Conjugacy Classes of p-Sylow Subgroups

I've run into a brick wall with a problem in Pinter's Book of Abstract Algebra where it attempts to guide the reader through a proof of the first Sylow theorem; if I take the result as given, I can follow the rest, but damned if I can get to this myself.

The relevant parts of the exercise read:

Let G be a finite group with a p-Sylow subgroup K. Let X be the set of the conjugates of K. For $C_1, C_2 \in X$, let $C_1 \sim C_2$ iff $C_1 = aC_2a^{-1}$ for some $a \in K$.

[$\sim$ is proved to be an equivalence relation on X without issue in part 1.]

Q2. For each $C \in X$, prove that the number of elements in the equivalence class [C] is a divisor of |K|. Conclude that for each $C \in X$, the number of elements in [C] is either 1 or some power of p.

The second part of the question I can see: K is p-Sylow by hypothesis, so $|K| = p^k$ for some integral k; its factors are $1,\ p\,\ p^2,\ \cdots,\ p^k$. Getting to this point is causing me grief, though.

I've been looking for some convenient bijection $\varphi$ from [C] to something involving K more directly, but have had no joy. I tried $\varphi: [C] \to K/J\ \varphi(kCk^{-1}) = Jk$ for some p-subgroup of K called J, but couldn't manage to show injectivity. I did find that $\varphi: [C] \to {N(C)k: k \in K}\ \varphi(kCk^{-1}) = N(C)k$ for C's normaliser in G N(C) was a bijection, but this didn't help much since I couldn't determine much about N(C) or the cosets N(C)k.

I've been trying to tie C to K a bit by noting that conjugate groups are isomorphic, but I've not found anywhere that could be applied here.

• The stabilizer of $C$ in $K$: $$S_K(C)=\{a\in K\mid aCa^{-1}=C\}$$ is easily seen to be a subgroup of $K$. But $|[C]|=[K:S_K(C)]$, and this is a factor of $|K|$. – Jyrki Lahtonen Jun 8 '14 at 19:44
• And a very warm welcome to Math.SE! If only all the first time posters were as meticulous as you are in describing their thoughts and giving context :-) – Jyrki Lahtonen Jun 8 '14 at 19:45
• Ahah; the book introduced group actions, orbits and stabilisers a while ago, but it introduced them in the context of straightforward permutations only. To make sure I follow it properly: conjugating some $C \in X$ is a permutation of the set, so the machinery of group actions still holds, and the orbit-stabiliser theorem can be applied to the situation? – CKA Jun 8 '14 at 20:10
• Correct. $K$ acts on $X$ by conjugation. The whole group $G$ also acts on $X$ by conjugation, but this time we restrict the action to $K$. – Jyrki Lahtonen Jun 8 '14 at 20:28

By orbit stabilizer theorem, $$|O_x|=|G:Stab(x)|$$ where $O_x$ denotes the orbit of $x$, from that sense your question is obvious but if you want to see this by map use the idea of the proof and
$$\phi :L(stab(x))\to O_x$$ by $\phi(rStab(x))=rx$ where $L(Stab(x))$ is the left cosets of $Stab(x)$ in $G$.
The hints provided by C. Pinter should be enough to solve the problem. As we know from exercise M2 that every conjugate of $$K$$ is also a p-Sylow subgroup of $$G$$, hence by the exercise I10*, the number of elements in $$X_C$$ or $$[C]$$ (using the notation of the question) is a divisor of $$|K|$$ where $$X_C = \{aCa^{-1}:a\in K\}$$ and $$C \in X$$.
Let $$K$$ be any subgroup of $$G$$, let $$K^{*}=\{Na:a\in K\}$$, and let $$X_K = \{aHa^{-1}:a \in K\}$$, then $$X_K$$ is in one-to-one correspondence with $$K^{*}$$. And the number of elements in $$X_K$$ is a divisor of $$|K|$$.