2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $|K|$. $\endgroup$ – Jyrki Lahtonen Jun 8 '14 at 19:44
  • $\begingroup$ 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 :-) $\endgroup$ – Jyrki Lahtonen Jun 8 '14 at 19:45
  • 1
    $\begingroup$ 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? $\endgroup$ – CKA Jun 8 '14 at 20:10
  • $\begingroup$ 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 $K$. $\endgroup$ – Jyrki Lahtonen Jun 8 '14 at 20:28
1
$\begingroup$

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 $G$.

here you can find details of the proof.

$\endgroup$
1
$\begingroup$

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 $C \in X$.

*Exercise I10:

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 $|K|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.