Sylow subgroup intersection of a certain index + F1 Trespasser

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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)|.

|O(P)|=1.

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}.

 

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One Response to Sylow subgroup intersection of a certain index + F1 Trespasser

  1. Pingback: Group of order 432 is not simple | Singapore Maths Tuition

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