could someone please explain the solution to me?
1) I don't understand why they know they could send epsilon to (epsilon)^i. 2) Did they assume the Galois group is Z/4Z because the order of Galois group has order 4 and so we could assume it is isomorphic to a group of order 4? 3) How did they calculate the "fourth line", i.e. (sigma_2)^4 = (sigma_1)(16) = ...?
Thank you very much!!
On
This is a special case of a more general result:
Theorem: Let $\zeta = e^{2\pi i/n}$ where $n$ is a positive integer and let $K = \mathbb{Q}(\zeta)$. Then the extension $K\supset \mathbb{Q}$ is Galois and the Galois group $\textrm{Gal}(K/\mathbb{Q})$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{\times}$ the group of reduced residue classes modulo $n$.
Your question corresponds to simpler case when $n$ is prime, say $p$. In this case we can see that the minimal polynomial of $\zeta$ is $$\Phi_{p}(x) = (x^{p} - 1)/(x - 1) = x^{p - 1} + x^{p - 2} + \dots + x + 1$$ and its roots are given by $\zeta, \zeta^{2}, \dots,\zeta^{p - 1}$. As noted in your question the automorphisms of $K$ which fix $\mathbb{Q}$ are those which map $\zeta$ to each of these roots and hence there are $(p - 1)$ such automorphisms.
To prove that the group of these automorphisms is cyclic we need to know that for any prime $p$ there is a positive integer $g$ such that the numbers $g, g^{2}, \dots, g^{p - 1}$ modulo $p$ are all different and lead to $1, 2, \dots, p - 1$ but not necessarily in that order. Such a number $g$ is called a primitive root of $p$. Thus the roots $\zeta, \zeta^{2}, \dots, \zeta^{p - 1}$ can be rewritten as $$\zeta^{g}, \zeta^{g^{2}}, \dots, \zeta^{g^{p - 1}}$$ Let $\sigma(\zeta) = \zeta^{g}$ be one non-identity automorphism of $K$ which fixes $\mathbb{Q}$. Then $\sigma^{i}(\zeta) = \zeta^{g^{i}}$ so that $\sigma, \sigma^{2}, \dots, \sigma^{p - 1}$ gives all the needed automorphisms. Thus the group of automorphisms of $K$ which fix $\mathbb{Q}$ is cyclic with generator $\sigma$.
In your case $p = 5$ and we can choose $g = 2$ as $2, 4, 8, 16$ mod $5$ is $2, 4, 3, 1$.
For general $n$ the Galois group $\textrm{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^{\times}$ is cyclic only when there exists a primitive root $g$ of $n$ and that happens only when $n = 2, 4, p^{k}, 2p^{k}$ where $p$ is odd prime and $k$ is any positive integer.
The notation in your question is confusing. $\sigma_{i}$ is defined by $\sigma_{i}(\zeta) = \zeta^{i}$. Then $$\sigma_{2}^{4}(\zeta) = \sigma_{2}^{3}(\zeta^{2}) = \zeta^{16} = \sigma_{1}(\zeta^{16}) = \zeta$$ and this is what he writes as $\sigma_{2}^{4} = \sigma_{1}(16)$. The equation $$\sigma_{2}^{4} = \sigma_{1}(16) = \sigma_{3}(8)\sigma_{2} = \sigma_{4}^{2}$$ is a shorthand notation for $$\sigma_{2}^{4}(\zeta) = \sigma_{1}(\zeta^{16}) = \sigma_{3}(\zeta^{8})\sigma_{2}(\zeta) = \sigma_{4}^{2}(\zeta)$$
You can factorize $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$, and the latter factor is irreducible and its roots over $\mathbb{C}$ are $\zeta^i$ for $1\le i\le 4$. Therefore the splitting field is indeed $\mathbb{Q}[\zeta]$ and it is easy to check that $\sigma:\zeta\rightarrow\zeta^i$ is indeed an automorphism (check the axioms if you like). Furthermore, these are the only automorphisms since $\zeta$ must be sent to another root of its minimal polynomial.
Admittedly the second part of the solution seems a bit strange, but one can probably reword it in the following way. Consider the automorphism $\sigma_i:\zeta\rightarrow\zeta^i$. Then $\sigma_1$ is the identity and we will calculate the order of $\sigma_2$. We have $\sigma_2^2=\sigma_4$ and $\sigma_2^3 = \sigma_3$, so the order can't be $2$ or $3$ and thus must be $4$. (You can check that $\sigma_2^4=\sigma_1$ if you like). Because the only other group of order $4$ (besides $\mathbb{Z}/4\mathbb{Z}$) doesn't have a n element of order $4$, we're done.