Sylow's second theorem explanation

I am reading the book Contemporary Abstract Algebra by Gallian and I can’t understand some specific part of the proof of Sylow’s second theorem that has been given here.

Theorem : If H is a subgroup of a finite group $G$ and $|H|$ is a power of a prime $p$ then $H$ is contained in some Sylow p-subgroup of $G$.

Proof : Let $K$ be a Sylow p-subgroup of $G$ and let $C = {K_1, K_2, … , K_n}$ with $K=K_i$ be the set of all conjugates of $K$ in $G$. Since conjugation is an automorphism, each element of $C$ is a Sylow p-subgroup of $G$. Let $S_C$ denote the group of all permutations of $C$. For each $g in G$, define $phi_g : C to C$ by $phi_g(K_i) = gK_ig^{-1}$. It is easy to show that each $phi_g in S_C$.

Now define $T:G to S_C$ by $T(g) = phi_g$. Since $phi_{gh}(K_i) = (gh)K_i(gh)^{-1} = g(hK_ih^{-1})g^{-1} = gphi_h(K_i)g^{-1} = phi_g(phi_h(K_i)) = (phi_gphi_h)(K_i)$, we have $phi_{gh} = phi_gphi_h$ , and therefore $T$ is a homomorphism from $G$ to $S_C$.

Next consider $T(H)$, the image of $H$ under $T$. Since $|H|$ is a power of $p$, so is $|T(H)|$. Thus, by the Orbit-Stabilizer Theorem, for each $i$, $|orb_{T(H)}(K_i)|$ divides $|T(H)|$, so that $|orb_{T(H)}(K_i)|$ is a power of $p$. Now we ask: Under what condition does $|orb_{T(H)}(K_i)| = 1$? Well, $|orb_{T(H)}(K_i)| = 1$ means that $phi_g(K_i) = gK_ig^{-1} = K_i$ for all $g in H$; that is, $|orb_{T(H)}(K_i)| = 1$ iff $Hle N(K_i)$. But the only elements of N(K_i) that have orders that are powers of $p$ are those of $K_i$. Thus, $|orb_{T(H)}(K_i)| = 1$ iff $H le K_i$.

So, to complete the proof, all we need to do is to show that for each $i, |orb_{T(H)}(K_i)| = 1$. Analogous to Theorem 24.1, we have $|C| = |G:N(K)|$. $ color {orange}{text{And since $|G:K| = |G:N(K)| |N(K):K|$} }$ is not divisible by $p$, neither is $|C|$. $color {brown}{text{Because the orbits partition $C$ , $|C|$ is the sum of powers of p. If no orbit has size 1, then p divides each summand}}$ and, therefore, $p$ divides $|C|$, which is a contradiction. Thus, there is an orbit of size $1$, and the proof is complete.

I can’t understand the two coloured lines. Especially about the second coloured line I don’t get that why orbits partition C, previously nowhere it’s written. And I only know about orbits as much as it’s given in this book which is just the definition and some examples. And the result that order of the orbit divides the order of the group.