Sylow subgroup intersection of a certain index + F1 Trespasser

Today, I read the news online, the latest news is that a man was strolling along the F1 track while the race was ongoing. Really unbelievable.

Also, recently our recommended books from Amazon for GAT/DSA preparation have been very popular with parents seeking preparation. Do check it out if your child is going for DSA soon.

Back to our topic on Sylow theory…

Let G be a finite group, where q is a prime divisor of G. Suppose that whenever Q_1 and Q_2 are two distinct Sylow q-subgroups of G, Q_1\cap Q_2 is a subgroup of Q_1 of index at least q^a. Prove that the number n_q of Sylow q-subgroups of G satisfies n_q\equiv 1\pmod {q^a}.

Proof: Let \Omega=\{Q_1,Q_2\dots,Q_n\} be the set of all Sylow q-subgroups of G. Fix P=Q_k\in \Omega. Consider the group action of P acting on \Omega by conjugation.

\phi:P\times\Omega\to\Omega, \phi_x(Q_i)=xQ_i x^{-1}

By Orbit-Stabilizer Theorem, |O(Q_i)|=|P|/|N_p(Q_i)|.

We claim that N_p(Q_i)=Q_i\cap P, since any element x outside of Q_i cannot normalise Q_i, since otherwise if x \neq Q_i, xQx^{-1}=Q_i, then \langle Q_i, x\rangle will be a larger q-subgroup of G than Q_i.

Thus, |O(Q_i)|=|P|/|Q_i\cap P|\geq q^a, i.e. q^a\mid |O(Q_i)|.


The orbits form a partition of \Omega, thus |\Omega|=1+\sum{|O(Q_i)|}, where the sum runs over all orbits other than O(P).

Thus, n_q\equiv 1\pmod {q^a}.


Author: mathtuition88

One thought on “Sylow subgroup intersection of a certain index + F1 Trespasser”

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 )

Google photo

You are commenting using your Google 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 )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.