6 \\ 2^3 \mbox{ and } 3 \nmid \left | G\right | & \Rightarrow & \left | 2^3 \right | = \left | 8 \right | = \left | G \right | = 12 \\ \vdots \end{array}$$ till $2^{12}$ – 2017-02-14
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    @JB-Franco the algorithm is right but there is a mistake since in this case $|G|=12$ and we know $3|12$ so $|8|=\frac{12}{3}=4$ – 2017-02-14
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    yes, a typo, the case is when $$2^7 \mbox{ and } 7 \nmid \left | G \right |$$ – 2017-02-14
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    @JB-Franco right – 2017-02-14
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    Another thing, since I know that in the set there are the elements ${1,2, \ldots, 12}$, and immediately I know that the order of $G$ is $12$. Since in the set, the first element that not divide $12$ is $5$, I can start just from $5$ to calculate the powers. The fact that $2$ is a generator and that I have discovered it soon, it is a question of luck! What do you think? – 2017-02-14
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    right, it's just luck; for example in the case of $\mathbb{Z}_7^*$, $\langle 2\rangle=\lbrace 1,2,4\rbrace$ so in this case $2$ isn't a generator – 2017-02-14
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    so, immediately, since in this last group there are, $\left \{1, \ldots, 6 \right \}$, i.e. the order is $6$, $5$ is a generator since $5$ not divide $6$. – 2017-02-14
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    @JB-Franco $5$ is a generator but I don't see where this reasoning come from. I can add that $3$ is also a generator. Besides as $5^2=4$ and as $4\in\langle 2\rangle$, using the reasoning of the beginning we know $|2|=3$ and $2^2=4$ and as $2\not| $ $3$, $4$ is a generator of $\langle 2\rangle$, so again as $4\in \langle 5\rangle\implies \langle 2\rangle=\langle 4\rangle \lt \langle 5\rangle$ and as $|5|$ must divide $6$, and is $>3$, is $6$, so $5$ is a generator – 2017-02-14
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    so, for discover one generator I have to make attempts starting from the first element on the group. I thought something like: "if an element in the group not divides the order of the group, then that element is one generator". – 2017-02-14
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    The order $\left | G \right | = 6$, the element $5$ is a generator. so basing on what said before, $5^4 \mbox{ and } 4 \nmid 6 \Rightarrow \left | 5^4 \right | = \left | 2 \right | = \left | G \right | = 6$, but we have said that $2$ is not generator and has order $3$. why this discrepance? – 2017-02-14
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    right, my mistake. I've updated the answer considering these cases and added a proof to avoid other mistake. In this case $\gcd (4,|G|)=\gcd (4,6)=2\implies |2|=\frac{6}{2}=3$ – 2017-02-15