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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 $\phi_{gh}(K_i) = (gh)K_i(gh)^{-1} = g(hK_ih^{-1})g^{-1} = g\phi_h(K_i)g^{-1} = \phi_g(\phi_h(K_i)) = (\phi_g\phi_h)(K_i)$, we have $\phi_{gh} = \phi_g\phi_h$ , 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 $H\le N(K_i)$. 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.

Itachi
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    The first colored line is just the degree theorem $(G:K)=(G:H)(H:K)$ by Lagrange. – Dietrich Burde Dec 16 '20 at 12:58
  • Okay thanks but the second one? – Itachi Dec 16 '20 at 12:59
  • Any subgroup of a p-Sylow subgroup is a p-subgroup, i.e. its order is some power of p. – Divide1918 Dec 16 '20 at 13:01
  • @Divide1918 okay but I don't get it what you're referring with this statement – Itachi Dec 16 '20 at 13:05
  • What is K? And what exactly is C? – Divide1918 Dec 16 '20 at 13:38
  • @Pritam Judging by the quotes you have shared here with us, the referenced textbook seems to be written in the nowadays typical sloppy, half-baked style. I would recommend using a more professional textbook. As to the second Sylow theorem itself, its proof relies on two crucial elements: 1) A fundamental lemma of group actions according to which given the action of a $p$-group $G$ on a finite set $A$, the cardinal of the subset of $G$-fixed elements is congruent with $|A|$ modulo $p$ and 2) a clever application of this fundamental lemma (to be cont.) – ΑΘΩ Dec 16 '20 at 13:44
  • @Pritam (cont.) this fundamental lemma to the canonical action of $p$-subgroup $H$ of arbitrary finite group $G$ to the left quotient $(G/P)_{\mathrm{s}}$, where $P$ is a certain fixed $p$-Sylow subgroup of $G$. By virtue of the fundamental lemma at 1) this action will have a fixed element, say the left congruence class of some $a \in G$ modulo $P$, and this can be equivalently expressed as $H \leqslant aPa^{-1}$. The conclusion is that any $p$-subgroup is included in the conjugate of a fixed $p$-Sylow subgroup and thus any two $p$-Sylow subgroups are conjugate. – ΑΘΩ Dec 16 '20 at 13:49
  • @ΑΘΩ I support what you said about the book. Well it's a really well known book that's why I was recommended and as my entrances are coming close so right now I can't read another book all-over again. But you tell me some name I'll read it later. And one thing, I haven't read Group actions, it's in Masters course here and I just finished my BSc. So I don't get your first comment. – Itachi Dec 16 '20 at 14:20
  • You realize that anyone without access to the book has no idea what $C$ is, what the action is, or what Theorem 24.1 says or how it is proven, right? – Arturo Magidin Dec 16 '20 at 16:22
  • @ArturoMagidin okay then I need to add the whole theorem... I'll do it as soon as possible... it's really long so I'll post it when I can. – Itachi Dec 16 '20 at 17:00
  • You need to say what the terms you are using mean. You may not need “the whole theorem”, but you certainly need to say what $C$ is and what the action is. – Arturo Magidin Dec 16 '20 at 18:03
  • @ArturoMagidin as I've stated in the comments that I haven't read or know anything about "Group actions" so I might need to upload the whole theorem otherwise there will be confusions... it's okay I'll do it...but I'll take my time...!! And just in short, $|C|$ is the set of all conjugate subgroups of $K$ , and what is $K$ ? $K$ is a sylow p-subgroup. – Itachi Dec 16 '20 at 18:18
  • @ArturoMagidin I need your help in somewhat similar question that you answered here https://math.stackexchange.com/q/164244/655021 . The statement I was given says if $G$ be a group and $H$ be it's subgroup then is it necessarily true that if $\frac{|G|}{|H|}$ is a prime number then $H$ is normal in $G$. Although in your answer there was written somewhere about "G acting" which I don't know the meaning of but I can guess that my statement is not true because it just says "prime" not least prime. Can you help me out with an counter example then? – Itachi Dec 16 '20 at 18:26
  • @Pritam: That is not what that statement says, and this is not the correct place to ask that question. And the statement you say is false: the subgroups of order $2$ have prime index in $S_3$. Are they normal? – Arturo Magidin Dec 16 '20 at 20:19
  • Let us continue this discussion in chat. – Itachi Dec 16 '20 at 20:33
  • @Pritam: I don’t do chat. I have no interest or desire of continuing this discussion. – Arturo Magidin Dec 16 '20 at 20:58
  • "@ArturoMagidin: Sorry for asking it here. I know that the question what you answered and the statement I wrote are different. I mentioned that I'm given this statement, and I need to check it's correctness. And thanks for the example I get it now." It showed to avoid many comments in comment section so I just posted the above and "last" comment in the chat. Neither I had any intention to continue it further nor I wanted to crowd here. Thought this last text is needed so I posted the comment. I understand that you don't do chats. – Itachi Dec 17 '20 at 07:43

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Let G be a group of permutation of a set S. for each i in S, let $${orb_G(i)={\{\phi(i)| \phi \in G\}}}$$

The set ${orb_G(i)}$ is a subset of S called the orbit of i under G.

so, orbit of i is everything that element is mapped to under the permutation. That is why it is subset of S. hence, it partitions the set S.

Now, intuitively if you think of a rubik's cube, the element at the top left corner of red color , it can move(permute) only to certain point in the rubik's cube, that is the orbit of that element (3x3x3) rubic's cube.

Also, you mentioned, no orbit has size 1 which means, any center color of rubik's cube can't move(permute) to any other position. so it has not orbit.

user863114
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