Help with Conrad's "Recognizing Galois Groups" I'm trying to do the proofs of theorems 2.1 and 2.2 in this article by Keith Conrad. I'll just quote the parts of theorem 2.1 I have problems with:

if $a \sim b$ then $ga \sim gb$... if $a \neq b$ then $(ab) \in G$ and its conjugate $g(ab)g^{-1} = (ga \space gb)$ is also in $G$.

I'm not sure what to ask here; my problem is only that I've never seen that the last equality is "legitimate". It's never dawned on me to e.g. calculate $(123)(23)(132)$ by $((123)2 \space (123)3)$. A reference to some elementary text would suffice as answer.

the action of $g$ provides a function $[a] \rightarrow [b]$ given by $x \mapsto gx$ (if $x \sim a$ then $gx \sim ga = b$) with inverse function $[b] \rightarrow [a]$ given by the action of $g^{-1}$. Therefore all equivalence classes have the same size.

For the inverse, does he mean $g^{-1}b=g^{-1}(ga) \sim g^{-1}(gx) = x$? As for the last assertion, it follows from there being a bijection between the equivalence classes, right?

The group $\langle g \rangle$ has order $p$, so its orbits in the action of $G$ on the equivalence classes each have size $1$ or $p$.

I don't even understand what this means, "its orbits in the action of $G$", what? As for the assertion itself it looks like that something that follows from a result in elementary group theory, but I'm unable to find (in my books) which result it might be, so, could you give a hint or tell what this follows from?

Therefore all $\langle g \rangle$-orbits have size $1$, so for every $a \in \{1,...,\}$ we have $[ga]=[a]$, which means $a \sim ga$.

All such orbits have size $1$, and since the orbit $id.[a]$ is always an orbit, this is the only orbit, whence $g[a]=[ga]=[a]$. Have I understood it correct?

Then $1 \sim 2 \sim 3 \sim ... \sim p,$

That $1 \sim 2$, for example means $(12) \in G$, but why does $G$ contain that proposition just because it contains $(1 \cdots p) = (12)(23) \cdots (p-1 \space p)$?
 A: 1) If $a \neq b \in \{1,\ldots,n\}$ and $g \in S_n$ then $g (ab) g^{-1}$ is conjugate to a transposition so must again be a transposition.  So the equality$g (ab) g^{-1} = (g(a) g(b))$ should not be suprising, and to check it we just need to verify that indeed the left hand side carries $g(a)$ to $g(b)$.  It does:
$g \circ (ab) \circ g^{-1} g(a) = g \circ (ab) a = g(b)$.
2) He is saying that if $ga = b$, then $x \mapsto gx$ gives a function from $[a]$ to $[b]$, $x \mapsto g^{-1} x$ gives a function from $[b]$ to $[a]$, and these functions are mutually inverse.  Thus multiplication by $g$ is a bijection from $[a]$ to $[b]$ so these two sets have the same size.
3) This refers to group actions, a very important concept in group theory but one which some undergraduate courses don't do enough with.  You should familiarize yourself with them before going on with the proof.  The precise question about why a group of order $p$ acting on a set has orbits of sizes $1$ or $p$ is a special case of the Orbit-Stabilizer Theorem.
4) "All $g$-orbits have size $1$" means that $g$ acts trivially on all the equivalence classes, so $[ga] = [a]$ for all $a$ and thus $(a g(a)) \in G$.  
5) Since $[ga] = [a]$, also $[g^2 a] = [ga]$ and similarly for all powers of $g$.  Since $g$ has order $p$ we get $a \sim ga \sim \ldots \sim g^{p-1} a$.  If we assume (via conjugation) that $g = (1...p)$, then this shows that $1 \sim 2 \sim \ldots \sim p$.  
