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In the context of abstract algebra and groups automorphism, I need help to understand the solution to this problem:

Show that $\operatorname{Aut}(C_{14})$ is a cyclic group.

Here's the solution (I'm directly translating from Greek handwritten notes so I hope there are no mistakes):

Since $C_{14}$ is cyclic with $14$ elements, $C_{14} \cong Z_{14}$. Hence, it suffice to show that $\operatorname{Aut}(\mathbb{Z}_{14})$ is cyclic. Also, $\mathbb{Z}_{14} = \{0, 1, 2, 3, \ldots 13\} = \langle 1 \rangle = \langle 3 \rangle = \langle 5 \rangle = \langle 9 \rangle = \langle 11 \rangle = \langle 13 \rangle$. Hence, there will be $6$ automorphism of $\mathbb{Z}_{14}$, specifically

$$f_1(1) = 1, \,\, f_2(1) = 3, \,\, f_3(1) = 5, \,\, f_4(1) = 9, \,\, f_5(1) = 11, \,\, f_6(1) = 13.$$

Therefore, $\left|\operatorname{Aut(\mathbb{Z}_{14})}\right| = 6$.

First the first part of the solution, if I understand the theory correctly, we are using two main "ingredients". The first one is that if a group $G$ is cyclic with $n$ elements, then $G$ is isomorphic to $(\mathbb{Z}_n,+)$. The second proposition we use is that for a group $G$ generated by $a$, i.e. $G = \langle a \rangle$, if $b$ is a generator of $G$, there is a unique automorphism $f:G \to G$ such that $f(a) = b$. Is my understanding correct so far?

Now for the second part of the solution which I really don't understand:

Immediately, $\operatorname{ord} f_1 = 1 |$ identity element of $(\operatorname{Aut(\mathbb{Z}_{14})}, \circ)$.

We have that $\operatorname{ord} f_2 | 6 \implies \operatorname{ord} f_2 = 2 \,\, \text{or} \,\, 3 \,\, \text{or} \,\, 6$.

Then, $f_2^2(1) = f_2(f_2(1)) = f_2(3) = f_2(1+1+1) = 3f_2(1) = 3.3 = 9$.

Then, $f_2^3(1) = f_2(f_2(3)) = 9 = 9f_2(1) = 9.3 = 27 = 13.$

Therefore, $\operatorname{ord} f_2 = 6 \left|\operatorname{Aut(\mathbb{Z}_{14})}\right|$ and $\operatorname{Aut(\mathbb{Z}_{14})}$ is cyclic, implying $\operatorname{Aut(\mathbb{C}_{14})}$ is cyclic.

I cannot understand anything from the second part of the solution. I think it is very misleading. He is using some divisibility argument and composition of functions but I don't understand what he is actually calculating, why it is necessary and how he is able to conclude from those calculations. I'm looking for an answer that will help me understand in detail the second part of the solution, possibly with the background theory that is needed to solve this kind of problem.

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    $$\text{Aut}(C_{14})\simeq \mathbb{Z}/(14\mathbb{Z})^* \simeq \mathbb{Z}/(7\mathbb{Z})^*$$ and the last group is cyclic since $7$ is an odd prime. – 2017-02-12
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    Or you may simply show that there is an element of $\text{Aut}(C^{14})$ generating the whole group of automorphisms. – 2017-02-12

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Your understanding of the first part looks correct.

For the second part, first recall that $\operatorname{Aut}(\mathbb Z_{14})$ is the group of automorphisms of $\mathbb Z_{14}$, where the group operation is composition of functions, so it should not be surprising to see that functions are being composed in the author's solution. The group identity is the identity map, which we have labeled $f_1$ in the first part of this problem.

The author uses Lagrange's theorem to observe that the order of $f_2$ in $\operatorname{Aut}(\mathbb Z_{14})$ must divide $|\operatorname{Aut}(\mathbb Z_{14})| = 6$, hence the order of $f_2$ must be one of $1,2,3,6$.

The order of $f_2$ certainly isn't $1$ because $f_2$ is not the identity (it does not map $1$ to itself).

Then the author explicitly computes $f_2^2(1)$ and $f_2^3(1)$, and since neither of these is $1$, it follows that neither $f_2^2$ nor $f_2^3$ is the identity, hence the order of $f_2$ is neither $2$ nor $3$.

(In fact, since $f_2^2(1) = 9$ and $f_4(1) = 9$, it follows that $f_2^2 = f_4$. Similarly, $f_2^3(1) = 13 = f_6(1)$, so $f_2^3 = f_6$.)

Therefore, the only remaining possibility is that the order of $f_2$ must be $6$, so $f_2$ generates $\operatorname{Aut}(\mathbb Z_{14})$, proving that the latter is cyclic.

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    Thank you very much for your explanation. It is extremely helpful. There is one key point in your answer which is not $100$% clear to me (although I get the intuition for it), specifically: why are we able to use Lagrange's theorem? The $f_n$'s are functions and I don't understand how we define the order of functions. Lagrange's theorem states that the order of a subgroup divides the order of the group. So are we considering the $f_n$'s as subgroups of $\operatorname{Aut}(\mathbb Z_{14})$? – 2017-02-12
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    @Elix Yes, the $f_n$'s are functions, but the beauty of group theory is that it works no matter what kind of elements we are dealing with. The $f_n$'s are elements of a finite group, so they must obey Lagrange's theorem: if $G$ is a finite group and $g$ is an element of $G$, then the order of $g$ must divide the order of $G$. Here, $G = \operatorname{Aut}(\mathbb Z_{14})$ and $g$ is any of the $f_n$'s. – 2017-02-12
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    @elix Regarding how we define the order of functions: we define this order the same way as we do for any element of any group. The order of an element $g$ is the smallest positive integer $k$ such $g^k$ equals the identity. In this problem, $g$ is one of the $f_n$'s, and $g^k$ means $f_n^k$, which is the composition of $f_n$ with itself $k$ times. And the identity is the identity map $f_1$. So, the order of $f_n$ is the smallest positive integer $k$ such that $f_n^k = f_1$. Note that this is an equality between functions, which means that $f_n^k(x)=f_1(x)=x$ for every $x \in \mathbb Z_{14}$. – 2017-02-12
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    @Elix Note that for **automorphisms** (as opposed to arbitrary functions), it suffices to check that $f_n^k(1) = f_1(1) = 1$, since this automatically implies that $f_n^k(x) = x$ for every $x \in \mathbb Z_{14}$. – 2017-02-12
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    I could not hope for a better answer! Everything is now perfectly clear. Thank you very much for your time, much appreciated. – 2017-02-12
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    @Elix My pleasure, I'm glad it was helpful. – 2017-02-12
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    @Elix One last remark regarding your question about Lagrange's theorem, in case it's not already clear. In any group $G$, the order of an element $g \in G$ is the same as the order of the subgroup $\langle g \rangle$ generated by $g$, which is why Lagrange's theorem (which constrains the order of a subgroup) applies to orders of elements as well. – 2017-02-12