In a finite group, if the number of Sylow subgroups is not 1 mod p² then there exist Sylow subgroups which intersect maximally

Let G be a finite group. Use the method of proof in Sylow’s Theorem to show that if n_p \neq 1 mod p^2, then there exist distinct Sylow p-subgroups P and Q such that [P : P \cap Q] = [Q : P \cap Q] = p.

Let \mathcal{S} = \{ P_1, P_2, \ldots, P_r \} be the Sylow p-subgroups of G.

Now let Q be any p-subgroup of G. Note that Q acts on \mathcal{S} by conjugation; we can then say \mathcal{S} = \bigcup_{P_i \in \mathcal{S}} Q \cdot P_i. Relabel the elements of \mathcal{S} so that P_i for 1 \leq i \leq s are a set of orbit representatives of the action of Q. Note that |Q \cdot P_i| = [Q : N_Q(P_i)] by the Orbit-Stabilizer Theorem, so that |Q \cdot P_i| = [Q : N_G(P_i) \cap Q] = [Q : P_i \cap Q] by Lemma 19 in the text. Thus n_p = \sum_{i=1}^s [Q : P_i \cap Q], where P_i ranges over a set of orbit representatives of the action of Q on \mathcal{S}.

We may let Q itself be a Sylow p-subgroup, and (without loss of generality and with a relabeling if needed) choose P_1 = Q. Then N_p = 1 + \sum_{i=2}^s [Q : P_i \cap Q]. Now because P_i \cap Q is a p-subgroup, [Q : P_i \cap Q] = p^{k_i} for some k_i. Modulo p^2, then, [Q : P_i \cap Q] \in \{0,p\}. If [Q : P_i \cap Q] = 0 (mod p^2) for all P_i, then we have n_p \equiv 1 \mod p^2, a contradiction. Thus there exists a Sylow p-subgroup P such that [Q : P \cap Q] = p.

Finally, note that [P : P \cap Q] = |P|/|P \cap Q| = |Q|/|P \cap Q| = [Q : P \cap Q] = p, by Lagrange and since |P| = |Q|.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: